Fundatii Izolate Cu Grinda de Echilibrare - Exemplu de Calcul

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


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


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⋅=:=


HTBF 0.5m:=

Df HTBF 20cm+>

Hinghet 0.9m:=

Df Hing 20cm+>



peff pacc<

peff

Ns2 Gf+

A2:=

Gf 0.4 Ns2⋅ 225.99 kN⋅=:=

A2 B2 L2⋅:=

pconv 400kN

m2

:=


m2

⋅=:=


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 :



L2 1.6m:=

B2 1.2m:=

H2 0.6m:=


Nf Ns2 Gf2+ Ggrinda.2+ GBCA.2+:=

Ns2 564.975 kN⋅=

γmed 21kN

m3

:= HCTS 0.5m:= Htotal Df HCTS+ 1.6m=:=


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


mB

0.913=:=

iγ 1

Tx.2


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


kN

m2

⋅=:=Rd

A'2

Nf2

A'2596.614

kN

m2

⋅=

Vd2

A'2

Rd2

A'2<

C. Stalpul S4 - fundatie izolata rigida


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⋅=:=


HTBF 0.5m:=

Df HTBF 20cm+>

Hinghet 0.9m:=

Df Hing 20cm+>



blocul din beton simplu :

peff pacc<

peff

Ns4 Gf+

A4:=

Gf 0.4 Ns4⋅ 451.98 kN⋅=:=

A4 B4 L4⋅:=

pconv 400kN

m2

:=


m2

⋅=:=


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=:=




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=:=


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


mB

0.902=:=

iγ 1

Tx.4


mB 1+

0.843=:=

ic iq

1 iq−( )Nc tan ϕ'd.arg( )⋅( )− 0.853=:=

γd.med

γmed

γγ21

kN

m3

⋅=:=

Df 1.4m=


m2

⋅=:=

Rd


kN

m2

⋅=:=Rd

A'4

Nf4

A'4611.288

kN

m2

⋅=

Vd4

A'4

Rd4

A'4<

D. Stalpul S3 - fundatie izolata rigida


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⋅=:=


HTBF 0.5m:=

Df HTBF 20cm+>

Hinghet 0.9m:=

Df Hing 20cm+>



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:=




L3 1.8m:= lc3 1.2m:=

B3 1.3m:= bc3 1 m⋅:=

H3 0.6m:= hc3 0.45m=


Nf Ns3 Gf3+ Gechil+ Ggrinda.3+:=

Ns3 564.975 kN⋅=

γmed 21kN

m3

:= Htotal Df HCTS+ 1.6m=:=


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


mB

0.926=:=

iγ 1

Tx.3


mB 1+

0.88=:=

ic iq

1 iq−( )Nc tan ϕ'd.arg( )⋅( )− 0.889=:=

γd.med

γmed

γγ21

kN

m3

⋅=:=

Df 1.1m=


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:



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⋅=:=


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⋅=:=




2− 0.76m=:=

T3.3 Asl.eff.3

fyd

3⋅<

Asl.nec.3

T3.3 3⋅

fydcm2⋅=:=

T3.3


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⋅⋅=:=


Alegem beton C20/25 si armaturi PC52:


fck

γc13.333 MPa⋅=:=


fyk

γs308.696 MPa⋅=:=

Φ1.max 28mm:=



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⋅=:=


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:



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⋅=:=




2− 0.548m=:=

T3.3 Asl.eff.3

fyd

3⋅<

Asl.nec.3

T3.3 3⋅

fyd1.136 cm

2⋅=:=


snh.3

B2 2 cnom⋅− 6 10⋅ mm−

521.28 cm⋅=:= < smax 25 cm⋅=

lcioc.3 15 10⋅ mm 15 cm⋅=:=



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

γc13.333 MPa⋅=:=


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⋅=:=


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⋅=:=


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⋅=:=


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=:=



⋅− ⋅−

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⋅⋅=:=




fck

γc13.333 MPa⋅=:=


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⋅=:=


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⋅=:=


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⋅=:=


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⋅=:=


⋅− ⋅−

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⋅=:=



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⋅⋅=



fck

γc13.333 MPa⋅=:=


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

⋅=:=


m2

⋅=:=

Stratul 7

α0.7 0.113:= h7 4.2m:=

σz7 α0.7 pnet.2⋅ 63.507kN

m2

⋅=:=


m2

⋅=:=

Stratul 8

α0.8 0.0625:= h8 4.8m:=

σz8 α0.8 pnet.2⋅ 35.126kN

m2

⋅=:=


m2

⋅=:=

Stratul 9

α0.9 0.0521:= h9 5.4m:=

σz9 α0.9 pnet.2⋅ 29.281kN

m2

⋅=:=


m2

⋅=:=

Stratul 10

α0.10 0.0484:= h10 6m:=

σz10 α0.10 pnet.2⋅ 27.201kN

m2

⋅=:=


m2

⋅=:=

Stratul 11

α0.11 0.0421:= h11 6.6m:=

σz11 α0.11 pnet.2⋅ 23.661kN

m2

⋅=:=


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:=

90 25+ 115=

115 76− 39=

Fundatii Izolate Cu Grinda de Echilibrare - Exemplu de Calcul

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Transcript of Fundatii Izolate Cu Grinda de Echilibrare - Exemplu de Calcul