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Button:=Caracteristicile straturilor:
Strat 1: Strat 2: Strat 3:
h2 6.6m:=h
1
4.8 m⋅:= h
3
1.00 m⋅:=
γ2 17.1kN
m3
:=γ1 18.1
kN
m3
:= γ3 17kN
m3
:=
c1 3.1kPa:= c2 19.5kPa:= c3 2.1kPa:=
ϕ1 19.7deg:= ϕ2 15deg:= ϕ3 11deg:=
Eforturi de calcul:
Nd 2000 kN⋅:=
Td 300kN:=
Vg 1400kN:=
Vc 600kN:=
He 0.9m:=
q0 He γ1⋅ 16.29mkN
m3
⋅=:= (suprasarcina de calcul)
Lcarosabil 7 m⋅:= Bcheson 3.00 m⋅:=
L
zidintors
0.3 m⋅:= H
cheson
0.8 m⋅ h
2
+ h
3
+ 8.4m=:=
Lcheson 7.00 m⋅:=
Coeficienti partiali de siguranta:
γG.stb 0.9:=- actiuni permanente: - favorabile
- nefavorabile γG.dst 1.1:=
- actiuni variabile: - favorabile γQ.std 0:=
- nefavorabile γQ.dst 1.5:=
γG 1:=- pentru actiuni sau efectul actiunilor: - actiuni permanente nefavorabile:
γQ 1.3:=- actiuni variabile nefavorabile:
- parametrii geotehnici: -unghi de frecare interna efectiv γϕ1 1.25:=
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-coeziunea efectiva γc1 1.25:=
γγ1 1.00:=-greutatea volumica
Valorile de calcul pentru caracterisiticile straturilor:Stratul 1: Stratul 2: Stratul 3:
γ1d
γ1
γγ1
18.1kN
m3
⋅=:= γ2d
γ2
γγ1
17.1kN
m3
⋅=:= γ3d
γ3
γγ1
17kN
m3
⋅=:=
c1d
c1
γc1
2.48 kPa⋅=:= c2d
c2
γc1
15.6 kPa⋅=:= c3d
c3
γc1
1.68 kPa⋅=:=
ϕ1d
ϕ1
γϕ1
15.76 deg⋅=:= ϕ2d
ϕ2
γϕ1
12 deg⋅=:= ϕ3d
ϕ3
γϕ1
8.8 deg⋅=:=
Calculul greutatii:
γbeton 25kN
m3
:=A1 3.195m
2:=
G1 A1 γbeton⋅ m⋅ 7⋅ 559.125 kN⋅=:=
A2 1.8m2
:=
G2 A2 γbeton⋅ m⋅ 7⋅ 315 kN⋅=:=
G3 3.18m
2
2⋅
7⋅
m γbeton⋅
18.56m
2
6.2⋅
m γbeton⋅+
3.99 10
3×
kN⋅=:=
A4 1.2 m2
⋅:=
G4 A4 γbeton⋅ 0.5⋅ m 2⋅ 30 kN⋅=:=
Coeficientii de presiuni active:
β 0deg:= δ1 0.5 ϕ1d⋅ 7.88 deg⋅=:= α1 0deg:=
k
a1
cos ϕ1d α1−( )
cos α1( )2
cos α1 δ1+( )⋅ 1sin ϕ1d δ1+( ) sin ϕ1d β−( )⋅
cos α1 ϕ1d+( ) cos β α1−( )⋅
+
2
⋅
0.544=:=
δ2 0.5 ϕ2d⋅ 6 deg⋅=:= β 0deg:= α2 0deg:=
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k a2
cos ϕ2d α2−( )
cos α2( )2
cos α2 δ2+( )⋅ 1sin ϕ2d δ2+( ) sin ϕ2d β−( )⋅
cos α2 ϕ2d+( ) cos β α2−( )⋅
+
2
⋅
0.623=:=
δ3 0.5 ϕ3d⋅ 4.4 deg⋅=:= β 0deg:= α3 0deg:=
k a3
cos ϕ3d α3−( )
cos α3( )2
cos α3 δ3+( )⋅ 1sin ϕ3d δ3+( ) sin ϕ3d β−( )⋅
cos α3 ϕ3d+( ) cos β α3−( )⋅
+
2
⋅
0.702=:=
Coeficientii de presiuni pasive:
δ1 0.5 ϕ1d⋅ 7.88 deg⋅=:= β 0deg:= δ1 0.5 ϕ1d⋅ 7.88 deg⋅=:= α1 0deg:=
k p1
cos ϕ1d α1+( )2
cos α1( )2
cos α1 δ1−( )⋅ 1sin ϕ1d δ1+( ) sin ϕ1d β−( )⋅
cos α1 δ1+( ) cos α1 β−( )⋅
−
22.093=:=
δ2
0.5 ϕ2d
⋅ 6 deg⋅=:= β 0deg:= α2 0deg:=
k p2
cos ϕ2d α2+( )2
cos α2( )2
cos α2 δ2−( )⋅ 1sin ϕ2d δ2+( ) sin ϕ2d β−( )⋅
cos δ2 α2+( ) cos α2 β−( )⋅
−
2
⋅
1.729=:=
δ3 0.5 ϕ3d⋅ 4.4 deg⋅=:= β 0deg:= α3 0deg:=
k p3
cos ϕ3d α3+( )2
cos α3( )2
cos α3 δ3−( )⋅ 1sin ϕ3d δ3+( ) sin ϕ3d β−( )⋅
cos δ3 α3+( ) cos α3 β−( )⋅
−
2
⋅
1.483=:=
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Ipoteza I (G1+G2+G3+G4):
Verificarea la sectiuni periculoase
xG.1 0.78m:=
xG.4 0.5m:=
V1 G1 G4+ 589.125 kN⋅=:=
M1 G1 xG.1⋅ G4 xG.4⋅− 421.118 kN m⋅⋅=:=
W1Bcheson Lcheson
2⋅
624.5 m
3⋅=:=
σmax
V1
Bcheson Lcheson⋅
M1
W1+ 45.242
kN
m2
⋅=:=
σmin
V1
Bcheson Lcheson⋅
M1
W1− 10.865
kN
m2
⋅=:=
Rezistenta betonului la compresiune
Aleg Beton C50/60
αcc 1:= γc 1.5:=f ck 50
N
mm2
:=
f cdαcc f ck ⋅
γc
3.333 104
×kN
m2
⋅=:=
σmax f cd≤ 1=
Rezistenta la intindere a betonului
f ct.k 5.3MPa:= f ct.k 5.3N
mm2
⋅=
αct 1:=f ctd
αct f ct.k ⋅
γc
3.533 103
×kN
m2
⋅=:=
σmin f ctd≤ 1=
NU NECESITA ARMATURA DE LEGATURA INTRE ELEVATIE SI RADIER CULEE
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Verificarea la rasturnare:
Sectiunea a-a:
- brate:
xG.1 0.78m:= xG.4 0.5m:=
- momente:
M1stb G1 xG.1⋅ G4 xG.4⋅+ 451.118 kN m⋅⋅=:=
M1dst 0kN m⋅:=
M1stb M1dst>
Sectiunea b-b:
- brate:
xG.1. 2.48m:= xG.4. 3.5m:= xG.2 1.5m:=
- momente:
M1stb. G1 xG.1⋅ G4 xG.4⋅+ G2 xG.2⋅+ 923.617 kN m⋅⋅=:=
M1dst. 0kN m⋅:=
M1stb M1dst>
Verificare la capacitate portanta:
Va G1 G2+ G4+ 904.125 kN⋅=:=
- brate:
xG1. 0.98m:= xG2. 0m:= xG4 2m:=
Ma G1 xG1.⋅ G2 xG2.⋅+ G4 xG4⋅+ 607.942 kN m⋅⋅=:=
ea
Ma
Va
0.672m=:=
B1 Bcheson 2 ea⋅− 1.655m=:=
L1 Lcheson:=
-aria efectiva: Aa B1 L1⋅ 11.586 m2
=:=
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Pef
Vc
Aa
5.179 104
× Pa=:=
- Factorii adimensionali pt calculul presiunii terenului de fundare:
-factori care tin de capaciatatea portanta:
Nq eπ tan ϕ3d( )⋅
tan 45degϕ3d
2+
2
⋅ 2.214=:=
Nc Nq 1−( )1
tan ϕ3d( )⋅ 7.841=:=
Nγ 2 Nq 1−( ) tan ϕ3d( )⋅ 0.376=:=
-factori care tin de inclinarea bazie fundatiei: ε 0:=
bq 1 ε tan ϕ3d( )⋅−( )2
1=:=
bγ bq:=
bc bq
1 bq−
Nc tan ϕ3d( )⋅
− 1=:=
-forma fundatiei:
sq 1B1
L1
sin ϕ3d( )⋅+ 1.036=:=
sγ 1 0.3B1
L1
⋅− 0.929=:=
sc
sq Nq⋅ 1−
Nq 1−
1.066=:=
-inclinarea incarcarii
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Hd 0kN:= (actiuni verticale).
m1
2B1−
L1
+
1B1−
L1
+
2.31=:=
iq 1Hd
Va A1 c3d⋅ 1tan ϕ3d deg⋅( )
⋅+
−
m1
1=:=
iγ 1Hd
Vg A1 c3d⋅1
tan ϕ3d deg⋅( )⋅+
−
m1 1+
1=:=
ic iq
1 iq−( )N
ctan ϕ
3d( )⋅
− 1=:=
Calculul suprasarcinii:
γmed
γ1 γ2+ γ3+( )3
17.4kN
m3
⋅=:=
Df h1 h2+ h3+ 12.4m=:=
Q1 Df γmed⋅ 215.76kN
m2
⋅=:=
Rd
A1
c3d Nc⋅ bc⋅ sc⋅ ic⋅ Q1 Nq⋅ bq⋅ sq⋅ iq⋅+ 0.5 γ3d⋅ m Nγ⋅ bγ⋅ sγ⋅ iγ⋅+ 511.951 kPa⋅=:=
Rd
A1
Vc Rd<
Ipoteza II (G1+G2+G4+Pa+Pp):
Verificarea la rasturnare:
Impingerea activa:
qk 0kN
m3
:=
pab1.0 γQ.dst qk ⋅ k a1⋅ 2 γG.dst⋅ c1d⋅ k a1⋅− 2.968−1
m2
kN⋅=:=
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Pab3H Pab3 cos2
3ϕ3d
⋅ 156.937 kN⋅=:=
Pab3V Pab3 sin2
3ϕ3d
⋅ 16.126 kN⋅=:=
zab3
h3 2 pab3.0⋅ pab3.1+( )⋅
pab3.0 pab3.1+
30.493m=:=
Impingerea pasiva:
h1' 2.2 m⋅:=
ppb1.0 γQ.dst qk ⋅ k p1⋅ 2 γG.dst⋅ c1d⋅ k p1⋅− 7.893−
1
m2 kN⋅=:=
ppb1.1 γQ.dst qk ⋅ γG.dst γ1d⋅ h1'⋅+( ) k p1⋅ 2 γG.dst⋅ c1d⋅ k p1⋅− 83.781
m2
kN⋅=:=
ppb2.0 γQ.dst qk ⋅ γG.dst γ1d⋅ h1'⋅+( ) k p2⋅ 2 γG.dst⋅ c2d⋅ k p2⋅− 30.621
m2
kN⋅=:=
ppb2.1 γQ.dst qk ⋅ γG.dst γ1d⋅ h1'⋅+ γG.dst γ2d⋅ h2⋅+( ) k p2⋅ 2 γG.dst⋅ c2d k p2⋅− 245.3281
m2
kN⋅=:=
ppb3.0 γQ.dst qk ⋅ γG.dst γ1d⋅ h1'⋅+ γG.dst γ2d⋅ h2⋅+( ) k p3⋅ 2 γG.dst⋅ c3d⋅ k p3⋅− 244.4921
m2
kN⋅=:=
ppb3.1 γQ qk ⋅ γG.dst γ1d⋅ h1'⋅+ γG.dst γ2d⋅ h2⋅+ γG.dst γ3d⋅ h3⋅+( ) k p3⋅ 2 γG.dst⋅ c3d⋅ k p3⋅−:=
ppb3.1 272.216 kN1
m2
⋅⋅=
Descompunerea impingerii dupa orizontala si verticala:
hc 0.21m:=
P
pb1
p
pb1.1
p
pb1.0
+
( )
h1
2
⋅ m⋅ 182.129 kN⋅=:=
Ppb1H Ppb1 cos2
3ϕ1d
⋅ 179.075 kN⋅=:=
Ppb1V Ppb1 sin2
3ϕ1d
⋅ 33.211 kN⋅=:=
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zpc11
3h1' hc−( ) 0.663m=:=
Ppb2 m ppb2.1 ppb2.0+( )h2
2⋅
⋅ 910.629 kN⋅=:=
Ppb2H Ppb2 cos2
3ϕ2d
⋅ 901.767 kN⋅=:=
Ppb2V Ppb2 sin2
3ϕ2d
⋅ 126.735 kN⋅=:=
zpb2
h2 2 ppb2.0⋅ ppb2.1+( )⋅
ppb2.0 ppb2.1+
32.444m=:=
Ppb3 m ppb3.1 ppb3.0+( )h
32
⋅
⋅ 258.354 kN⋅=:=
Ppb3H Ppb3 cos2
3ϕ3d
⋅ 257.001 kN⋅=:=
Ppb3V Ppb3 sin2
3ϕ3d
⋅ 26.407 kN⋅=:=
zpb3
h3 2 ppb3.0⋅ ppb3.1+( )⋅
ppb3.0 ppb3.1+
30.491m=:=
Verificarea la rasturnare
Sectiunea a-a
- brate:
xG1 0.78 m⋅:= xab1 13 m⋅:= xT 2.9m:=
xG4. 1.80 m⋅:= yab1 0.09 m⋅:=
- momente:
M2stb G1 xG1⋅ G4 xG4.⋅+ Pab1V xab1⋅+ 746.123 kN m⋅⋅=:=
M2dstb Pab1H yab1⋅ 9.557 kN m⋅⋅=:=
M2dst M2stb<
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Sectiunea b-b
- brate:
xG1.. 2.48 m⋅:= xab1. 3 m⋅:= xpb1. 0 m⋅:= yab1. 9 m⋅:= ypb1. 8.14 m⋅:=
xG2.. 1.5m:= xab2. 3 m⋅:= xpb2. 0 m⋅:= yab2. 3.71 m⋅:= ypb2. 3.43 m⋅:=
xG3. 1.5 m⋅:= xab3. 3 m⋅:= xpb3. 0 m⋅:= yab3. 0.49 m⋅:= ypb3. 0.49 m⋅:=
xG4.. 3.5 m⋅:= xT. 11.81 m⋅:=
- momente:
M2stb. G1 xG1..⋅ G2 xG2..⋅+ G3 xG3.⋅+ G4 xG4.⋅+ Pab1V xab1.⋅+ Pab2V xab2.⋅+ Pab3V xab3.⋅+:=
M2stb. 8.201 103
× kN m⋅⋅=
M2dstb. Pab1H yab1.⋅ Pab2H yab2.⋅+ Pab3H yab3.⋅+ Td xT.⋅+ Ppb1H ypb1.⋅−( )Ppb2H− ypb2.⋅ Ppb3H ypb3.⋅−( )+
...:=
M2dstb. 1.624 103
× kN m⋅⋅=
M2dsb. M2stb<
Verificarea la capacitate portanta
Impingerea activa:
pab1.0. γQ qk ⋅
k a1⋅
2 γG⋅
c1d⋅
k a1⋅−
3.658−
1
m2 kN
⋅=:=
pab1.1. γQ qk ⋅ γG γ1d⋅ h1⋅+( ) k a1⋅ 2 γG⋅ c1d⋅ k a1⋅− 43.6051
m2
kN⋅=:=
pab2.0. γQ qk ⋅ γG γ1d⋅ h1⋅+( ) k a2⋅ 2 γG⋅ c2d⋅ k a2⋅− 29.5121
m2
kN⋅=:=
pab2.1. γQ qk ⋅ γG γ1d⋅ h1⋅+ γG γ2d⋅ h2⋅+( ) k a2⋅ 2 γG⋅ c2d k a2⋅− 99.8441
m2
kN⋅=:=
pab3.0. γQ qk ⋅ γG γ1d⋅ h1⋅+ γG γ2d⋅ h2⋅+( ) k a3⋅ 2 γG⋅ c3d⋅ k a3⋅− 137.4521
m2
kN⋅=:=
pab3.1. γQ qk ⋅ γG γ1d⋅ h1⋅+ γG γ2d⋅ h2⋅+ γG γ3d⋅ h3⋅+( ) k a3⋅ 2 γG⋅ c3d⋅ k a3⋅− 149.3911
m2
kN⋅=:=
Descompunerea impingerii dupa orizontala si verticala:
hc. 0.3m:=
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Pab1. pab1.1. pab1.0.+( )h1
2⋅
m⋅ 95.872 kN⋅=:=
Pab1H. Pab1. cos2
3
ϕ1d
⋅ 94.265 kN⋅=:=
Pab1V. Pab1. sin2
3ϕ1d
⋅ 17.482 kN⋅=:=
zab1.1
3h1 hc.−( )⋅ 1.5m=:=
Pab2. m pab2.1. pab2.0.+( )h2
2⋅
⋅ 426.873 kN⋅=:=
Pab2H. Pab2. cos2
3
ϕ2d
⋅ 422.719 kN⋅=:=
Pab2V. Pab2. sin2
3ϕ2d
⋅ 59.409 kN⋅=:=
zab2.
h2 2 pab2.0.⋅ pab2.1.+( )⋅
pab2.0. pab2.1.+
32.702m=:=
Pab3. m pab3.1. pab3.0.+( )h3
2⋅
⋅ 143.421 kN⋅=:=
Pab3H.
Pab3.
cos2
3ϕ
3d
⋅ 142.67 kN⋅=:=
Pab3V. Pab3. sin2
3ϕ3d
⋅ 14.66 kN⋅=:=
zab3.
h3 2 pab3.0.⋅ pab3.1.+( )⋅
pab3.0. pab3.1.+
30.493m=:=
Impingerea pasiva:
ppb1.0. 2− γG⋅ c1d⋅ k p1⋅ 7.176−1
m2
kN⋅=:=
ppb1.1. γG γ1d⋅ h1'⋅( ) k p1⋅ 2 γG⋅ c1d⋅ k p1⋅− 76.1641
m2
kN⋅=:=
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ppb2.0. γG γ1d⋅ h1'⋅( ) k p2⋅ 2 γG⋅ c2d⋅ k p2⋅− 27.8371
m2
kN⋅=:=
ppb2.1. γG γ1d⋅ h1'⋅ γG γ2d⋅ h2⋅+( ) k p2⋅ 2 γG⋅ c2d k p2⋅− 223.0251
m
2kN⋅=:=
ppb3.0. γG γ1d⋅ h1'⋅ γG γ2d⋅ h2⋅+( ) k p3⋅ 2 γG⋅ c3d⋅ k p3⋅− 222.2661
m2
kN⋅=:=
ppb3.1. γG γ1d⋅ h1'⋅ γG γ2d⋅ h2⋅+ γG γ3d⋅ h3⋅+( ) k p3⋅ 2 γG⋅ c3d⋅ k p3⋅− 247.4691
m2
kN⋅=:=
Descompunerea impingerii dupa orizontala si verticala:
hc. 0.21m:=
P
pb1.
p
pb1.1.
p
pb1.0.
+
( )
h1
2
⋅ m⋅ 165.572 kN⋅=:=
Ppb1H. Ppb1. cos2
3ϕ1d
⋅ 162.796 kN⋅=:=
Ppb1V. Ppb1. sin2
3ϕ1d
⋅ 30.192 kN⋅=:=
zpb1.1
3h1' hc.−( )⋅ 0.663m=:=
Ppb2. m ppb2.1. ppb2.0.+( )h2
2⋅
⋅ 827.844 kN⋅=:=
Ppb2H. Ppb2. cos2
3ϕ2d
⋅ 819.788 kN⋅=:=
Ppb2V. Ppb2. sin2
3ϕ2d
⋅ 115.214 kN⋅=:=
zpb2.
h2 2 ppb2.0.⋅ ppb2.1.+( )⋅
ppb2.0. ppb2.1.+
32.444m=:=
Ppb3. m ppb3.1. ppb3.0.+( )h3
2⋅
⋅ 234.868 kN⋅=:=
Ppb3H. Ppb3. cos2
3ϕ3d
⋅ 233.637 kN⋅=:=
Ppb3V. Ppb3. sin2
3ϕ3d
⋅ 24.007 kN⋅=:=
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zpb3.
h3 2 ppb3.0.⋅ ppb3.1.+( )⋅
ppb3.0. ppb3.1.+
30.491m=:=
- brate:
xG1. 0.98 m⋅:= xab1.. 1.5 m⋅:= xpb1.. 1.5 m⋅:= yab1.. 9.01 m⋅:= ypb1.. 8.17 m⋅:=
xG4.. 2 m⋅:= xab2.. 1.5 m⋅:= xpb2.. 1.5 m⋅:= yab2.. 3.71 m⋅:= ypb2.. 3.43 m⋅:=
xG2. 0m:= xab3.. 1.5 m⋅:= xpb3.. 1.5 m⋅:= yab3.. 0.49 m⋅:= ypb3.. 0.49 m⋅:=
xG3 0m:= xT. 11.81 m⋅:=
Mb G1 xG1.⋅ G4 xG4.⋅+ Pab1V xab1..⋅+ Pab2V xab2..⋅+ Pab3V xab3..⋅+ Ppb1V xpb1..⋅−( )Ppb2V− xpb2..⋅ Ppb3V xpb3..⋅− Pab1H yab1..⋅− Pab2H yab2..⋅−( )+
...
Pab3H− yab3..⋅ Ppb1H ypb1..⋅+ Ppb2H ypb2..⋅+ Ppb2H ypb3..⋅+ Td xT.⋅−( )+
...:=
Mb 829.59− kN m⋅⋅=
Vb G1 G2+ G3+ G4+ Pab1V+ Pab2V+ Pab3V+ Ppb1V+ Ppb2V+ Ppb3V+ 5.181 103
× kN⋅=:=
-excentricitatea:
eb
Mb
Vb0.16− m=:=
B1b Bcheson 2 eb⋅− 3.32 m=:=
L1 Lcheson:=
-aria efectiva: Ab B1b L1⋅ 23.242 m2
=:=
Pef
Vb
Ab
222.939 kPa⋅=:=
- Factorii adimensionali pt calculul presiunii terenului de fundare:
-factori care tin de capaciatatea portanta:
Nq. eπ tan ϕ3d( )⋅
tan 45degϕ3d
2+
2
⋅ 2.214=:=
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Nc. Nq. 1−( )1
tan ϕ3d( )⋅ 7.841=:=
Nγ. 2 Nq. 1−( ) tan ϕ3d( )⋅ 0.376=:=
-factori care tin de inclinarea bazie fundatiei: ε 0:=
bq. 1 ε tan ϕ3d( )⋅−( )2
1=:=
bγ. bq.:=
bc. bq
1 bq.−
Nc tan ϕ3d( )⋅
− 1=:=
-forma fundatiei:
sq. 1B1b
L1
sin ϕ3d( )⋅+ 1.073=:=
sγ. 1 0.3B1b
L1
⋅− 0.858=:=
sc.
sq. Nq.⋅ 1−
Nq. 1−
1.132=:=
-inclinarea incarcarii
-daca H actioneaza pe directia lui B'=>m1.
2 B1−
L1
+
1B1−
L1
+
2.31=:=
Hd. Pab1H Pab2H+ Pab3H+ Ppb1H− Ppb2H− Ppb3H− Td+ 309.732− kN⋅=:=.
iq. 1Hd.
Vb Ab c3d⋅
1
tan ϕ3d deg⋅( )⋅+
−
m1.
1.037=:=
iγ. 1Hd
Vb Ab c3d⋅1
tan ϕ3d deg⋅( )⋅+
−
m1. 1+
1=:=
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ic. iq.
1 iq.−( )Nc. tan ϕ3d( )⋅
− 1.067=:=
Calculul suprasarcinii:
γmed.
γ1 γ2+ γ3+( )3
17.4kN
m3
⋅=:=
Df. h1 h2+ h3+ 12.4 m=:=
Q1. Df γmed⋅ 215.76kN
m2
⋅=:=
Rd
Ab
c3d Nc.⋅ bc.⋅ sc.⋅ ic.⋅ Q1 Nq.⋅ bq.⋅ sq. iq.⋅+ 0.5 γ3d⋅ m Nγ.⋅ bγ.⋅ sγ.⋅ iγ.⋅+ 549.726 kPa⋅=:=
Rd
Ab
Vb Rd< - verifica
Verificarea la sectiuni periculoase:
Sectiune A-A
xG.1 0.78m:=
xG.4 0.5m:=
P1 3.31m 0.5⋅ m 31.03⋅kN
m2
51.355 kN⋅=:=
P1H P1 cos α1 δ1+( )⋅ 50.87 kN⋅=:=
P1V P1 sin α1 δ1+( )⋅ 7.041 kN⋅=:=
P2 0.8m 0.5⋅ m 25.44⋅kN
m2
10.176 kN⋅=:=
P2H P2 cos α1 δ1+( )⋅ 10.08 kN⋅=:=
P2V P2 sin α1 δ1+( )⋅ 1.395 kN⋅=:=
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V2 G1 G4+ P1V+ P2V+ 597.561 kN⋅=:=
M2 G1 xG1⋅ G4 xG.4⋅− P1H 1.13⋅ m+ P2V 1.25⋅ m+ P2H 0.2667⋅ m− P1V 0⋅ m− 477.656 kN m⋅⋅=:=
W2Bcheson Lcheson
2⋅
624.5 m
3⋅=:=
σmax
V2
Bcheson Lcheson⋅
M2
W2+ 47.951
kN
m2
⋅=:=
σmin
V2
Bcheson Lcheson⋅
M2
W2− 8.959
kN
m2
⋅=:=
Rezistenta betonului la compresiune
Aleg Beton C50/60
f ck 50N
mm2
:=αcc 1:= γc 1.5:=
f cd
αcc f ck ⋅
γc
3.333 104
×kN
m2
⋅=:=
σmax f cd≤ 1=
Rezistenta la intindere a betonului
f ct.k 5.3MPa:= f ct.k 5.3N
mm2
⋅=
αct 1:=f ctd
αct f ct.k ⋅
γc
3.533 103
×kN
m2
⋅=:=
σmin f ctd≤ 1=
NU NECESITA ARMATURA DE LEGATURA INTRE ELEVATIE SI RADIER CULEE
Ipoteza III (G1+G2+G4+Pa+Pp+q+Vg):
Verificarea la rasturnare:
Impingerea activa:
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pac1.0 γQ.dst q0⋅ k a1⋅ 2 γG.dst⋅ c1d⋅ k a1⋅− 9.2691
m2
kN⋅=:=
pac1.1 γQ.dst q0⋅ γG.dst γ1d⋅ h1⋅+( ) k a1⋅ 2 γG.dst⋅ c1d⋅ k a1⋅− 61.2581
m2
kN⋅=:=
pac2.0 γQ.dst q0⋅ γG.dst γ1d⋅ h1⋅+( ) k a2⋅ 2 γG.dst⋅ c2d⋅ k a2⋅− 47.691
m2
kN⋅=:=
pac2.1 γQ.dst q0⋅ γG.dst γ1d⋅ h1⋅+ γG.dst γ2d⋅ h2⋅+( ) k a2⋅ 2 γG.dst⋅ c2d k a2⋅− 125.0551
m2
kN⋅=:=
pac3.0 γQ.dst q0⋅ γG.dst γ1d⋅ h1⋅+ γG.dst γ2d⋅ h2⋅+( ) k a3⋅ 2 γG.dst⋅ c3d⋅ k a3⋅− 168.3571
m2
kN⋅=:=
pac3.1 γQ.dst q0⋅ γG.dst γ1d⋅ h1⋅+ γG.dst γ2d⋅ h2⋅+ γG.dst γ3d⋅ h3⋅+( ) k a3⋅ 2 γG.dst⋅ c3d⋅ k a3⋅−:=
pac3.1 181.489 kN1
m2
⋅⋅=
Descompunerea impingerii dupa orizontala si verticala:
Pac1 pac1.1 pac1.0+( )h1
2⋅
m⋅ 169.265 kN⋅=:=
Pac1H Pac1 cos2
3ϕ1d
⋅ 166.427 kN⋅=:=
Pac1V Pac1 sin
2
3 ϕ1d
⋅
30.865 kN⋅=:=
zac1
h1 2 pac1.0⋅ pac1.1+( )⋅
pac1.0 pac1.1+
31.81m=:=
Pac2 m pac2.1 pac2.0+( )h2
2⋅
⋅ 570.061 kN⋅=:=
Pac2H Pac2 cos 23
ϕ2d
⋅ 564.513 kN⋅=:=
Pac2V Pac2 sin2
3ϕ2d
⋅ 79.337 kN⋅=:=
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zac2
h2 2 pac2.0⋅ pac2.1+( )⋅
pac2.0 pac2.1+
32.807m=:=
Pac3 m pac3.1 pac3.0+( ) h32
⋅
⋅ 174.923 kN⋅=:=
Pac3H Pac3 cos2
3ϕ3d
⋅ 174.007 kN⋅=:=
Pac3V Pac3 sin2
3ϕ3d
⋅ 17.88 kN⋅=:=
zac3
h3 2 pac3.0⋅ pac3.1+( )⋅
pac3.0 pac3.1+
30.494m=:=
Impingerea pasiva:
h1.' 2.2 m⋅:=
ppc1.0 γQ.dst q0⋅ k p1⋅ 2 γG.dst⋅ c1d⋅ k p1⋅− 43.2471
m2
kN⋅=:=
ppc1.1 γQ.dst q0⋅ γG.dst γ1d⋅ h1'⋅+( ) k p1⋅ 2 γG.dst⋅ c1d⋅ k p1⋅− 134.921
m2
kN⋅=:=
ppc2.0
γQ.dst
q0
⋅ γG.dst
γ1d
⋅ h1'
⋅+
( )k
p2⋅ 2 γ
G.dst⋅ c
2d⋅ k
p2⋅− 72.88
1
m2
kN⋅=:=
ppc2.1 γQ.dst q0⋅ γG.dst γ1d⋅ h1'⋅+ γG.dst γ2d⋅ h2⋅+( ) k p2⋅ 2 γG.dst⋅ c2d k p2⋅− 287.5871
m2
kN⋅=:=
ppc3.0 γQ.dst q0⋅ γG.dst γ1d⋅ h1'⋅+ γG.dst γ2d⋅ h2⋅+( ) k p3⋅ 2 γG.dst⋅ c3d⋅ k p3⋅− 280.7191
m2
kN⋅=:=
ppc3.1 γQ.dst q0⋅ γG.dst γ1d⋅ h1'⋅+ γG.dst γ2d⋅ h2⋅+ γG.dst γ3d⋅ h3⋅+( ) k p3⋅ 2 γG.dst⋅ c3d⋅ k p3⋅−:=
ppc3.1 308.442 kN1
m2
⋅⋅=
Descompunerea impingerii dupa orizontala si verticala:
Ppc1 ppc1.1 ppc1.0+( )h1
2⋅ m⋅ 427.602 kN⋅=:=
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Ppc1H Ppc1 cos2
3ϕ1d
⋅ 420.432 kN⋅=:=
Ppc1V Ppc1 sin2
3ϕ1d
⋅ 77.973 kN⋅=:=
zpc1.
h1' 2 ppc1.0⋅ ppc1.1+( )⋅
ppc1.0 ppc1.1+
30.911m=:=
Ppc2 m ppc2.1 ppc2.0+( )h2
2⋅
⋅ 1.19 10
3× kN⋅=:=
Ppc2H Ppc2 cos2
3ϕ2d
⋅ 1.178 103
× kN⋅=:=
Ppc2V Ppc2 sin 23
ϕ2d ⋅ 165.552 kN⋅=:=
zpc2
h2 2 ppc2.0⋅ ppc2.1+( )⋅
ppc2.0 ppc2.1+
32.645m=:=
Ppc3 m ppc3.1 ppc3.0+( )h3
2⋅
⋅ 294.581 kN⋅=:=
Ppc3H Ppc3 cos2
3
ϕ3d
⋅ 293.038 kN⋅=:=
Ppc3V Ppc3 sin2
3ϕ3d
⋅ 30.11 kN⋅=:=
zpc3
h3 2 ppc3.0⋅ ppc3.1+( )⋅
ppc3.0 ppc3.1+
30.492m=:=
Sectiunea a-a
- brate:
xG1 0.78 m⋅:= xac1 1.3 m⋅:= xpc1 0 m⋅:= xVg 0.75m:=
xG4 1.80 m⋅:= yac1 0.41 m⋅:= ypc1 0 m⋅:=
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- momente:
M3stb G1 xG1⋅ G4 xG4⋅+ Pac1V xac1⋅+ Ppc1V xpc1⋅− Vg xVg⋅+ 1.58 103
× kN m⋅⋅=:=
M3dstb Pac1H yac1⋅
Ppc1H ypc1⋅−
68.235 kN m⋅⋅=:=
M3dst M3stb< - verifica
Sectiunea b-b
- brate:
xG1. 2.48 m⋅:= xac1. 3 m⋅:= xpc1. 0 m⋅:= yac1. 9.32 m⋅:= ypc1. 8.42 m⋅:=
xG2. 1.5m:= xac2. 3 m⋅:= xpc2. 0 m⋅:= yac2. 3.77 m⋅:= ypc2. 3.63 m⋅:=
xG3. 1.5 m⋅:= xac3. 3 m⋅:= xpc3. 0 m⋅:= yac3. 0.49 m⋅:= ypc3. 0.49 m⋅:=
xG4. 3.5 m⋅:=xVg. 2.45 m⋅:=
- momente:
M3stb. G1 xG1.⋅ G2 xG2.⋅+ G3 xG3.⋅+ G4 xG4.⋅+ Pac1V xac1.⋅+ Pac2V xac2.⋅+ Pac3V xac3.⋅+
Ppc1V− xpc1.⋅ Ppc2V xpc2.⋅− Ppc3V xpc3.⋅−( ) Vg xVg.⋅++
...:=
M3stb. 1.176 104
× kN m⋅⋅=
M3dstb. Pac1H yac1.⋅
Pac2H yac2.⋅+
Pac3H yac3.⋅+
Ppc1H ypc1.⋅−
Ppc2H ypc2.⋅−
Ppc3H ypc3.⋅−:=
M3dstb. 4.195− 103
× kN m⋅⋅=
M3dst. M3stb.< - verifica
Verificarea la capacitate portanta:
Impingerea activa:
pac1.0. γQ q0⋅ k a1⋅ 2 γG⋅ c1d⋅ k a1⋅− 7.8621
m2
kN⋅=:=
pac1.1. γQ q0⋅ γG γ1d⋅ h1⋅+( ) k a1⋅ 2 γG⋅ c1d⋅ k a1⋅− 55.1261
m2
kN⋅=:=
pac2.0. γQ q0⋅ γG γ1d⋅ h1⋅+( ) k a2⋅ 2 γG⋅ c2d⋅ k a2⋅− 42.7091
m2
kN⋅=:=
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pac2.1. γQ q0⋅ γG γ1d⋅ h1⋅+ γG γ2d⋅ h2⋅+( ) k a2⋅ 2 γG⋅ c2d k a2⋅− 113.0411
m2
kN⋅=:=
pac3.0. γQ q0⋅ γG γ1d⋅ h1⋅+ γG γ2d⋅ h2⋅+( ) k a3⋅ 2 γG⋅ c3d⋅ k a3⋅− 152.3241
m
2kN⋅=:=
pac3.1. γQ q0⋅ γG γ1d⋅ h1⋅+ γG γ2d⋅ h2⋅+ γG γ3d⋅ h3⋅+( ) k a3⋅ 2 γG⋅ c3d⋅ k a3⋅− 164.2621
m2
kN⋅=:=
Descompunerea impingerii dupa orizontala si verticala:
Pac1. pac1.1. pac1.0.+( )h1
2⋅
m⋅ 151.17 kN⋅=:=
Pac1H. Pac1. cos2
3ϕ1d
⋅ 148.636 kN⋅=:=
Pac1V. Pac1. sin2
3ϕ1d
⋅ 27.566 kN⋅=:=
zac1.
h1 2 pac1.0.⋅ pac1.1.+( )⋅
pac1.0. pac1.1.+
31.8m=:=
Pac2. m pac2.1. pac2.0.+( )h2
2⋅
⋅ 513.974 kN⋅=:=
Pac2H. Pac2. cos 2
3ϕ2d
⋅ 508.972 kN⋅=:=
Pac2V. Pac2. sin2
3ϕ2d
⋅ 71.531 kN⋅=:=
zac2.
h2 2 pac2.0.⋅ pac2.1.+( )⋅
pac2.0. pac2.1.+
32.803m=:=
Pac3. m pac3.1. pac3.0.+( )h
32
⋅
⋅ 158.293 kN⋅=:=
Pac3H. Pac3. cos2
3ϕ3d
⋅ 157.464 kN⋅=:=
Pac3V. Pac3. sin2
3ϕ3d
⋅ 16.18 kN⋅=:=
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zac3.
h3 2 pac3.0.⋅ pac3.1.+( )⋅
pac3.0. pac3.1.+
30.494m=:=
Impingerea pasiva:
ppc1.0. γQ.dst q0⋅ k p1⋅ 2 γG⋅ c1d⋅ k p1⋅− 43.9651
m2
kN⋅=:=
ppc1.1. γQ.dst q0⋅ γG γ1d⋅ h1'⋅+( ) k p1⋅ 2 γG⋅ c1d⋅ k p1⋅− 127.3041
m2
kN⋅=:=
ppc2.0. γQ.dst q0⋅ γG γ1d⋅ h1'⋅+( ) k p2⋅ 2 γG⋅ c2d⋅ k p2⋅− 70.0961
m2
kN⋅=:=
ppc2.1. γQ.dst q0⋅
γG γ1d⋅
h1'⋅+
γG γ2d⋅
h2⋅+
( ) k p2⋅
2 γG⋅
c2d k p2⋅−
265.285
1
m2 kN
⋅=:=
ppc3.0. γQ.dst q0⋅ γG γ1d⋅ h1'⋅+ γG γ2d⋅ h2⋅+( ) k p3⋅ 2 γG⋅ c3d⋅ k p3⋅− 258.4921
m2
kN⋅=:=
ppc3.1. γQ.dst q0⋅ γG γ1d⋅ h1'⋅+ γG γ2d⋅ h2⋅+ γG γ3d⋅ h3⋅+( ) k p3⋅ 2 γG⋅ c3d⋅ k p3⋅− 283.6961
m2
k ⋅=:=
Descompunerea impingerii dupa orizontala si verticala:
Ppc1. ppc1.1. ppc1.0.+( )h1
2⋅ m⋅ 411.044 kN⋅=:=
Ppc1H. Ppc1. cos 23
ϕ1d
⋅ 404.153 kN⋅=:=
Ppc1V. Ppc1. sin2
3ϕ1d
⋅ 74.954 kN⋅=:=
zp.c1.
h1' 2 ppc1.0.⋅ ppc1.1.+( )⋅
ppc1.0. ppc1.1.+
30.922m=:=
Ppc2. m ppc2.1. ppc2.0.+( )h2
2⋅
⋅ 1.107 10
3× kN⋅=:=
Ppc2H. Ppc2. cos2
3ϕ2d
⋅ 1.096 103
× kN⋅=:=
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Ppc2V. Ppc2. sin2
3ϕ2d
⋅ 154.031 kN⋅=:=
zpc2.
h2 2 ppc2.0.⋅ ppc2.1.+( )⋅
ppc2.0. ppc2.1.+
32.66 m=:=
Ppc3. m ppc3.1. ppc3.0.+( )h3
2⋅
⋅ 271.094 kN⋅=:=
Ppc3H. Ppc3. cos2
3ϕ3d
⋅ 269.674 kN⋅=:=
Ppc3V. Ppc3. sin2
3ϕ3d
⋅ 27.71 kN⋅=:=
zpc3.
h3
2 ppc3.0.
⋅ ppc3.1.
+
( )⋅
ppc3.0. ppc3.1.+
30.492m=:=
- brate:
xG1.. 0.98 m⋅:= xac1.. 1.5 m⋅:= xpc1.. 1.5 m⋅:= yac1.. 9.32 m⋅:= ypc1.. 8.43 m⋅:=
xG4.. 2 m⋅:= xac2.. 1.5 m⋅:= xpc2.. 1.5 m⋅:= yac2.. 3.81 m⋅:= ypc2.. 3.65 m⋅:=
xac3.. 1.5 m⋅:= xpc3.. 1.5 m⋅:= yac3.. 0.49 m⋅:= ypc3.. 0.49 m⋅:=xVg.. 0.95 m⋅:=
Mc G1 xG1..⋅ G4 xG4..⋅+ Pac1V xac1..⋅+ Pac2V xac2..⋅+ Pac3V xac3..⋅+ Ppc1V xpc1..⋅−
Ppc2V xpc2..⋅ Ppc3V xpc3..⋅− Pac1H yac1..⋅− Pac2H yac2..⋅− Pac3H yac3..⋅−( )+
...
Ppc1H ypc1..⋅ Ppc2H ypc2..⋅+ Ppc2H ypc3..⋅+ Vg xVg..⋅++
...
:=
Mc 6.85 103
× kN m⋅⋅=
Vc. Vg G1+ G2+ G3+ G4+ Pac1V Pac2V+ Pac3V+ Ppc1V+ Ppc2V++ Ppc3V+ 6.696 103
× ⋅=:=
-excentricitatea:
ec
Mc
Vc.1.023m=:=
B1c Bcheson 2 ec⋅− 0.954m=:=
L1 Lcheson:=
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-aria efectiva: Ac B1c L1⋅ 6.677m2
=:=
Pef
Vc.
Ac
1.003 103
× kPa⋅=:=
- Factorii adimensionali pt calculul presiunii terenului de fundare:
-factori care tin de capaciatatea portanta:
Nq.. eπ tan ϕ3d( )⋅
tan 45degϕ3d
2+
2
⋅ 2.214=:=
Nc.. Nq.. 1−( )1
tan ϕ3d( )⋅ 7.841=:=
Nγ.. 2 Nq.. 1−( ) tan ϕ3d( )⋅ 0.376=:=
-factori care tin de inclinarea bazie fundatiei: ε 0:=
bq.. 1 ε tan ϕ3d( )⋅−( )2
1=:=
bγ.. bq..:=
bc.. bq
1 bq..−
Nc.. tan ϕ3d( )⋅
− 1=:=
-forma fundatiei:
sq.. 1B1c
L1
sin ϕ3d( )⋅+ 1.021=:=
sγ.. 1 0.3B1c
L1
⋅− 0.959=:=
s
c..
sq Nq⋅ 1−
Nq 1−
1.066=:=
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:=
-inclinarea incarcarii
-daca H actioneaza pe directia lui B'=>m1..
2B1−
L1
+
1 B1−
L1
+
2.31=:=
Hd.. Pac1H Pac2H+ Pac3H+ Ppc1H− Ppc2H− Ppc3H− Td+ 686.489− kN⋅=:=.
iq.. 1Hd
Vg Ac c3d⋅1
tan ϕ3d deg⋅( )⋅+
−
m1
1=:=
iγ.. 1Hd
Vg Ac c3d⋅1
tan ϕ3d deg⋅( )⋅+
−
m1 1+
1=:=
ic.. iq
1 iq−( )Nc tan ϕ3d( )⋅
− 1=:=
Calculul suprasarcinii:
γmed..
γ1 γ2+ γ3+( )3
17.4kN
m3
⋅=:=
Df.. h1 h2+ h3+ 12.4 m=:=
Q1.. Df γmed⋅ 215.76kN
m2
⋅=:=
Rd
Ac
c3d Nc⋅ bc⋅ sc⋅ ic⋅ Q1 Nq⋅ bq⋅ sq⋅ iq⋅+ 0.5 γ3d⋅ m Nγ⋅ bγ⋅ sγ⋅ iγ⋅+ 511.951 kPa⋅=:=
Rd
Ac
Vc Rd< - verifica
Verificarea la Sectiuni periculoase:
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xG.1 0.78m:=
xG.4 0.5m:=
P1
3.31m 0.5⋅ m 51.03⋅kN
m2
84.455 kN⋅=:=
P1H P1 cos α1 δ1+( )⋅ 83.657 kN⋅=:=
P1V P1 sin α1 δ1+( )⋅ 11.579 kN⋅=:=
P2 0.8m 0.5⋅ m 45.44⋅kN
m2
18.176 kN⋅=:=
P2H P2 cos α1 δ1+( )⋅ 18.004 kN⋅=:=
P2V P2 sin α1 δ1+( )⋅ 2.492 kN⋅=:= :=
V3 G1 G4+ P1V+ P2V+ 603.196 kN⋅=:=
M3 G1 xG1⋅ G4 xG.4⋅− P1H 1.13⋅ m+ P2V 1.25⋅ m+ P2H 0.2667⋅ m− P1V 0⋅ m− Vg 0.55⋅ m+ 1.284 1×=:=
W3Bcheson Lcheson
2⋅
624.5 m
3⋅=:=
σmax
V3
Bcheson Lcheson⋅
M3
W2+ 81.13
kN
m2
⋅=:=
σmin
V3
Bcheson Lcheson⋅
M3
W3− 23.683−
kN
m2
⋅=:=
Rezistenta betonului la compresiune
Aleg Beton C50/60
f ck 50N
mm2
:=αcc 1:= γc 1.5:=
f cd
αcc f ck ⋅
γc
3.333 104
×kN
m2
⋅=:=
σmax f cd≤ 1=
Rezistenta la intindere a betonului
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f ct.k 5.3MPa:= f ct.k 5.3N
mm2
⋅=
αct 1:=
f ctd
αct f ct.k ⋅
γc 3.533 10
3×
kN
m2
⋅=:=
σmin f ctd≤ 1=
NU NECESITA ARMATURA DE LEGATURA INTRE ELEVATIE SI RADIER CULEE
Ipoteza IV (G1+G2+G3+G4+Pa+Pp+Vc+Vg+Td):
Verificarea la rasturnare:
Impingerea activa:
qk. 0kN
m3
:=
pad1.0 γQ.dst qk ⋅ k a1⋅ 2 γG.dst⋅ c1d⋅ k a1⋅− 4.024−1
m2
kN⋅=:=
pad1.1 γQ.dst qk ⋅ γG.dst γ1d⋅ h1⋅+( ) k a1⋅ 2 γG.dst⋅ c1d⋅ k a1⋅− 47.9661
m2
kN⋅=:=
pad2.0 γQ.dst qk ⋅
γG.dst γ1d⋅
h1⋅+
( ) k a2⋅
2 γG.dst⋅
c2d⋅
k a2⋅−
32.463
1
m2 kN
⋅=:=
pad2.1 γQ.dst qk ⋅ γG.dst γ1d⋅ h1⋅+ γG.dst γ2d⋅ h2⋅+( ) k a2⋅ 2 γG.dst⋅ c2d k a2⋅− 109.8281
m2
kN⋅=:=
pad3.0 γQ.dst qk ⋅ γG.dst γ1d⋅ h1⋅+ γG.dst γ2d⋅ h2⋅+( ) k a3⋅ 2 γG.dst⋅ c3d⋅ k a3⋅− 151.1981
m2
kN⋅=:=
pad3.1 γQ.dst qk ⋅ γG.dst γ1d⋅ h1⋅+ γG.dst γ2d⋅ h2⋅+ γG.dst γ3d⋅ h3⋅+( ) k a3⋅ 2 γG.dst⋅ c3d⋅ k a3⋅−:=
pad3.1 164.33 kN1
m2
⋅⋅=
Descompunerea impingerii dupa orizontala si verticala:
hc'. 0.32m:=
Pad1 pad1.1 pad1.0+( )h1
2⋅
m⋅ 105.459 kN⋅=:=
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Pad1H Pad1 cos2
3ϕ1d
⋅ 103.691 kN⋅=:=
Pad1V Pad1 sin2
3
ϕ1d
⋅ 19.231 kN⋅=:=
zad11
3h1 hc'−( )⋅ 1.493m=:=
Pad2 m pad2.1 pad2.0+( )h2
2⋅
⋅ 469.56 kN⋅=:=
Pad2H Pab2 cos2
3ϕ2d
⋅ 464.991 kN⋅=:=
Pad2V Pab2 sin2
3ϕ2d
⋅ 65.35 kN⋅=:=
zad2
h2 2 pad2.0⋅ pad2.1+( )⋅
pad2.0 pad2.1+
32.702m=:=
Pad3 m pad3.1 pad3.0+( )h3
2⋅
⋅ 157.764 kN⋅=:=
Pad3H
Pad3
cos2
3ϕ
3d
⋅ 156.937 kN⋅=:=
Pad3V Pad3 sin2
3ϕ3d
⋅ 16.126 kN⋅=:=
zad3
h3 2 pad3.0⋅ pad3.1+( )⋅
pad3.0 pad3.1+
30.493m=:=
Impingerea pasiva:
h.1.' 2.2 m⋅:=
ppd1.0 γQ.dst qk ⋅ k p1⋅ 2 γG.dst⋅ c1d⋅ k p1⋅− 7.893−1
m2
kN⋅=:=
ppd1.1 γQ.dst qk ⋅ γG.dst γ1d⋅ h1'⋅+( ) k p1⋅ 2 γG.dst⋅ c1d⋅ k p1⋅− 83.781
m2
kN⋅=:=
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ppd2.0 γQ.dst qk ⋅ γG.dst γ1d⋅ h1'⋅+( ) k p2⋅ 2 γG.dst⋅ c2d⋅ k p2⋅− 30.621
m2
kN⋅=:=
ppd2.1 γQ.dst qk ⋅ γG.dst γ1d⋅ h1'⋅+ γG.dst γ2d⋅ h2⋅+( ) k p2⋅ 2 γG.dst⋅ c2d k p2⋅− 245.3281
m
2kN⋅=:=
ppd3.0 γQ.dst qk ⋅ γG.dst γ1d⋅ h1'⋅+ γG.dst γ2d⋅ h2⋅+( ) k p3⋅ 2 γG.dst⋅ c3d⋅ k p3⋅− 244.4921
m2
kN⋅=:=
ppd3.1 γQ qk ⋅ γG.dst γ1d⋅ h1'⋅+ γG.dst γ2d⋅ h2⋅+ γG.dst γ3d⋅ h3⋅+( ) k p3⋅ 2 γG.dst⋅ c3d⋅ k p3⋅−:=
ppd3.1 272.216 kN1
m2
⋅⋅=
Descompunerea impingerii dupa orizontala si verticala:
h'c. 0.21m:=
Ppd1 ppd1.1 ppd1.0+( )h1
2⋅ m⋅ 182.129 kN⋅=:=
Ppd1H Ppd1 cos2
3ϕ1d
⋅ 179.075 kN⋅=:=
Ppd1V Ppd1 sin2
3ϕ1d
⋅ 33.211 kN⋅=:=
zpd11
3h1' hc−( ) 0.663m=:=
Ppd2 m ppd2.1 ppd2.0+( )h2
2⋅
⋅ 910.629 kN⋅=:=
Ppd2H Ppd2 cos2
3ϕ2d
⋅ 901.767 kN⋅=:=
Ppd2V Ppb2 sin2
3ϕ2d
⋅ 126.735 kN⋅=:=
z
pd2
h2 2 ppd2.0⋅ ppd2.1+( )⋅
ppd2.0 ppd2.1+
3
2.444m=:=
Ppd3 m ppd3.1 ppd3.0+( )h3
2⋅
⋅ 258.354 kN⋅=:=
Ppd3H Ppd3 cos2
3ϕ3d
⋅ 257.001 kN⋅=:=
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Ppd3V Ppd3 sin2
3ϕ3d
⋅ 26.407 kN⋅=:=
zpd3
h3 2 ppd3.0⋅ ppd3.1+( )⋅
ppd3.0 ppd3.1+
30.491m=:=
Sectiunea a-a
- brate:
xG..1 0.78 m⋅:= xad.1 13 m⋅:= xT.. 2.9m:=
xG..4. 1.80 m⋅:= yad.1 0.09 m⋅:= xN 0.75m:=
- momente:
M4stb G1 xG1⋅ G4 xG4.⋅+ Pad1V xad.1⋅+ Nd xN⋅+ 2.291 103
× kN m⋅⋅=:=
M4dstb Pad1H yad.1⋅ Td xT⋅+ 879.332 kN m⋅⋅=:=
M4dst M4stb< - verifica
Sectiunea b-b
- brate:
xG.1.. 2.48 m⋅:= xad1. 3 m⋅:= xpd1. 0 m⋅:= yad1. 9 m⋅:= ypd1. 8.14 m⋅:=
xG.2.. 1.5m:= xad2. 3 m⋅:= xpd2. 0 m⋅:= yad2. 3.71 m⋅:= ypd2. 3.43 m⋅:=
xG.3. 1.5 m⋅:= xad3. 3 m⋅:= xpd3. 0 m⋅:= yad3. 0.49 m⋅:= ypd3. 0.49 m⋅:=
xG.4.. 3.5 m⋅:=
xT... 11.81 m⋅:= xN. 2.45m:=
- momente:
M4stb. G1 xG1..⋅ G2 xG2..⋅+ G3 xG3.⋅+ G4 xG4.⋅+ Pad1V xad1.⋅+ Pad2V xad2.⋅+ Pad3V xad3.⋅+ N+:=
M4stb. 8.912 10
3
× kN m⋅⋅=
M4dstb. Pad1H yad1.⋅ Pad2H yad2.⋅+ Pad3H yad3.⋅+ Td xT.⋅+ Ppd1H ypd1.⋅− Ppd2H ypd2.⋅−
Ppd3H ypd3.⋅ 1−⋅( )+
...:=
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M4dstb. 1.602 103
× kN m⋅⋅=
M4dst M4stb< - verifica
Verificarea la capacitate portanta
Impingerea activa:
pad1.0. γQ qk ⋅ k a1⋅ 2 γG⋅ c1d⋅ k a1⋅− 3.658−1
m2
kN⋅=:=
pad1.1. γQ qk ⋅ γG γ1d⋅ h1⋅+( ) k a1⋅ 2 γG⋅ c1d⋅ k a1⋅− 43.6051
m2
kN⋅=:=
pad2.0. γQ qk ⋅ γG γ1d⋅ h1⋅+( ) k a2⋅ 2 γG⋅ c2d⋅ k a2⋅− 29.5121
m2
kN⋅=:=
pad2.1. γQ qk ⋅ γG γ1d⋅ h1⋅+ γG γ2d⋅ h2⋅+( ) k a2⋅ 2 γG⋅ c2d k a2⋅− 99.8441
m2
kN⋅=:=
pad3.0. γQ qk ⋅ γG γ1d⋅ h1⋅+ γG γ2d⋅ h2⋅+( ) k a3⋅ 2 γG⋅ c3d⋅ k a3⋅− 137.4521
m2
kN⋅=:=
pad3.1. γQ qk ⋅ γG γ1d⋅ h1⋅+ γG γ2d⋅ h2⋅+ γG γ3d⋅ h3⋅+( ) k a3⋅ 2 γG⋅ c3d⋅ k a3⋅− 149.3911
m2
kN⋅=:=
Descompunerea impingerii dupa orizontala si verticala:
hc.. 0.3m:=
Pad1. pad1.1. pad1.0.+( )h1
2⋅
m⋅ 95.872 kN⋅=:=
Pad1H. Pad1. cos2
3ϕ1d
⋅ 94.265 kN⋅=:=
Pad1V. Pad1. sin2
3ϕ1d
⋅ 17.482 kN⋅=:=
zad1.1
3h1 hc.−( )⋅ 1.53 m=:=
Pad2. m pad2.1. pad2.0.+( )
h2
2⋅
⋅ 426.873 kN⋅=:=
Pad2H. Pad2. cos2
3ϕ2d
⋅ 422.719 kN⋅=:=
Pad2V. Pad2. sin2
3ϕ2d
⋅ 59.409 kN⋅=:=
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zad2.
h2 2 pad2.0.⋅ pad2.1.+( )⋅
pad2.0. pad2.1.+
32.702m=:=
Pad3. m pad3.1. pad3.0.+( ) h32
⋅
⋅ 143.421 kN⋅=:=
Pad3H. Pad3. cos2
3ϕ3d
⋅ 142.67 kN⋅=:=
Pad3V. Pad3. sin2
3ϕ3d
⋅ 14.66 kN⋅=:=
zad3.
h3 2 pad3.0.⋅ pad3.1.+( )⋅
pad3.0. pad3.1.+
30.493m=:=
Impingerea pasiva:
ppd1.0. 2− γG⋅ c1d⋅ k p1⋅ 7.176−1
m2
kN⋅=:=
ppd1.1. γG γ1d⋅ h1'⋅( ) k p1⋅ 2 γG⋅ c1d⋅ k p1⋅− 76.1641
m2
kN⋅=:=
ppd2.0. γG γ1d⋅ h1'⋅( ) k p2⋅ 2 γG⋅ c2d⋅ k p2⋅− 27.8371
m2
kN⋅=:=
ppd2.1. γG γ1d⋅ h1'⋅ γG γ2d⋅ h2⋅+( ) k p2⋅ 2 γG⋅ c2d k p2⋅− 223.0251
m2
kN⋅=:=
ppd3.0. γG γ1d⋅ h1'⋅ γG γ2d⋅ h2⋅+( ) k p3⋅ 2 γG⋅ c3d⋅ k p3⋅− 222.2661
m2
kN⋅=:=
ppd3.1. γG γ1d⋅ h1'⋅ γG γ2d⋅ h2⋅+ γG γ3d⋅ h3⋅+( ) k p3⋅ 2 γG⋅ c3d⋅ k p3⋅− 247.4691
m2
kN⋅=:=
Descompunerea impingerii dupa orizontala si verticala:
h.c. 0.21m:=
Ppd1. ppd1.1. ppd1.0.+( )
h1
2⋅ m⋅ 165.572 kN⋅=:=
Ppd1H. Ppd1. cos2
3ϕ1d
⋅ 162.796 kN⋅=:=
Ppd1V. Ppd1. sin2
3ϕ1d
⋅ 30.192 kN⋅=:=
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zpd1.1
3h1' hc.−( )⋅ 0.663m=:=
Ppd2. m ppd2.1. ppd2.0.+( )h2
2
⋅
⋅ 827.844 kN⋅=:=
Ppd2H. Ppd2. cos2
3ϕ2d
⋅ 819.788 kN⋅=:=
Ppd2V. Ppd2. sin2
3ϕ2d
⋅ 115.214 kN⋅=:=
zpd2.
h2 2 ppd2.0.⋅ ppd2.1.+( )⋅
ppd2.0. ppd2.1.+
32.444m=:=
Ppd3. m ppd3.1. ppd3.0.+( )
h3
2⋅
⋅ 234.868 kN⋅=:=
Ppd3H. Ppd3. cos2
3ϕ3d
⋅ 233.637 kN⋅=:=
Ppd3V. Ppd3. sin2
3ϕ3d
⋅ 24.007 kN⋅=:=
zpd3.
h3 2 ppd3.0.⋅ ppd3.1.+( )⋅
ppd3.0. ppd3.1.+
30.491m=:=
- brate:
xG.1. 0.98 m⋅:= xad1.. 1.5 m⋅:= xpd1.. 1.5 m⋅:= yad1.. 9.01 m⋅:= ypd1.. 8.17 m⋅:=
xG.4.. 2 m⋅:= xad2.. 1.5 m⋅:= xpd2.. 1.5 m⋅:= yad2.. 3.71 m⋅:= ypd2.. 3.43 m⋅:=
xG2. 0m:= xad3.. 1.5 m⋅:= xpd3.. 1.5 m⋅:= yad3.. 0.49 m⋅:= ypd3.. 0.49 m⋅:=
xG3 0m:= x.T.. 11.81 m⋅:= xN.. 0.95m:=
Md G1 xG1.⋅ G4 xG4.⋅+ Pad1V xad1..⋅+ Pad2V xad2..⋅+ Pad3V xad3..⋅+ Ppd1V xpd1..⋅−
Ppd2V− xpd2..⋅ Ppd3V xpd3..⋅− Pad1H yad1..⋅− Pad2H yad2..⋅− Pad3H yad3..⋅−( )+
...
Ppd1H ypd1..⋅ Ppd2H ypd2..⋅+ Ppd2H ypd3..⋅+ Td xT.⋅− Nd xN.⋅++
...
:=
Md 4.982 103
× kN m⋅⋅=
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Vd G1 G2+ G3+ G4+ Pad1V+ Pad2V+ Pad3V+ Ppd1V+ Ppd2V+ Ppd3V+ Nd+ 7.181 103
× k ⋅=:=
-excentricitatea:
edMdVd
0.694m=:=
B1d Bcheson 2 eb⋅− 3.32 m=:=
L1 Lcheson:=
-aria efectiva: Ad B1d L1⋅ 23.242 m2
=:=
Pef
Vd
Ad
308.972 kPa⋅=:=
- Factorii adimensionali pt calculul presiunii terenului de fundare:
-factori care tin de capaciatatea portanta:
N.q. eπ tan ϕ3d( )⋅
tan 45degϕ3d
2+
2
⋅ 2.214=:=
N.c. Nq. 1−( )1
tan ϕ3d( )⋅ 7.841=:=
N.γ.
2 Nq.
1−
( )tan ϕ
3d( )⋅ 0.376=:=
-factori care tin de inclinarea bazie fundatiei: ε 0:=
b.q. 1 ε tan ϕ3d( )⋅−( )2
1=:=
b.γ. bq.:=
b.c. bq
1 bq.−
Nc tan ϕ3d( )⋅
− 1=:=
-forma fundatiei:
s.q. 1B1d
L1
sin ϕ3d( )⋅+ 1.073=:=
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s.γ. 1 0.3B1d
L1
⋅− 0.858=:=
s.c.
sq. Nq.⋅ 1−
Nq. 1−
1.132=:=
-inclinarea incarcarii
-daca H actioneaza pe directia lui B'=>m.1.
2B1−
L1
+
1B1−
L1
+
2.31=:=
H.d. Pad1H Pad2H+ Pad3H+ Ppd1H− Ppd2H− Ppd3H− Td+ 312.224− kN⋅=:=.
i.q. 1 H.d.
Vb Ad c3d⋅1
tan ϕ3d deg⋅( )⋅+
−
m1.
1.037=:=
i.γ. 1H.d.
Vb Ad c3d⋅1
tan ϕ3d deg⋅( )⋅+
−
m1. 1+
1.053=:=
i.c. iq.
1 iq.−( )Nc. tan ϕ3d( )⋅
− 1.067=:=
Calculul suprasarcinii:
γ.med
γ1 γ2+ γ3+( )3
17.4kN
m3
⋅=:=
D.f h1 h2+ h3+ 12.4 m=:=
Q.1 Df γmed⋅
215.76
kN
m2
⋅=:=
Rd
Ad
c3d Nc.⋅ bc.⋅ sc.⋅ ic.⋅ Q1 Nq.⋅ bq.⋅ sq. iq.⋅+ 0.5 γ3d⋅ m Nγ.⋅ bγ.⋅ sγ.⋅ iγ.⋅+ 549.726 kPa⋅=:=
Rd
Ad
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Vd Rd< - verifica
Verificare sectiuni periculoase:
xG1 51.68cm 0.517m=:=
xG4 50cm 0.5m=:=
P1 3.4m 0.5⋅ m 51.27⋅kN
m2
87.159 kN⋅=:= P1H P1 cos α1 δ1+( )⋅ 86.336 kN⋅=:=
P1V P1 sin α1 δ1+( )⋅ 11.949 kN⋅=:=
P2 0.8m 0.5⋅ m 25.44⋅
kN
m2 10.176 kN⋅=:=
P2H P2 cos α1 δ1+( )⋅ 10.08 kN⋅=:=
P2V P2 sin α1 δ1+( )⋅ 1.395 kN⋅=:=
V4b G1 G4+ P1V+ P2V+ Vg+ Vc+ 2.602 103
× kN⋅=:=
M4b G1 xG1⋅ G4 xG4⋅− P1H 1.13⋅ m+ P2V 1.25⋅ m+ P2H 0.2667⋅ m−
P1V 0⋅ m 0.55m Vg Vc+( )⋅+ Td 2.77⋅ m++
...:=
M4b 2.302 103× kN m⋅⋅=
M'4b G1 xG1⋅ G4 xG4⋅+ Ppd3V. 1.13⋅ m− Ppd3V. 1.25⋅ m+ Ppd1V. 0.2667⋅ m+
Ppd3V. 0⋅ m 0.75m Vg Vc+( )⋅+ Td 2.90⋅ m−+
...:=
M'4b 944.889 kN m⋅⋅=
W4bBcheson Lcheson
2⋅
624.5 m
3⋅=:=
σmax
V4b
Bcheson Lcheson⋅
M4b
W4b+ 217.869
kN
m2
⋅=:=
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σmax1
V4b
Bcheson Lcheson⋅
M'4b
W4b+ 162.494
kN
m2
⋅=:=
σmin
V4b
Bcheson Lcheson⋅
M4b
W4b− 29.985 kN
m2
⋅=:=
σmin1
V4b
Bcheson Lcheson⋅
M'4b
W4b− 85.36
kN
m2
⋅=:=
Rezistenta betonului la compresiune
Aleg Beton C50/60
f ck 50 N
mm2
:=αcc 1:= γc 1.5:=
f cd
αcc f ck ⋅
γc
3.333 104
×kN
m2
⋅=:=
σmax f cd≤ 1=σmax1 f cd≤ 1=
Rezistenta la intindere a betonului
f ct.k 5.3MPa:= f ct.k 5.3N
mm2
⋅=
αct 1:=f ctd
αct f ct.k ⋅
γc
3.533 103
×kN
m2
⋅=:=
σmin f ctd≤ 1= σmin1 f ctd≤ 1=
NU NECESITA ARMATURA DE LEGATURA INTRE ELEVATIE SI RADIER CULEE