Fundatii izolate
1. Date intrare :
n 19:=
a 0.1:=
bst 50cm:=dimensiunile stalpiilor
hst 50cm:=
bBCA 30cm:=dimensiunile peretiilor din BCA
HBCA 3m:=
CTS 0.4 a+( )m− 0.5− m=:= cota terenului sistematizat
Stratificatia terenului :
0.00..... 0.50− Umplutura
0.5....− 3.50− Argila prafoasa, cafenie , plastic consistenta
γk.arg 18.2kN
m3
:= warg 22%:= Ip.arg 21%:= Ic.arg 0.70:= ϕ'k.arg 12°:=
earg 0.69:= Earg 12000kN
m2
:= c'k.arg 50 0.15n+( )kPa 52.85 kPa⋅=:=
3.5....− 10.0− Nisip argilos, galben cafeniu
γk.nis 19kN
m3
:= wnis 18%:= Ip.nis 11%:= enis 0.65:= ϕ'k.nis 24°:=
Enis 14000kN
m2
:= c'k.nis 5 0.2n+( )kPa 8.8 kPa⋅=:=
Incarcarile caracteristice :
pk 15 0.1n+( )kPa 16.9 kPa⋅=:=
qk 8 0.1n+( )kPa 9.9 kPa⋅=:=
fundatiile se vor proiecta pentru CP 3 : A1 + M2 + R3 , conform SR EN 1997
valorile coeficientilor de siguranta pt CP 3 :
pentru actiuni :
γG 1.35:= actiuni permanente
γQ 1.5:= actiuni variabile
pentru parametri geotehnici :
γϕ 1.25:= unghiul de frecare interna
γc 1.25:= coeziune efectiva in conditii drenate
γγ 1:= greutate volumica
pentru rezistente :
γR 1:= capacitate portanta
Incarcarile de calcul :
pd γG pk⋅ 22.815kN
m2
⋅=:=
qd γQ qk⋅ 14.85kN
m2
⋅=:=
argila :
ϕ'd.arg ϕ'k.arg γϕ⋅ 15 °⋅=:=
c'd.arg c'k.arg γc⋅ 66.063kN
m2
⋅=:=
γd.arg γk.arg γγ⋅ 18.2kN
m3
⋅=:=
nisip :
ϕ'd.nis ϕ'k.nis γϕ⋅ 30 °⋅=:=
c'd.nis c'k.nis γc⋅ 11kN
m2
⋅=:=
γd.nis γk.arg γγ⋅ 18.2kN
m3
⋅=:=
II. Proiectarea fundatiilor :
A. Stalpul S1 - fundatie izolata elastica
Valorile eforturilor sectionale :
Aaf.s1 6m 5⋅ m 30 m2⋅=:=
Ns1 1500kN:=
Mx1 180kN m⋅:= Pe directia cu 500mm
My1 161kN m⋅:= Pe directia cu 400mm
Tx.1 63kN:=
Ty.1 50kN:=
1. Alegerea adancimii de fundare :
HTBF 0.5m:=
Df HTBF 20cm+>
Hinghet 0.9m:=
Df Hing 20cm+>
vom alege : Df 1.0m:=
2. Predimensionare :
peff pacc<
peff
Ns1 Gf+
A1:=
Gf 0.1 Ns1⋅ 150 kN⋅=:=
A1 B1 L1⋅:=
pconv 400kN
m2
:=
pacc 0.8 pconv⋅ 320kN
m2
⋅=:=
la limita : peff pacc:= rezulta : A1
1.4Ns1
pacc6.563m
2=:=
L1
B1
lst
bst÷
hst
bst1=
L1 1.25 B1⋅:=
B1.nec
1.1Ns1
1 pacc⋅2.271m=:=
alegem : B1 2m:=
L1 2m:=
Hmin 30cm:=
H
L0.3:=
H1.nec 0.3 L1⋅ 0.6m=:=
alegem : H1 0.8m:=
3.Verificarea stalpului :
Verificarea se va face conform cazului de proiectare CP2
valorile coeficientilor de siguranta pt CP2 :
pentru parametri geotehnici :
γϕ 1:= unghiul de frecare interna
γc 1:= coeziune efectiva in conditii drenate
γγ 1:= greutate volumica
pentru rezistente :
γR 1.4:= capacitate portanta
argila :
ϕ'd.arg ϕ'k.arg γϕ⋅ 12 °⋅=:=
c'd.arg c'k.arg γc⋅ 52.85kN
m2
⋅=:=
γd.arg γk.arg γγ⋅ 18.2kN
m3
⋅=:=
nisip :
ϕ'd.nis ϕ'k.nis γϕ⋅ 24 °⋅=:=
c'd.nis c'k.nis γc⋅ 8.8kN
m2
⋅=:=
γd.nis γk.arg γγ⋅ 18.2kN
m3
⋅=:=
Facandu - se verificarea stalpului cu dimensiunile fundatiei alese in predimensionare s-a constatat cafundatia este supradimensionata. Prin urmare vom alege :
L1 1.9m:=
B1 1.9m:=H1 0.8m:=
solicitarile la talpa funatiei :
Nf Ns1 Gf1+:=
Ns1 1.5 103× kN⋅=
γmed 23kN
m3
:= Htotal Df 1m=:=
Gf1 L1 B1⋅ Htotal⋅ γmed⋅ 83.03 kN⋅=:=
Nf1 Ns1 Gf1+ 1.583 103× kN⋅=:=
ML.f1 Mx1 Tx.1 H1⋅+ 230.4 kN m⋅⋅=:=
MB.f1 My1 Ty.1 H1⋅+ 201 kN m⋅⋅=:=
eL1
ML.f1
Nf114.554 cm⋅=:= <
L1
631.667 cm⋅=
eB1
MB.f1
Nf112.697 cm⋅=:= <
B1
631.667 cm⋅=
L'1 L1 2 eL1⋅− 1.609m=:=
B'1 B1 2 eB1⋅− 1.646m=:=
A'1 L'1 B'1⋅ 2.648 m2⋅=:=
Vd
A'
Nf1
A'1597.739
kN
m2
⋅=:=Vd
A'
Conditii drenate :
Nq eπ tan ϕ'd.arg( )
tan 45° ϕ'd.arg+( )⋅ 3.003=:=
Nc Nq 1−( ) 1
tan ϕ'd.arg( )⋅ 9.421=:=
Nγ 2 Nq 1−( ) tan ϕ'd.arg( )⋅ 0.851=:=
α 0:= rezulta : bq 1:= bc 1:= bγ 1:=
sq 1
B'1
L'1sin ϕ'd.arg( )⋅+ 1.213=:=
sγ 1 0.3
B'1
L'1⋅− 0.693=:=
sc
sq Nq⋅ 1−
Nq 1−1.319=:=
mB
2
B'1
L'1+
1
B'1
L'1+
1.494=:=
iq 1
Tx.1
Nf1 A'1 c'd.arg⋅ cot ϕ'd.arg( )⋅+−
mB
0.958=:=
iγ 1
Tx.1
Nf1 A'1 c'd.arg⋅ cot ϕ'd.arg( )⋅+−
mB 1+
0.931=:=
ic iq
1 iq−( )Nc tan ϕ'd.arg( )⋅( )− 0.937=:=
γd.med
γmed
γγ23
kN
m3
⋅=:=
Df 1m=
q' γd.med Df⋅ 23kN
m2
⋅=:=
Rd1
A'1c'd.arg Nc⋅ bc⋅ sc⋅ ic⋅ q' Nq⋅ bq⋅ sq⋅ iq⋅+ 0.5 γd.arg⋅ B'1⋅ Nγ⋅ bγ⋅ sγ⋅ iγ⋅+ 704.134
kN
m2
⋅=:=Rd1
A'1accept
Nf1
A'1597.739
kN
m2
⋅= efectiva
Vd1
A'1
Rd1
A'1< peff pacc<
B. Stalpul S2 - fundatie izolata elastica
Valorile eforturilor sectionale :
Aaf.s2 5m 3⋅ m 15 m2⋅=:=
Ns2 Aaf.s2 pd qd+( )⋅ 564.975 kN⋅=:=
Mx2 0.2m Ns2⋅ 112.995 kN m⋅⋅=:= Tx.2 0.1 Ns2⋅ 56.498 kN⋅=:=
My2 0.1m Ns2⋅ 56.498 kN m⋅⋅=:= Ty.2 0.05 Ns2⋅ 28.249 kN⋅=:=
1. Alegerea adancimii de fundare :
HTBF 0.5m:=
Df HTBF 20cm+>
Hinghet 0.9m:=
Df Hing 20cm+>
vom alege : Df 1.1m:=
2. Predimensionare :
peff pacc<
peff
Ns2 Gf+
A2:=
Gf 0.4 Ns2⋅ 225.99 kN⋅=:=
A2 B2 L2⋅:=
pconv 400kN
m2
:=
pacc 0.8 pconv⋅ 320kN
m2
⋅=:=
la limita : peff pacc:= rezulta : A2
1.4Ns2
pacc2.472m
2=:=
L2
B2
lst
bst÷
hst
bst1=
L2 1.25 B2⋅:=
B2.nec
1.4Ns2
1.25 pacc⋅1.406m=:=
alegem : B2 1.6m:=
L2 B2 1.25⋅ 2m=:=
Hmin 30cm:=
H
L0.3:=
H2.nec 0.3 L2⋅ 0.6m=:=
alegem : H2 0.6m:=
3. Verificarea stalpului :
Verificarea se va face conform cazului de proiectare CP2
Facandu - se verificarea stalpului cu dimensiunile fundatiei alese in predimensionare s-a constatat cafundatia este supradimensionata. Prin urmare vom alege :
L2 1.6m:=
B2 1.2m:=
H2 0.6m:=
solicitarile la talpa funatiei :
Nf Ns2 Gf2+ Ggrinda.2+ GBCA.2+:=
Ns2 564.975 kN⋅=
γmed 21kN
m3
:= HCTS 0.5m:= Htotal Df HCTS+ 1.6m=:=
Gf2 L2 B2⋅ Htotal⋅ γmed⋅ 64.512 kN⋅=:=
vom alege grinda de fundare de sectiune : bgr 50cm:= hgr 50cm:= γbeton 25kN
m3
:=
Ggrinda.2 bgr hgr⋅ 5 m⋅ bst−( )⋅ γbeton⋅ 28.125 kN⋅=:=
bBCA 30 cm⋅= HBCA 3m= γBCA 8kN
m3
:=peretele din BCA are dimensiunile :
GBCA.2 bBCA HBCA⋅ 5⋅ m γBCA⋅ 36 kN⋅=:=
Nf2 Ns2 Gf2+ Ggrinda.2+ GBCA.2+ 693.612 kN⋅=:=
ML.f2 Mx2 Tx.2 H2⋅+ 146.893 kN m⋅⋅=:=
MB.f2 My2 Ty.2 H2⋅+ 73.447 kN m⋅⋅=:=
eL2
ML.f2
Nf221.178 cm⋅=:= <
L2
626.667 cm⋅=
eB2
MB.f2
Nf210.589 cm⋅=:= <
B2
620 cm⋅=
L'2 L2 2 eL2⋅− 1.176m=:=
B'2 B2 2 eB2⋅− 0.988m=:=
A'2 L'2 B'2⋅ 1.163 m2⋅=:=
Vd
A'
Nf2
A'2596.614
kN
m2
⋅=:=Vd
A'
Conditii drenate :
Nq 3.003=
Nc Nq 1−( ) 1
tan ϕ'd.arg( )⋅ 9.421=:=
Nγ 2 Nq 1−( ) tan ϕ'd.arg( )⋅ 0.851=:=
α 0:= rezulta : bq 1:= bc 1:= bγ 1:=
sq 1
B'2
L'2sin ϕ'd.arg( )⋅+ 1.175=:=
sγ 1 0.3
B'2
L'2⋅− 0.748=:=
sc
sq Nq⋅ 1−
Nq 1−1.262=:=
mB
2
B'2
L'2+
1
B'2
L'2+
1.543=:=
iq 1
Tx.2
Nf2 A'2 c'd.arg⋅ cot ϕ'd.arg( )⋅+−
mB
0.913=:=
iγ 1
Tx.2
Nf2 A'2 c'd.arg⋅ cot ϕ'd.arg( )⋅+−
mB 1+
0.86=:=
−( )
ic iq
1 iq−( )Nc tan ϕ'd.arg( )⋅( )− 0.869=:=
γd.med
γmed
γγ21
kN
m3
⋅=:=
Df 1.1m=
q' γd.med Df⋅ 23.1kN
m2
⋅=:=
Rd
A'2c'd.arg Nc⋅ bc⋅ sc⋅ ic⋅ q' Nq⋅ bq⋅ sq⋅ iq⋅+ 0.5 γd.arg⋅ B'2⋅ Nγ⋅ bγ⋅ sγ⋅ iγ⋅+ 625.299
kN
m2
⋅=:=Rd
A'2
Nf2
A'2596.614
kN
m2
⋅=
Vd2
A'2
Rd2
A'2<
C. Stalpul S4 - fundatie izolata rigida
Valorile eforturilor sectionale :
Aaf.s4 6m 5⋅ m 30 m2⋅=:=
Ns4 Aaf.s4 pd qd+( )⋅ 1.13 103× kN⋅=:=
Mx4 0.2m Ns4⋅ 225.99 kN m⋅⋅=:= Tx.4 0.1 Ns4⋅ 112.995 kN⋅=:=
My4 0.1m Ns4⋅ 112.995 kN m⋅⋅=:= Ty.4 0.05 Ns4⋅ 56.498 kN⋅=:=
1. Alegerea adancimii de fundare :
HTBF 0.5m:=
Df HTBF 20cm+>
Hinghet 0.9m:=
Df Hing 20cm+>
vom alege : Df 1.1m:=
2. Predimensionare :
blocul din beton simplu :
peff pacc<
peff
Ns4 Gf+
A4:=
Gf 0.4 Ns4⋅ 451.98 kN⋅=:=
A4 B4 L4⋅:=
pconv 400kN
m2
:=
pacc 0.8 pconv⋅ 320kN
m2
⋅=:=
la limita : peff pacc:= rezulta : A4
1.4Ns4
pacc4.944m
2=:=
L4
B4
lst
bst÷
hst
bst1=
L4 1.25 B4⋅:=
B4.nec
1.4Ns4
1.25 pacc⋅1.989m=:=
alegem : B4 2m:=
L4 B4 1.25⋅ 2.5m=:=
Cuzinetul din beton armat
lc
L
bc
B⋅=
lc
L0.4 0.5÷:=
lc4 0.5 L4⋅ 1.25 m⋅=:=
alegem : lc4 1.3m:=
bc4 0.5 B4⋅ 1m=:=
tg α( ) tg αadm( )>
Hmin 40cm:=
tgα.adm 1.5:=
tg α( )H4
B4 bc4−
2
tga.adm>:=
la limita :
H4.nec
B4 bc4−
2tgα.adm⋅ 0.75m=:=
tg α( )H4
L4 lc4−
2
tgα.adm>:=
la limita :
H4.nec
L4 lc4−
2tgα.adm⋅ 0.9m=:=
vom executa blocul de beton in 2 trepte :
H4.1 50cm:= H4.2 40cm:= H4 0.9 m⋅:=
hc.min 30cm:=
hc
lc0.25>
hc.4.nec 0.25 lc4⋅ 0.325 m⋅=:=
alegem : hc4 0.45m:=
tgβ
hc4
bc4 bst−
2
1.8=:=
tgβ
hc4
lc4 hst−
2
1.125=:=
Grinda de echiibrare :
hechil hc4 0.45m=:=
bechil hechil 0.45m=:=
3. Verificarea stalpului :
Verificarea se va face conform cazului de proiectare CP2
Facandu - se verificarea stalpului cu dimensiunile fundatiei alese in predimensionare s-a constatat cafundatia este supradimensionata. Prin urmare vom alege :
L4 2m:= lc4 1m:=
B4 1.6m:= bc4 0.8 m⋅:= Df :=
H4 0.75m:= hc4 0.45m=
H4.1 0.4m:= H4.2 0.35m:=
solicitarile la talpa fundatiei :
+ +:=
Nf Ns4 Gf4+ Gechil+:=
Ns4 1.13 103× kN⋅=
Df 0.2m H4+ hc4+ 1.4m=:=
γmed 21kN
m3
:= Htotal Df 1.4m=:=
Gf4 L4 B4⋅ Htotal⋅ γmed⋅ 94.08 kN⋅=:=
Gechil bechil hechil⋅ 2.5mbc4
2−
⋅ γbeton⋅ 10.631 kN⋅=:=
Nf4 Ns4 Gf4+ Gechil+ 1.235 103× kN⋅=:=
ML.f4 Mx4 Tx.2 H4 hc4+( )⋅+ 293.787 kN m⋅⋅=:=
MB.f4 My4 Ty.4 H4 hc4+( )⋅+ Gechil
2.5mbc4
2−
2⋅− 169.629 kN m⋅⋅=:=
eL4
ML.f4
Nf423.795 cm⋅=:= <
L4
633.333 cm⋅=
eB4
MB.f4
Nf413.739 cm⋅=:= <
B4
626.667 cm⋅=
L'4 L4 2 eL4⋅− 1.524m=:=
B'4 B4 2 eB4⋅− 1.325m=:=
A'4 L'4 B'4⋅ 2.02 m2⋅=:=
Vd
A'
Nf4
A'4611.288
kN
m2
⋅=:=Vd
A'
Conditii drenate :
Nq 3.003=
Nc Nq 1−( ) 1
tan ϕ'd.arg( )⋅ 9.421=:=
Nγ 2 Nq 1−( ) tan ϕ'd.arg( )⋅ 0.851=:=
α 0:= rezulta : bq 1:= bc 1:= bγ 1:=
sq 1
B'4
L'4sin ϕ'd.arg( )⋅+ 1.181=:=
sγ 1 0.3
B'4
L'4⋅− 0.739=:=
sc
sq Nq⋅ 1−
Nq 1−1.271=:=
mB
2
B'4
L'4+
1
B'4
L'4+
1.535=:=
iq 1
Tx.4
Nf4 A'4 c'd.arg⋅ cot ϕ'd.arg( )⋅+−
mB
0.902=:=
iγ 1
Tx.4
Nf4 A'4 c'd.arg⋅ cot ϕ'd.arg( )⋅+−
mB 1+
0.843=:=
ic iq
1 iq−( )Nc tan ϕ'd.arg( )⋅( )− 0.853=:=
γd.med
γmed
γγ21
kN
m3
⋅=:=
Df 1.4m=
q' γd.med Df⋅ 29.4kN
m2
⋅=:=
Rd
A'4c'd.arg Nc⋅ bc⋅ sc⋅ ic⋅ q' Nq⋅ bq⋅ sq⋅ iq⋅+ 0.5 γd.arg⋅ B'4⋅ Nγ⋅ bγ⋅ sγ⋅ iγ⋅+ 640.193
kN
m2
⋅=:=Rd
A'4
Nf4
A'4611.288
kN
m2
⋅=
Vd4
A'4
Rd4
A'4<
D. Stalpul S3 - fundatie izolata rigida
Valorile eforturilor sectionale :
Aaf.s3 6m 2.5⋅ m 15 m2⋅=:=
Ns3 Aaf.s3 pd qd+( )⋅ 564.975 kN⋅=:=
Mx3 0.2m Ns3⋅ 112.995 kN m⋅⋅=:= Tx.3 0.1 Ns3⋅ 56.498 kN⋅=:=
My3 0.1m Ns3⋅ 56.498 kN m⋅⋅=:= Ty.3 0.05 Ns3⋅ 28.249 kN⋅=:=
1. Alegerea adancimii de fundare :
HTBF 0.5m:=
Df HTBF 20cm+>
Hinghet 0.9m:=
Df Hing 20cm+>
vom alege : Df 1.1m:=
2. Predimensionare :
blocul din beton simplu :
L3
B3
lst
bst÷
hst
bst1=
L3 1.25 B3⋅:= B3
vom impune o excentricitate : e3 40cm:=
B3 20cm e3+( ) 2⋅ 1.2m=:=
L3 B3 1.25⋅ 1.5m=:=
Cuzinetul din beton armat
lc
L
bc
B⋅=
lc
lc
L0.5 0.65÷:=
lc
L
lc3 0.5 L3⋅ 0.75 m⋅=:=
alegem : lc3 0.8m:=
bc3 0.5 B3⋅ 0.6m=:=
tg α( ) tg αadm( )>
Hmin 40cm:=
tgα.adm 1.5:=
tg α( )H3
B3 bc3−
2
tga.adm>:=H3
la limita :
H3.nec
B3 bc3−
2tgα.adm⋅ 0.45m=:=
tg α( )H4
L4 lc4−
2
tga.adm>:= tga.adm
la limita :
H3.nec
L3 lc3−
2tgα.adm⋅ 0.525m=:=
H3 0.6 m⋅:=
hc.min 30cm:=
hc
lc0.25>
hc.3.nec 0.25 lc3⋅ 0.2 m⋅=:=
:=
alegem : hc3 0.45m:=
tgβ
hc3
bc3 bst−
2
9=:=
tgβ
hc3
lc3 hst−
2
3=:=
Grinda de echilibrare :
hechil hc4 0.45m=:=
bechil hechil 0.45m=:=
Grinda de fundare :
hgr 0.4m:=
bgr 0.4m:=
3. Verificarea stalpului :
Verificarea se va face conform cazului de proiectare CP2
Facandu - se verificarea stalpului cu dimensiunile fundatiei alese in predimensionare s-a constatat cafundatia este supradimensionata. Prin urmare vom alege :
L3 1.8m:= lc3 1.2m:=
B3 1.3m:= bc3 1 m⋅:=
H3 0.6m:= hc3 0.45m=
solicitarile la talpa funatiei :
Nf Ns3 Gf3+ Gechil+ Ggrinda.3+:=
Ns3 564.975 kN⋅=
γmed 21kN
m3
:= Htotal Df HCTS+ 1.6m=:=
Gf3 L3 B3⋅ Htotal⋅ γmed⋅ 78.624 kN⋅=:=
Gechil bechil hechil⋅ 2.5m 0.2m+ bc3−( )⋅ γbeton⋅ 8.606 kN⋅=:=
⋅ −( )⋅ ⋅ ⋅=:=
Ggrinda.3 hgr bgr⋅ 6m hst−( )⋅ γbeton⋅ 22 kN⋅=:=
GBCA.3 bBCA HBCA⋅ 5m hst−( )⋅ γBCA⋅ 32.4 kN⋅=:=
Calculam reactiunea aparuta in fundatia stalpului S3 :
ΣMO4 0:=
My3 Ns3 Gf3+( ) 5⋅ m+ Tx.3 H3 hc3+( )⋅+ Gf3 Gechil+( ) 5m e3−( )⋅+ MB.f4+ R3 5m e3−( )⋅−
R3
My3 Ns3 Ggrinda.3+ GBCA.3+( ) 5⋅ m+ Ty.3 H3 hc3+( )⋅+
Gf3 Gechil+( ) 5m e3−( )⋅ MB.f4++
...
5m e3−816.07 kN⋅=:=
R3 816.07 kN⋅=
Nf3 Ns3 Gf3+ Gechil+ Ggrinda.3+ GBCA.3+ 706.605 kN⋅=:=
ML.f3 Mx3 Tx.3 H3 hc3+( )⋅+ 172.317 kN m⋅⋅=:=
eL3
ML.f3
Nf324.387 cm⋅=:= <
L3
630 cm⋅=
L'3 L3 2 eL3⋅− 1.312m=:=
B'3 B3:=
A'3 L'3 B3⋅ 1.706 m2⋅=:=
Vd
A'
Nf3
A'3414.201
kN
m2
⋅=:=Vd
A'
Conditii drenate :
Nq 3.003=
Nc Nq 1−( ) 1
tan ϕ'd.arg( )⋅ 9.421=:=
Nγ 2 Nq 1−( ) tan ϕ'd.arg( )⋅ 0.851=:=
α 0:= rezulta : bq 1:= bc 1:= bγ 1:=
sq 1
B3
L'3sin ϕ'd.arg( )⋅+ 1.206=:=
sγ 1 0.3
B3
L'3⋅− 0.703=:=
sc
sq Nq⋅ 1−
Nq 1−1.309=:=
mB
2
B3
L'3+
1
B3
L'3+
1.502=:=
iq 1
Tx.3
Nf3 A'3 c'd.arg⋅ cot ϕ'd.arg( )⋅+−
mB
0.926=:=
iγ 1
Tx.3
Nf3 A'3 c'd.arg⋅ cot ϕ'd.arg( )⋅+−
mB 1+
0.88=:=
ic iq
1 iq−( )Nc tan ϕ'd.arg( )⋅( )− 0.889=:=
γd.med
γmed
γγ21
kN
m3
⋅=:=
Df 1.1m=
q' γd.med Df⋅ 23.1kN
m2
⋅=:=
Rd
A'3c'd.arg Nc⋅ bc⋅ sc⋅ ic⋅ q' Nq⋅ bq⋅ sq⋅ iq⋅+ 0.5 γd.arg⋅ B3⋅ Nγ⋅ bγ⋅ sγ⋅ iγ⋅+ 662.934
kN
m2
⋅=:=Rd
A'3
R3
A'3478.368
kN
m2
⋅=
ΣMO3 0:=
Nf4 5m e3−( )⋅ MB.f4− Ns3 Gf3+( ) e3⋅− My3− Tx.3 H3 hc3+( )⋅− R4 5m e3−( )⋅−
R4
Nf4 5m e3−( )⋅ MB.f4− Ns3 Ggrinda.3+ GBCA.3+( ) e3⋅−
My3− Ty.3 H3 hc3+( )⋅−+
...
5m e3−1.125 10
3× kN⋅=:=
Nf4 1.235 103× kN⋅=
R4 Nf4<
IV. Armarea fundatiilor
IV.A. Armarea fundatiei elastice aferente stalpului S1
Nf1 Ns1 1.5 103× kN⋅=:=
ML.f1 230.4 kN m⋅⋅=
MB.f1 201 kN m⋅⋅=
A1 L1 B1⋅ 3.61m2=:=
eL1
ML.f1
Nf115.36 cm⋅=:= <
L1
60.317m=
eB1
MB.f1
Nf113.4 cm⋅=:= <
B1
60.317m=
Pe directia L:
p1
Nf1
A11
6 eL1⋅
L1−
6 eB1⋅
B1−
⋅ 38.14kN
m2
⋅=:=
p2
Nf1
A11
6 eL1⋅
L1+
6 eB1⋅
B1+
⋅ 792.885kN
m2
⋅=:=
lcons
L1 hst−
20.7 m⋅=:=
p0
L1 lcons−
L1p2 p1−( )⋅ p1+ 514.821
kN
m2
⋅=:=
M1.1 p0 lcons⋅( )lcons
2⋅ p2 p0−( )
lcons
2⋅
2
3⋅ lcons⋅+
B1⋅ 325.942 kN m⋅⋅=:=
Pe directia B:
bcons
B1 bst−
20.7m=:=
pmed
p1 p2+
2415.512
kN
m2
⋅=:=
M2.2 pmed bcons⋅bcons
2⋅
L1⋅ 193.421 kN m⋅⋅=:=
Armatura marca 1 pe directia L:
Alegem beton C20/25 si armaturi PC52:
fck 20MPa:= γc 1.5:= fcd
fck
γc13.333 MPa⋅=:=
fyk 500MPa:= γs 1.15:= fyd
fyk
γs434.783 MPa⋅=:=
Φ1.max 20mm:=
cnom Φ1.max 10mm+ 30 mm⋅=:=
d1 H1 cnom−Φ1.max
2− 0.76m=:=
μ1
M1.1
B1 d12⋅ fcd⋅
0.022=:=
ω1 1 1 2μ1−− 0.023=:=
Asl.nec.1 ω1 B1⋅ d1⋅fcd
fyd⋅ 9.976 cm
2⋅=:=
Asl.eff.1 11.30cm2:=Alegem 10Φ12 cu
d1 0.76m=
As.min 0.0013 B1⋅ d1⋅ 18.772 cm2⋅=:=
As.max 0.04A1 1.444 103× cm
2⋅=:=
ql
Asl.eff.1
B1 d1⋅7.825 10
4−×=:=
ql.min 0.075% 7.5 104−×=:=
⋅− ⋅−
snh.1
B1 2 cnom⋅− 10 12⋅ mm−
919.111 cm⋅=:= < smax 25cm:=
lcioc.1 15 16⋅ mm 24 cm⋅=:=
Armatura marca 2 pe directia B:
cnom Φ1.max 10mm+ 30 mm⋅=:=
d2 H1 cnom−Φ1.max
2− 0.76m=:=
μ2
M2.2
L1 d22⋅ fcd⋅
0.013=:=
ω2 1 1 2μ2−− 0.013=:=
Asl.nec.1 ω2 L1⋅ d2⋅fcd
fyd⋅ 5.893 cm
2⋅=:=
Asl.eff.2 7.85cm2:=Alegem 10Φ10 cu
As.min 0.0013 L1⋅ d2⋅ 18.772 cm2⋅=:=
As.max 0.04A1 1.444 103× cm
2⋅=:=
ql
Asl.eff.2
L1 d2⋅5.436 10
4−×=:=
ql.min 0.075% 7.5 104−×=:=
snh.2
L1 2 cnom⋅− 9 14⋅ mm−
821.425 cm⋅=:= < smax 25 cm⋅=
lcioc.2 15 14⋅ mm 21 cm⋅=:=
Armatura marca 3 pe directia L:
cnom Φ1.max 10mm+ 30 mm⋅=:=
d3 H1 cnom−Φ1.max
2− 0.76m=:=
T3.3 Asl.eff.3
fyd
3⋅<
Asl.nec.3
T3.3 3⋅
fydcm2⋅=:=
T3.3
Asl.eff.3 5.49cm2:=Alegem 7Φ10 cu
snh.3
B1 2 cnom⋅− 7 10⋅ mm−
629.5 cm⋅=:= < smax 25 cm⋅=
lcioc.3 15 10⋅ mm 15 cm⋅=:=
Armatura marca 3` pe directia B :
alegem : 9ϕ10 cu Asl.eff.3' 7.06cm2:=
snh.2
L1 2 cnom⋅− 9 10⋅ mm−
821.875 cm⋅=:= < smax 25cm:=
lcioc.3' 15 10⋅ mm 15 cm⋅=:=
IV.B. Armarea fundatiei elastice aferente stalpului S2
Nf2 Ns2 564.975 kN⋅=:=
ML.f2 146.893 kN m⋅⋅=
MB.f2 73.447 kN m⋅⋅=
A2 L2 B2⋅ 1.92m2=:=
eL2
ML.f2
Nf226 cm⋅=:= <
L2
60.267m=
eB2
MB.f2
Nf213 cm⋅=:= <
B2
60.2m=
Pe directia L:
p1
Nf2
A21
6 eL2⋅
L1−
6 eB2⋅
B2−
⋅ 138.611−kN
m2
⋅=:= intindere
p2
Nf2
A21
6 eL2⋅
L2+
6 eB2⋅
B2+
⋅ 772.427kN
m2
⋅=:=
−
a3
p1−
p1− p2+L2⋅ 0.243m=:=lcons
L2 hst−
20.55 m⋅=:=
p0
L1 a3− lcons−
L1 a3−p2⋅ 515.972
kN
m2
⋅=:=
M1.1 p0 lcons⋅( )lcons
2⋅ p2 p0−( )
lcons
2⋅
2
3⋅ lcons⋅+
B2⋅ 124.68 kN m⋅⋅=:=
T3.3
p1 a3⋅
2B2⋅ 20.246 kN⋅=:=
Pe directia B:
bcons
B2 bst−
20.35m=:=
pmed
p1 p2+
2316.908
kN
m2
⋅=:=
M2.2 pmed bcons⋅bcons
2⋅
L2⋅ 31.057 kN m⋅⋅=:=
Armatura marca 1 pe directia L:
Alegem beton C20/25 si armaturi PC52:
fck 20MPa:= γc 1.5:= fcd
fck
γc13.333 MPa⋅=:=
fyk 355MPa:= γs 1.15:= fyd
fyk
γs308.696 MPa⋅=:=
Φ1.max 28mm:=
cnom Φ1.max 10mm+ 38 mm⋅=:=
d1 H2 cnom−Φ1.max
2− 0.548m=:=
μ1
M1.1
B1 d12⋅ fcd⋅
0.016=:=
ω1 1 1 2μ1−− 0.017=:=
Asl.nec.1 ω1 B2⋅ d1⋅fcd
fyd⋅ 4.694 cm
2⋅=:=
Asl.eff.1 9.24cm2:=Alegem 6Φ14 cu
As.min 0.0013 B2⋅ d1⋅ 8.549 cm2⋅=:=
As.max 0.04A2 768 cm2⋅=:=
ql
Asl.eff.1
B2 d1⋅1.405 10
3−×=:=
ql.min 0.075% 7.5 104−×=:=
snh.1
B2 2 cnom⋅− 6 14⋅ mm−
520.8 cm⋅=:= < smax 25cm:=
lcioc.1 15 14⋅ mm 21 cm⋅=:=
Armatura marca 2 pe directia B:
cnom Φ1.max 10mm+ 38 mm⋅=:=
d2 H2 cnom−Φ1.max
2− 0.548m=:=
μ2
M2.2
L2 d22⋅ fcd⋅
4.848 103−×=:=
ω2 1 1 2μ2−− 4.86 103−×=:=
Asl.nec.2 ω2 L1⋅ d2⋅fcd
fyd⋅ 2.185 cm
2⋅=:=
Asl.eff.2 12.32cm2:= din conditii impuse de distanta dintre armaturi
Alegem 8Φ14 cu
As.min 0.0013 L2⋅ d2⋅ 11.398 cm2⋅=:=
As.max 0.04A2 768 cm2⋅=:=
ql
Asl.eff.2
L2 d2⋅1.405 10
3−×=:=
ql.min 0.075% 7.5 104−×=:=
⋅− ⋅−
snh.2
L2 2 cnom⋅− 8 14⋅ mm−
720.171 cm⋅=:= < smax 25 cm⋅=
lcioc.2 15 14⋅ mm 21 cm⋅=:=
Armatura marca 3 pe directia L:
cnom Φ1.max 10mm+ 38 mm⋅=:=
d3 H2 cnom−Φ1.max
2− 0.548m=:=
T3.3 Asl.eff.3
fyd
3⋅<
Asl.nec.3
T3.3 3⋅
fyd1.136 cm
2⋅=:=
Asl.eff.3 4.71cm2:=Alegem 6Φ10 cu
snh.3
B2 2 cnom⋅− 6 10⋅ mm−
521.28 cm⋅=:= < smax 25 cm⋅=
lcioc.3 15 10⋅ mm 15 cm⋅=:=
Armatura marca 3` pe directia B :
alegem : 7ϕ10 cu Asl.eff.3' 5.49cm2:=
snh.3
L2 2 cnom⋅− 7 10⋅ mm−
624.233 cm⋅=:= < smax 25cm:=
lcioc.3' 15 10⋅ mm 15 cm⋅=:=
IV.C. Armarea fundatiei rigide aferente stalpului S3
Nc3 Ns3 564.975 kN⋅=:=
ML.c3 Mx3 Tx.3 hc3⋅+ 138.419 kN m⋅⋅=:=
MB.c3 My3 Ty.3 hc3⋅+ 69.209 kN m⋅⋅=:=
Ac3 bc3 lc3⋅ 1.2m2=:=
eL.c3
ML.c3
Nc324.5 cm⋅=:=
eB.c3
MB.c3
Nc312.25 cm⋅=:=
directie L:
Pc1.l
Nc3
Ac31 6
eL.c3
lc3⋅−
⋅ 105.933−kN
m2
⋅=:=
Pc2.l
Nc3
Ac31 6
eL.c3
lc3+
1.048 103×kN
m2
⋅=:=
lcons
lc3 hst−
235 cm⋅=:= a3
Pc1.l− m⋅
Pc1.l− Pc2.l+0.092m=:=
deoarece Pc1<0 , atunci consideram ca aceasta va lua valoare 0 iar :
clc3
2eL.c3− 35.5 cm⋅=:=
Pc2.l
2 Nc3⋅
3 bc3⋅ c⋅1.061 10
3×kN
m2
⋅=:=
Pc0.l
lc3 a3− lcons−
lc3 a3−Pc2.l⋅ 725.886
kN
m2
⋅=:=
M1.1 Pc0.l lcons⋅( )lcons
2⋅ Pc2.l Pc0.l−( )
lcons
2⋅
2
3⋅ lcons⋅+
bc3⋅ 58.144 kN m⋅⋅=:=
T3.3
Pc1.l a3⋅
2bc3⋅ 4.864 kN⋅=:=
directia B:
Pc1.b
Nc3
Ac31 6
eB.c3
bc3⋅−
⋅ 124.765kN
m2
⋅=:=
Pc2.b
Nc3
Ac31 6
eB.c3
bc3+
816.86kN
m2
⋅=:=
bcons bc3 bst− 50 cm⋅=:=
Pc0.b
Pc2.b Pc1.b−
2
bst
bc3 bst−⋅
Pc1.b Pc2.b+
2+ 816.86
kN
m2
⋅=:=
− −( ) ⋅
M2.2 Pc1.b bcons⋅bcons
2⋅ Pc0.b bcons⋅( )
bcons
2⋅+
Pc2.b Pc0.b− Pc1.b−( ) bcons⋅
2
2
3⋅ bcons⋅+
lc3⋅:=
M2.2 128.767 kN m⋅⋅=
Armatura marca 1 pe directia L :
alegem clasa de beton C20/25 , si otel PC 52 , rezultand :
fck 20MPa:= γc 1.5:= fcd
fck
γc13.333 MPa⋅=:=
fyk 355MPa:= γs 1.15:= fyd
fyk
γs308.696 MPa⋅=:=
cnom 5cm:=
Φ1.max 28 mm⋅=
d1 hc3 cnom−Φ1.max
2− 0.386m=:=
μ1
M1.1
bc3 d12⋅ fcd⋅
0.029=:=
ω1 1 1 2μ1−− 0.03=:=
Asl.nec.1 ω1 bc3⋅ d1⋅fcd
fyd⋅ 4.953 cm
2⋅=:=
alegem : 5ϕ12 cu Asl.eff.1 5.65cm2:=
As.min 0.0013 bc3⋅ d1⋅ 5.018 cm2⋅=:=
As.max 0.04Ac3 480 cm2⋅=:=
ql
Asl.eff.1
bc3 d1⋅1.464 10
3−×=:=
snh.1
bc3 2 cnom⋅− 5 12⋅ mm−
421 cm⋅=:= < smax 25cm:=
lcioc.1 15 12⋅ mm 18 cm⋅=:=
Armatura marca 2 pe directia B :
cnom 5 cm⋅=
d2 hc3 cnom− 12mm−Φ1.max
2− 0.374m=:=
μ2
M2.2
lc3 d22⋅ fcd⋅
0.058=:=
ω2 1 1 2μ2−− 0.059=:=
Asl.nec.2 ω2 lc3⋅ d2⋅fcd
fyd⋅ 11.494 cm
2⋅=:=
alegem : 7ϕ18 cu Asl.eff.2 17.78cm2:=
As.min 0.0013 lc3⋅ d2⋅ 5.834 cm2⋅=:=
As.max 0.04Ac3 480 cm2⋅=:=
snh.2
lc3 2 cnom⋅− 7 18⋅ mm−
616.233 cm⋅=:= < smax 25cm:=
lcioc.2 15 18⋅ mm 27 cm⋅=:=
Armatura marca 3 pe directia L :
d3 hc3 cnom− Φ1.max− 0.372m=:=
T3.3 Asl.eff.3
fyd
3⋅<
la limita :
Asl.nec.3 T3.33
fyd⋅ 0.273 cm
2⋅=:=
alegem : 6ϕ10 cu Asl.eff.3 4.71cm2:=
snh.3
bc3 2 cnom⋅− 6 10⋅ mm−
516.8 cm⋅=:= < smax 25cm:=
lb.rqd 58cm500
355⋅ 81.69 cm⋅=:=α1 1:= α2 1:= α3 1:= α4 1:= α5 1:=
S355 (echivalentul lui PC52)lbd.3 α1 α2⋅ α3⋅ α4⋅ α5⋅ lb.rqd⋅ 0.817m=:=
Armatura marca 3` pe directia B :
alegem : 6ϕ10 cu Asl.eff.3' 4.71cm2:=
⋅− ⋅−
snh.2
lc3 2 cnom⋅− 6 10⋅ mm−
520.8 cm⋅=:= < smax 25cm:=
lcioc.3' 15 10⋅ mm 15 cm⋅=:=
α1 1:= α2 1:= α3 1:= α4 1:= α5 1:= lb.rqd 58cm500
355⋅ 81.69 cm⋅=:=
lbd.3' α1 α2⋅ α3⋅ α4⋅ α5⋅ lb.rqd⋅ 0.817m=:= S355 (echivalentul lui PC52)
IV.D. Armarea fundatiei rigide aferente stalpului S4
Nc4 Ns4 1.13 103× kN⋅=:=
ML.c4 Mx4 Tx.4 hc4⋅+ 276.838 kN m⋅⋅=:=
MB.c4 My4 Ty.4 hc4⋅+ 138.419 kN m⋅⋅=:=
Ac4 bc4 lc4⋅ 0.8m2=:=
eL.c4
ML.c4
Nc424.5 cm⋅=:=
eB.c4
MB.c4
Nc412.25 cm⋅=:=
directie L:
Pc1.l
Nc4
Ac41 6
eL.c4
lc4⋅−
⋅ 663.846−kN
m2
⋅=:=
Pc2.l
Nc4
Ac41 6
eL.c4
lc4+
3.489 103×kN
m2
⋅=:=
lcons
lc4 hst−
225 cm⋅=:= a3
Pc1.l− m⋅
Pc1.l− Pc2.l+0.16m=:=
deoarece Pc1<0 , atunci consideram ca aceasta va lua valoare 0 iar :
clc4
2eL.c4− 25.5 cm⋅=:=
Pc2
2 Nc4⋅
3 bc4⋅ c⋅3.693 10
3×kN
m2
⋅=:=
Pc0.l
lc3 a3− lcons−
lc3 a3−Pc2.l⋅ 2.65 10
3×kN
m2
⋅=:=
M1.1 Pc0.l lcons⋅( )lcons
2⋅ Pc2 Pc0.l−( )
lcons
2⋅
2
3⋅ lcons⋅+
bc4⋅ 83.629 kN m⋅⋅=:=
T3.3
Pc1.l a3⋅
2bc4⋅ 42.45 kN⋅=:=
directia B:
bcons
bc4 bst−
215 cm⋅=:=
Pmed
Pc1.l Pc2.l+
21.412 10
3×kN
m2
⋅=:=
M2.2 Pmed bcons⋅( )bcons
2⋅
lc3⋅ 19.068 kN m⋅⋅=:=
Armatura marca 1 pe directia L :
alegem clasa de beton C20/25 , si otel PC 52 , rezultand :
fck 20MPa:= γc 1.5:= fcd
fck
γc13.333 MPa⋅=:=
fyk 355MPa:= γs 1.15:= fyd
fyk
γs308.696 MPa⋅=:=
cnom 5cm:=
Φ1.max 28 mm⋅=
d1 hc4 cnom−Φ1.max
2− 0.386m=:=
μ1
M1.1
bc4 d12⋅ fcd⋅
0.053=:=
ω1 1 1 2μ1−− 0.054=:=
Asl.nec.1 ω1 bc4⋅ d1⋅fcd
fyd⋅ 7.213 cm
2⋅=:=
alegem : 5ϕ14 cu Asl.eff.1 7.70cm2:=
As.min 0.0013 bc4⋅ d1⋅ 4.014 cm2⋅=:=
As.max 0.04Ac4 320 cm2⋅=:=
ql
Asl.eff.1
bc4 d1⋅2.494 10
3−×=:=
⋅− ⋅−
snh.1
bc4 2 cnom⋅− 5 14⋅ mm−
415.75 cm⋅=:= < smax 25cm:=
lcioc.1 15 14⋅ mm 21 cm⋅=:=
Armatura marca 2 pe directia B :
cnom 5 cm⋅=
d2 hc4 cnom− 12mm−Φ1.max
2− 0.374m=:=
μ2
M2.2
lc4 d22⋅ fcd⋅
0.01=:=
ω2 1 1 2μ2−− 0.01=:=
Asl.nec.2 ω2 lc4⋅ d2⋅fcd
fyd⋅ 1.66 cm
2⋅=:=
alegem : 5ϕ12 cu Asl.eff.2 5.65cm2:=
As.min 0.0013 lc4⋅ d2⋅ 4.862 cm2⋅=:=
As.max 0.04Ac4 320 cm2⋅=:=
snh.2
lc4 2 cnom⋅− 5 12⋅ mm−
421 cm⋅=:= < smax 25cm:=
lcioc.2 15 12⋅ mm 18 cm⋅=:=
Armatura marca 3 pe directia L :
d3 hc4 cnom− Φ1.max− 0.372m=:=
T3.3 Asl.eff.3
fyd
3⋅<
la limita :
Asl.nec.3 T3.33
fyd⋅ 2.382 cm
2⋅=:=
alegem : 4ϕ10 cu Asl.eff.3 3.14cm2:=
⋅− ⋅−
snh.3
bc4 2 cnom⋅− 4 10⋅ mm−
322 cm⋅=:= < smax 25cm:=
α1 1:= α2 1:= α3 1:= α4 1:= α5 1:= lb.rqd 0.817m=
lbd.3 α1 α2⋅ α3⋅ α4⋅ α5⋅ lb.rqd⋅ 81.69 cm⋅=:=
Armatura marca 3` pe directia B :
alegem : 6ϕ10 cu Asl.eff.3' 4.71cm2:=
snh.2
lc3 2 cnom⋅− 6 10⋅ mm−
520.8 cm⋅=:= < smax 25cm:=
lcioc.3' 15 10⋅ mm 15 cm⋅=:=
α1 1:= α2 1:= α3 1:= α4 1:= α5 1:= lb.rqd 0.817m=
lbd.3' α1 α2⋅ α3⋅ α4⋅ α5⋅ lb.rqd⋅ 81.69 cm⋅=:=
V. Armarea grinzii de echilibrare
deschiderea de calcul a grinzii : lgrinda 4.7m:=
hechil 0.45m=
bechil 0.45m=
humpl 0.3m:=
γbeton 25kN
m3
⋅= γu 19kN
m3
:=
qechil bechil hechil⋅ γbeton⋅ 5.063kN
m⋅=:=
qumpl 6m humpl⋅ γu⋅ 34.2kN
m⋅=:=
e3 0.4m=ΣMo3 0:=
R4
qechil
lgrinda2
2⋅ qumpl
lgrinda2
2⋅+ MB.c3− MB.c4− Nc3 e3⋅−
lgrinda7.66 10
3−× kN⋅=:=
R3
Nc3 lgrinda e3+( )⋅ MB.c3+ MB.c4+ qechil
lgrinda2
2⋅+ qumpl
lgrinda2
2⋅+
lgrinda749.501 kN⋅=:=
M1 MB.c3 69.209 kN m⋅⋅=:=
Mo3 MB.c3 Nc3 e3⋅+ 295.199 kN m⋅⋅=:=
Mcamp MB.c3 Nc3
lgrinda
2e3+
⋅+ qechil
lgrinda
2⋅
lgrinda
4⋅+
qumpl
lgrinda
2⋅
lgrinda
4⋅ R3
lgrinda
2⋅−+
... 30.023− kN m⋅⋅=:=
Mreazem 155.594kN m⋅:=
Mcamp 30.023− kN m⋅⋅=
alegem clasa de beton C20/25 , si otel PC 52 , rezultand :
fck 20MPa:= γc 1.5:= fcd
fck
γc13.333 MPa⋅=:=
fyk 355MPa:= γs 1.15:= fyd
fyk
γs308.696 MPa⋅=:=
Dimensionarea la moment incovoietor :
Armarea pe camp :
Φ1.max 28 mm⋅=
cnom Φ1.max 10mm+ 3.8 cm⋅=:=
d hechil cnom−Φ1.max
2− 0.398m=:=
μMcamp
bechil d2⋅ fcd⋅
0.032=:=
ω 1 1 2μ−− 0.032=:=
Asl.nec.gr ω bechil⋅ d⋅fcd
fyd⋅ 2.484 cm
2⋅=:=
alegem : 2ϕ14 cu Asl.eff.camp 3.08cm2:=
As.min 0.0013 bechil⋅ d⋅ 2.328 cm2⋅=:=
As.max 0.04bechil hechil⋅ 81 cm2⋅=:=
ql
Asl.eff.camp
bechil d⋅1.72 10
3−×=:=
snh bechil 2 cnom⋅− 2 14⋅ mm− 34.6 cm⋅=:= > snh.min 20cm:=
lcioc 15 14⋅ mm 21 cm⋅=:=
Armarea pe reazem :
Φ1.max 28 mm⋅=
cnom Φ1.max 10mm+ 3.8 cm⋅=:=
d hechil cnom−Φ1.max
2− 0.398m=:=
μMreazem
bechil d2⋅ fcd⋅
0.164=:=
ω 1 1 2μ−− 0.18=:=
Asl.nec.gr ω bechil⋅ d⋅fcd
fyd⋅ 13.916 cm
2⋅=:=
alegem : 3Φ25 cu Asl.eff.reazem 14.73cm2:=
As.min 0.0013 bechil⋅ d⋅ 2.328 cm2⋅=:=
As.max 0.04bechil hechil⋅ 81 cm2⋅=:=
ql
Asl.eff.reazem
bechil d⋅8.224 10
3−×=:=
snh
bechil 2 cnom⋅− 3 25⋅ mm−
214.95 cm⋅=:= > snh.min 20cm:=
lcioc 15 25⋅ mm 37.5 cm⋅=:=
lbd
lgrinda
41.175m=:= lungimea de ancorare
α6 1.5:= lb.rqd 1.35m:=
l0 α1 α2⋅ α3⋅ α4⋅ α5⋅ α6⋅ lb.rqd⋅ 2.025m=:= lungimea de suprapunere
Pentru zona de suprapunere din camp vom folosi armatura constructiva 2ϕ10
Dimensionarea la forta taietoare:
qw.min 0.08
fck
MPa
fyk
MPa
⋅ 1.008 103−×=:=
ν1 0.6:= αcw 1:= ctgθ 1.75:= fywd fyd 308.696 MPa⋅=:=
distanta maxima dintre etrieri:
longitudinal : sl.max 0.75 d⋅ 29.85 cm⋅=:=
transversal : st.max 0.75d 29.85 cm⋅=:=
Reazem 1
VED.1 164.232kN:= zona in care grinda nu mai este incastrata in cuzinet
CRD.c0.18
γc0.12=:=
K 1200mm
d+ 1.709=:= ρsl
Asl.eff.reazem
bechil d⋅8.224 10
3−×=:=
VRD.c CRD.c K⋅ 100 ρsl⋅fck
MPa⋅
1
3
⋅
bechil⋅ d⋅ MPa⋅ 93.404 kN⋅=:= VED< trebuie etrieri
νmin 0.035 K
3
2⋅fck
MPa⋅ 0.35=:=
VRD.min νmin
bechil
mm⋅
d
mm⋅ N⋅ 62.625 kN⋅=:= VRD.c<
z 0.9d 35.82 cm⋅=:=
VRD.max
αcw bechil⋅ z⋅ ν1⋅ fcd⋅
ctgθ1
ctgθ+
555.486 kN⋅=:=
distanta dintre etrieri:
alegem etrieri 2ϕ8 Asw 1.005cm2:=
snec
Asw z⋅ fywd⋅ ctgθ⋅
VED.111.841 cm⋅=:=
seff.r1 11cm:=
qw.eff
Asw
seff.r1 bechil⋅2.03 10
3−×=:= qw.min>
alegem etrieri ϕ8/150mm
VRD.s
Asw
seff.r1z⋅ fywd⋅ ctgθ⋅ 176.794 kN⋅=:= VED.r1>
VRD.max<
Reazem
VED.2 12.09kN:=
CRD.c0.18
γc0.12=:=
K 1200mm
d+ 1.709=:= ρsl
Asl.eff.camp
bechil d⋅1.72 10
3−×=:=
VRD.c CRD.c K⋅ 100 ρsl⋅fck
MPa⋅
1
3
⋅
bechil⋅ d⋅ MPa⋅ 55.439 kN⋅=:= VED> nu trebuie
etrieri
alegem etrieri constructivi ϕ8/250mm
VI. Calculul tasarii :
vom face calculul pentru stalpul S1
L1 1.9m= L1
B11=
B1 1.9m=
peff 582.911kN
m2
:=
γmed.1 18.2kN
m3
:= γmed.2 19kN
m3
:=
Df 1.1m:=
pnet.1 peff γmed.1 Df⋅− 562.891kN
m2
⋅=:= pnet.2 peff γmed.2 Df⋅− 562.011kN
m2
⋅=:=
Stratul 1
α0.1 0.963:= h1 0.6m:=
σz1 α0.1 pnet.1⋅ 542.064kN
m2
⋅=:=
σgz.1.20% 0.2 γmed.1⋅ h1⋅ 2.184kN
m2
⋅=:=
Stratul 2
α0.2 0.6728:= h2 1.2m:=
σz2 α0.2 pnet.1⋅ 378.713kN
m2
⋅=:=
σgz.2.20% 0.2 γmed.1⋅ h2⋅ 4.368kN
m2
⋅=:=
Stratul 3
α0.3 0.396:= h3 1.8m:=
σz3 α0.3 pnet.1⋅ 222.905kN
m2
⋅=:=
σgz.3.20% 0.2 γmed.1⋅ h3⋅ 6.552kN
m2
⋅=:=
Stratul 4
α0.4 0.268:= h4 2.4m:=
σz4 α0.4 pnet.1⋅ 150.855kN
m2
⋅=:=
σgz.4.20% 0.2 γmed.1⋅ h4⋅ 8.736kN
m2
⋅=:=
Stratul 5
α0.5 0.1337:= h5 3m:=
σz5 α0.5 pnet.2⋅ 75.141kN
m2
⋅=:=
σgz.5.20% 0.2 γmed.1 h4⋅ γmed.2 h5 h4−( )⋅+ ⋅ 11.016kN
m2
⋅=:=
Stratul 6
α0.6 0.1273:= h6 3.6m:=
σz6 α0.6 pnet.2⋅ 71.544kN
m2
⋅=:=
σgz.6.20% 0.2 γmed.1 h4⋅ γmed.2 h6 h4−( )⋅+ ⋅ 13.296kN
m2
⋅=:=
Stratul 7
α0.7 0.113:= h7 4.2m:=
σz7 α0.7 pnet.2⋅ 63.507kN
m2
⋅=:=
σgz.7.20% 0.2 γmed.1 h4⋅ γmed.2 h7 h4−( )⋅+ ⋅ 15.576kN
m2
⋅=:=
Stratul 8
α0.8 0.0625:= h8 4.8m:=
σz8 α0.8 pnet.2⋅ 35.126kN
m2
⋅=:=
σgz.8.20% 0.2 γmed.1 h4⋅ γmed.2 h8 h4−( )⋅+ ⋅ 17.856kN
m2
⋅=:=
Stratul 9
α0.9 0.0521:= h9 5.4m:=
σz9 α0.9 pnet.2⋅ 29.281kN
m2
⋅=:=
σgz.9.20% 0.2 γmed.1 h4⋅ γmed.2 h9 h4−( )⋅+ ⋅ 20.136kN
m2
⋅=:=
Stratul 10
α0.10 0.0484:= h10 6m:=
σz10 α0.10 pnet.2⋅ 27.201kN
m2
⋅=:=
σgz.10.20% 0.2 γmed.1 h4⋅ γmed.2 h10 h4−( )⋅+ ⋅ 22.416kN
m2
⋅=:=
Stratul 11
α0.11 0.0421:= h11 6.6m:=
σz11 α0.11 pnet.2⋅ 23.661kN
m2
⋅=:=
σgz.11.20% 0.2 γmed.1 h4⋅ γmed.2 h11 h4−( )⋅+ ⋅ 24.696kN
m2
⋅=:=
σz11 σgz.11.20%<
β 0.8:=
seff βσz1 σz2+ σz3+ σz4+( ) 0.6⋅ m
12000kPa
σz5 σz6+ σz7+ σz8+
σz9 σz10+ σz11++
...
0.6⋅ m
14000kPa+
⋅ 6.294 cm⋅=:=
sadm 8cm:=
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