Aleg beton clasa C25/30 pentru careRezistenta la compresiune pe cilindrii
fck 25MPa:=
Rezistenta de calcul la compresiune
αcc coeficient care ia in considerare efectele de lunga durata si efectele
defavorabile rezultate din modul de aplicare al incarcarilor
αcc 1:=
γc coeficient partial de siguranta pentru beton
γc 1.5:=
fcd
αcc fck⋅
γc
16.667 MPa⋅=:=
fctm 2.6MPa:=
Clasa de expunere XC1,mediu uscat sau permanent ud,beton in interiorul cladirilor
cu imiditate in aer scazuta.Beton submersibil in apa.
Clasa de consistenta S4 pentru elemente sau monolitizari cu armaturi
dese sau dificultati de compactare,elemente cu sectiuni reduse
fyk 500MPa:=Aleg otel S500
γs 1.15:=
fyd
fyk
γs
434.783 MPa⋅=:=
Stratul de acoperire cu beton:
bgr 30cm:=
hgr 45cm:=
Acoperirea cu beton nominala este definita ca acoperirea minima
plus toleranta admisa cnom cmin ∆ctol+=
cmin max cmin.b cmin.dur ∆cdur.γ+ ∆cdur.st− ∆cdur.add−, 10mm, ( )=
unde cmin,b este acoperirea minima datorita conditiilor de aderenta
cmin,dur este acoperirea minima datorita conditiilor de mediu
∆cdur.γ este coeficientul suplimentar de siguranta,in lipsa unor
norme speciale se poate lua 0.
∆cdur.st este o reducere a acoperirii minime datorita folosirii otelurilor
neoxidante,in cazurile obisnuite este 0
∆cdur.add este o reducere a acoperirii minime datorita protectiei suplimentare
valoarea recomandata este 0.
cmin.b 25mm:=
cmin.dur 10mm:=
∆cdur.γ 0:=
∆cdur.st 0:=
∆cdur.add 0:=
∆ctol 10mm:=Recomandat
cmin max cmin.b cmin.dur ∆cdur.γ+ ∆cdur.st− ∆cdur.add−, 10mm, ( ) 25 mm⋅=:=
cnom cmin ∆ctol+ 35 mm⋅=:=
Inaltimea utila a sectiunii
Φsl 28mm:=
Φsw 8mm:=
d1 hgr cnom− Φsw−Φsl
2− 393 mm⋅=:=
Dimensionarea armaturilor pentru grinda transversala:
Campul 1:
Momente in reazemul 1:
M1.st.sup 108kN m⋅:=
M1.st.sup.red 90.7kN m⋅:=
μlim 0.372:=
μM1.st.sup.red
bgr hgr2
⋅ fcd⋅
0.09=:= μ μlim< 1= rezulta armare simpla
kr1
M1.st.sup.red
bgr d12
⋅ fck⋅
0.078=:=
zd1
21 1 3.53 kr1⋅−+( )⋅ 363.652 mm⋅=:= z 0.95 d1⋅< 1=
As1.sup
M1.st.sup.red
fyd z⋅5.737 cm
2⋅=:= aleg 2ϕ16 si 1ϕ18 Aseff.1.st.sup 6.56cm
2:=
Procent minim de armare:
Asmin 0.5fctm
fyk
⋅ bgr⋅ d1⋅ 3.065 cm2
⋅=:=
Procent maxim de armare:
Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2
⋅=:=
Asmin Aseff.1.st.sup≤ Asmax≤ 1=
M1.st.inf.red 2.3kN m⋅:=
Placa fiind situata in zona comprimata a grinzii,aceasta se dimensioneaza ca o sectiune
''T'' simplu armata.Latimea activa de placa trebuie sa indeplineasca urmatoarele conditii:
L1 3.90m:=
hpl 13m:=
bgr 0.3m=
L0 0.7 L1⋅ 2.73m=:=
b1 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.1 0.2 b1⋅ 0.1 L0⋅+ 0.633m=:=
b2 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.2 0.2 b2⋅ 0.1 L0⋅+ 0.633m=:=
beff min beff.1 beff.2+ bgr+ b1 b2+ bgr+, ( ) 1.566m=:=
Se verifica pozitia axei neutre cu relatia:
rezulta axa neutra intersecteaza placa astfel se calculeaza ca
o sectiune ''T''.
beff
bgr
5> 1=
μM1.st.inf.red
beff d12
⋅ fcd⋅
0.001=:= rezulta ω 0.0101:=
As1.st.inf ω beff⋅ d1⋅fcd
fyd
⋅ 2.383 cm2
⋅=:= aleg 2ϕ16 Aseff.1.st.inf 4.02cm2
:=
Asmin Aseff.1.st.inf≤ Asmax≤ 1=
M1.dr.sup 120.7kN m⋅:=
M1.dr.sup.red 97.5kN m⋅:=
μlim 0.372:=
μM1.dr.sup.red
bgr hgr2
⋅ fcd⋅
0.096=:= μ μlim< 1= rezulta armare simpla
kr1
M1.dr.sup.red
bgr d12
⋅ fck⋅
0.084=:=
z 0.95 d1⋅< 1=z
d1
21 1 3.53 kr1⋅−+( )⋅ 361.241 mm⋅=:=
As2.sup
M1.dr.sup.red
fyd z⋅6.208 cm
2⋅=:= aleg 2ϕ16 si 1ϕ18 Aseff.1.dr.sup 6.56cm
2:=
Procent minim de armare:
Asmin 0.5fctm
fyk
⋅ bgr⋅ d1⋅ 3.065 cm2
⋅=:=
Procent maxim de armare:
Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2
⋅=:=
Asmin Aseff.1.dr.sup≤ Asmax≤ 1=
M1.dr.inf.red 8.2kN m⋅:=
Placa fiind situata in zona comprimata a grinzii,aceasta se dimensioneaza ca o sectiune
''T'' simplu armata.Latimea activa de placa trebuie sa indeplineasca urmatoarele conditii:
L1 3.90m:=
hpl 13m:=
bgr 0.3m=
L0 0.7 L1⋅ 2.73m=:=
b1 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.1 0.2 b1⋅ 0.1 L0⋅+ 0.633m=:=
b2 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.2 0.2 b2⋅ 0.1 L0⋅+ 0.633m=:=
beff min beff.1 beff.2+ bgr+ b1 b2+ bgr+, ( ) 1.566m=:=
Se verifica pozitia axei neutre cu relatia:
rezulta axa neutra intersecteaza placa astfel se calculeaza ca
o sectiune ''T''.
beff
bgr
5> 1=
μM1.dr.inf.red
beff d12
⋅ fcd⋅
0.002=:= rezulta ω 0.0101:=
As1.dr.inf ω beff⋅ d1⋅fcd
fyd
⋅ 2.383 cm2
⋅=:= aleg 2ϕ16 Aseff1.dr.inf 4.02cm2
:=
Asmin Aseff1.dr.inf≤ Asmax≤ 1=
Moment in campul 1:
M1.cent 62.1kN m⋅:=
Placa fiind situata in zona comprimata a grinzii,aceasta se dimensioneaza ca o sectiune
''T'' simplu armata.Latimea activa de placa trebuie sa indeplineasca urmatoarele conditii:
L1 3.90m:=
hpl 13m:=
bgr 0.3m=
L0 0.7 L1⋅ 2.73m=:=
b1 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.1 0.2 b1⋅ 0.1 L0⋅+ 0.633m=:=
b2 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.2 0.2 b2⋅ 0.1 L0⋅+ 0.633m=:=
beff min beff.1 beff.2+ bgr+ b1 b2+ bgr+, ( ) 1.566m=:=
Se verifica pozitia axei neutre cu relatia:
rezulta axa neutra intersecteaza placa astfel se calculeaza ca
o sectiune ''T''.
beff
bgr
5> 1=
μM1.cent
beff d12
⋅ fcd⋅
0.015=:= rezulta ω 0.0151:=
As1.cent ω beff⋅ d1⋅fcd
fyd
⋅ 3.562 cm2
⋅=:= aleg 2ϕ16 Aseff.1.cent 4.02cm2
:=
Asmin Aseff.1.cent≤ Asmax≤ 1=
Campul 2:
Momente in reazemul 1:
M2.st.sup 120.7kN m⋅:=
M2.st.sup.red 97.5kN m⋅:=
μlim 0.372:=
μ μlim< 1= rezulta armare simplaμ
M2.st.sup.red
bgr hgr2
⋅ fcd⋅
0.096=:=
kr1
M2.st.sup.red
bgr d12
⋅ fck⋅
0.084=:=
zd1
21 1 3.53 kr1⋅−+( )⋅ 361.241 mm⋅=:= z 0.95 d1⋅< 1=
As.2.sup
M2.st.sup.red
fyd z⋅6.208 cm
2⋅=:= aleg 2ϕ16 si 1ϕ18 Aseff.2.st.sup 6.56cm
2:=
Procent minim de armare:
Asmin 0.50fctm
fyk
⋅ bgr⋅ d1⋅ 3.065 cm2
⋅=:=
Procent maxim de armare:
Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2
⋅=:=
Asmin Aseff.2.st.sup≤ Asmax≤ 1=
M2.st.inf.red 44.9kN m⋅:=
Placa fiind situata in zona comprimata a grinzii,aceasta se dimensioneaza ca o sectiune
''T'' simplu armata.Latimea activa de placa trebuie sa indeplineasca urmatoarele conditii:
L1 3.90m:=
hpl 13m:=
bgr 0.3m=
L0 0.7 L1⋅ 2.73m=:=
b1 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.1 0.2 b1⋅ 0.1 L0⋅+ 0.633m=:=
b2 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.2 0.2 b2⋅ 0.1 L0⋅+ 0.633m=:=
beff min beff.1 beff.2+ bgr+ b1 b2+ bgr+, ( ) 1.566m=:=
Se verifica pozitia axei neutre cu relatia:
rezulta axa neutra intersecteaza placa astfel se calculeaza ca
o sectiune ''T''.
beff
bgr
5> 1=
μM2.st.inf.red
beff d12
⋅ fcd⋅
0.011=:= rezulta ω 0.011:=
As2.st.inf ω beff⋅ d1⋅fcd
fyd
⋅ 2.595 cm2
⋅=:= aleg 2ϕ16 Aseff.2.st.inf 4.02cm2
:=
Asmin Aseff.2.st.inf≤ Asmax≤ 1=
M2.dr.sup 120.8kN m⋅:=
M2.dr.sup.red 74.9kN m⋅:=
μlim 0.372:=
μ μlim< 1= rezulta armare simplaμ
M2.dr.sup.red
bgr hgr2
⋅ fcd⋅
0.074=:=
kr1
M2.dr.sup.red
bgr d12
⋅ fck⋅
0.065=:=
zd1
21 1 3.53 kr1⋅−+( )⋅ 369.124 mm⋅=:= z 0.95 d1⋅< 1=
As2.dr.sup
M2.dr.sup.red
fyd z⋅4.667 cm
2⋅=:= aleg 2ϕ16 si 1ϕ18 Aseff.2.dr.sup 6.56cm
2:=
Procent minim de armare:
Asmin 0.5fctm
fyk
⋅ bgr⋅ d1⋅ 3.065 cm2
⋅=:=
Procent maxim de armare:
Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2
⋅=:=
Asmin Aseff.2.dr.sup≤ Asmax≤ 1=
M2.dr.inf.red 44.9kN m⋅:=
Placa fiind situata in zona comprimata a grinzii,aceasta se dimensioneaza ca o sectiune
''T'' simplu armata.Latimea activa de placa trebuie sa indeplineasca urmatoarele conditii:
L1 3.90m:=
hpl 13m:=
bgr 0.3m=
L0 0.7 L1⋅ 2.73m=:=
b1 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.1 0.2 b1⋅ 0.1 L0⋅+ 0.633m=:=
b2 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.2 0.2 b2⋅ 0.1 L0⋅+ 0.633m=:=
beff min beff.1 beff.2+ bgr+ b1 b2+ bgr+, ( ) 1.566m=:=
Se verifica pozitia axei neutre cu relatia:
rezulta axa neutra intersecteaza placa astfel se calculeaza ca
o sectiune ''T''.
beff
bgr
5> 1=
μM2.dr.inf.red
beff d12
⋅ fcd⋅
0.011=:= rezulta ω 0.011:=
As2.dr.inf ω beff⋅ d1⋅fcd
fyd
⋅ 2.595 cm2
⋅=:= aleg 2ϕ16 Aseff.2.dr.inf. 4.02cm2
:=
Asmin Aseff.2.dr.inf.≤ Asmax≤ 1=
Moment in campul 2:
M2.cent 14.4kN m⋅:=
Placa fiind situata in zona comprimata a grinzii,aceasta se dimensioneaza ca o sectiune
''T'' simplu armata.Latimea activa de placa trebuie sa indeplineasca urmatoarele conditii:
L1 3.90m:=
hpl 13m:=
bgr 0.3m=
L0 0.7 L1⋅ 2.73m=:=
b1 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.1 0.2 b1⋅ 0.1 L0⋅+ 0.633m=:=
b2 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.2 0.2 b2⋅ 0.1 L0⋅+ 0.633m=:=
beff min beff.1 beff.2+ bgr+ b1 b2+ bgr+, ( ) 1.566m=:=
Se verifica pozitia axei neutre cu relatia:
rezulta axa neutra intersecteaza placa astfel se calculeaza ca
o sectiune ''T''.
beff
bgr
5> 1=
μM2.cent
beff d12
⋅ fcd⋅
0.004=:= rezulta ω 0.0101:=
As2.cent ω beff⋅ d1⋅fcd
fyd
⋅ 2.383 cm2
⋅=:= aleg 2ϕ16 Aseff.2.cent 4.02cm2
:=
Asmin Aseff.2.cent≤ Asmax≤ 1=
Campul 3:
Momente in reazemul 1:
M3.st.sup 120.8kN m⋅:=
M3.st.sup.red 97.6kN m⋅:=
μlim 0.372:=
μM3.st.sup.red
bgr hgr2
⋅ fcd⋅
0.096=:= μ μlim< 1= rezulta armare simpla
kr1
M3.st.sup.red
bgr d12
⋅ fck⋅
0.084=:=
zd1
21 1 3.53 kr1⋅−+( )⋅ 361.206 mm⋅=:= z 0.95 d1⋅< 1=
As.3.sup
M3.st.sup.red
fyd z⋅6.215 cm
2⋅=:= aleg 2ϕ16 si 1ϕ18 Aseff.3.st.sup 6.56cm
2:=
Procent minim de armare:
Asmin 0.5fctm
fyk
⋅ bgr⋅ d1⋅ 3.065 cm2
⋅=:=
Procent maxim de armare:
Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2
⋅=:=
Asmin Aseff.3.st.sup≤ Asmax≤ 1=
M3.st.inf.red 8.2kN m⋅:=
Placa fiind situata in zona comprimata a grinzii,aceasta se dimensioneaza ca o sectiune
''T'' simplu armata.Latimea activa de placa trebuie sa indeplineasca urmatoarele conditii:
L1 3.90m:=
hpl 13m:=
bgr 0.3m=
L0 0.7 L1⋅ 2.73m=:=
b1 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.1 0.2 b1⋅ 0.1 L0⋅+ 0.633m=:=
b2 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.2 0.2 b2⋅ 0.1 L0⋅+ 0.633m=:=
beff min beff.1 beff.2+ bgr+ b1 b2+ bgr+, ( ) 1.566m=:=
Se verifica pozitia axei neutre cu relatia:
rezulta axa neutra intersecteaza placa astfel se calculeaza ca
o sectiune ''T''.
beff
bgr
5> 1=
μM3.st.inf.red
beff d12
⋅ fcd⋅
0.002=:= rezulta ω 0.0101:=
As3.st.inf ω beff⋅ d1⋅fcd
fyd
⋅ 2.383 cm2
⋅=:= aleg 2ϕ16 Aseff3.st.inf 4.02cm2
:=
Asmin Aseff3.st.inf≤ Asmax≤ 1=
M3.dr.sup 108.1kN m⋅:=
M3.dr.sup.red 90.9kN m⋅:=
μlim 0.372:=
μ μlim< 1= rezulta armare simplaμ
M3.dr.sup.red
bgr hgr2
⋅ fcd⋅
0.09=:=
kr1
M3.dr.sup.red
bgr d12
⋅ fck⋅
0.078=:=
z 0.95 d1⋅< 1=z
d1
21 1 3.53 kr1⋅−+( )⋅ 363.582 mm⋅=:=
As3.dr.sup
M3.dr.sup.red
fyd z⋅5.75 cm
2⋅=:= aleg 2ϕ16 si 1ϕ18 Aseff.3.dr.sup 6.56cm
2:=
Procent minim de armare:
Asmin 0.5fctm
fyk
⋅ bgr⋅ d1⋅ 3.065 cm2
⋅=:=
Procent maxim de armare:
Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2
⋅=:=
Asmin Aseff.3.dr.sup≤ Asmax≤ 1=
M3.dr.inf.red 20.8kN m⋅:=
Placa fiind situata in zona comprimata a grinzii,aceasta se dimensioneaza ca o sectiune
''T'' simplu armata.Latimea activa de placa trebuie sa indeplineasca urmatoarele conditii:
L1 3.90m:=
hpl 13m:=
bgr 0.3m=
L0 0.7 L1⋅ 2.73m=:=
b1 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.1 0.2 b1⋅ 0.1 L0⋅+ 0.633m=:=
b2 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.2 0.2 b2⋅ 0.1 L0⋅+ 0.633m=:=
beff min beff.1 beff.2+ bgr+ b1 b2+ bgr+, ( ) 1.566m=:=
Se verifica pozitia axei neutre cu relatia:
rezulta axa neutra intersecteaza placa astfel se calculeaza ca
o sectiune ''T''.
beff
bgr
5> 1=
μM3.dr.inf.red
beff d12
⋅ fcd⋅
0.005=:= rezulta ω 0.0101:=
As3.dr.inf ω beff⋅ d1⋅fcd
fyd
⋅ 2.383 cm2
⋅=:= aleg 2ϕ16 Aseff3.dr.inf 4.02cm2
:=
Asmin Aseff3.dr.inf≤ Asmax≤ 1=
Moment in campul 3:
M3.cent 61.9kN m⋅:=
Placa fiind situata in zona comprimata a grinzii,aceasta se dimensioneaza ca o sectiune
''T'' simplu armata.Latimea activa de placa trebuie sa indeplineasca urmatoarele conditii:
L1 3.90m:=
hpl 13m:=
bgr 0.3m=
L0 0.7 L1⋅ 2.73m=:=
b1 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.1 0.2 b1⋅ 0.1 L0⋅+ 0.633m=:=
b2 0.5 L1 bgr−( )⋅ 1.8m=:=
beff.2 0.2 b2⋅ 0.1 L0⋅+ 0.633m=:=
beff min beff.1 beff.2+ bgr+ b1 b2+ bgr+, ( ) 1.566m=:=
Se verifica pozitia axei neutre cu relatia:
rezulta axa neutra intersecteaza placa astfel se calculeaza ca
o sectiune ''T''.
beff
bgr
5> 1=
μM3.cent
beff d12
⋅ fcd⋅
0.015=:= rezulta ω 0.0151:=
As3.cent ω beff⋅ d1⋅fcd
fyd
⋅ 3.562 cm2
⋅=:= aleg 2ϕ16 Aseff3.cent 4.02cm2
:=
Asmin Aseff3.cent≤ Asmax≤ 1=
Armare transversala la forta taietoare:
-pentru fiecare grinda se calculeaza doua valori ale fortelor taietoare de proiectare,
corespunzatoare valorilor maxime ale momentelor pozitive si negative care se dezvolta
la extremitati.
Mdb.i γRb. MRb.i⋅ min 1ΣMRc
ΣMRb
,
⋅=
γRb 1.20:=factor de suprarezistenta
MRb.1 0.9 As.eff.1⋅ d⋅ fyd⋅=
MRb.2 0.9 As.eff.2⋅ d⋅ fyd⋅=
Deschiderea A-B:
-seismul actioneaza de la stanga la dreapta
Momentul capabil pentru 2Φ16
As.eff.1 4.02cm2
:= η 1:= λ 0.8:=
xAs.eff.1 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.044m=:=
zreal d1λ x⋅( )
2− 375.522 mm⋅=:=
Mcap.1 As.eff.1 fyd⋅ zreal⋅ 65.635 kN m⋅⋅=:=
Momentul capabil pentru 2ϕ16 si 1ϕ18 :
As.eff.2 6.56cm2
:= η 1:= λ 0.8:=
xAs.eff.2 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.071m=:=
zreal d1λ x⋅( )
2− 364.478 mm⋅=:=
Mcap.2 As.eff.2 fyd⋅ zreal⋅ 103.956 kN m⋅⋅=:=
Momentul capabil petntru 5Φ18:
As.eff.3 12.8cm2
:= η 1:= λ 0.8:=
xAs.eff.3 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.139m=:=
zreal d1λ x⋅( )
2− 337.348 mm⋅=:=
Mcap.3 As.eff.3 fyd⋅ zreal⋅ 187.741 kN m⋅⋅=:=
ΣMRb Mcap.1 Mcap.2+ 169.59 kN m⋅⋅=:= moment capabil grinda
ΣMRc 2 Mcap.3⋅ 375.483 kN m⋅⋅=:= moment capabil stalpi care intra in nod
ΣMRc
ΣMRb
2.214=
Mdb.i γRb. MRb.i⋅ min 1ΣMRc
ΣMRb
,
⋅=
γRb 1.20:= factor de suprarezistenta
MRb.1 As.eff.1 d1⋅ fyd⋅ 68.69 kN m⋅⋅=:= aa ???
MRb.2 As.eff.2 d1⋅ fyd⋅ 112.09 kN m⋅⋅=:=
Mdb.1 γRb MRb.1⋅ 1⋅ 82.427 kN m⋅⋅=:=
Mdb.2 γRb MRb.2⋅ 1⋅ 134.509 kN m⋅⋅=:=
L1 6m:= bst.marginal 45cm:= bst.central 55cm:= γb.a 25kN
m3
:= hpl 13cm:=
Ppardoseala 1.4kN
m2
:=lcl L1
bst.marginal
2
bst.central
2+
− 5.5m=:=
Agr 3.9m 5.5⋅ m 21.45m2
=:=
U1 2kN
m2
:= incarcare utila din plansee
U2 0.8kN
m2
:= incarcare utila din pereti despartitori
qU1 U2+( )Agr
lcl
10.92kN
m⋅=:=
gPpardoseala hpl γb.a⋅+( ) Agr⋅
lcl
bgr hgr⋅ γb.a⋅+ 21.51kN
m⋅=:=
ψ2 0.3:=
VEd.st
Mdb.1 Mdb.2+
lcl
g ψ2 q⋅+( ) lcl⋅
2+
107.604 kN⋅=:=
-seismul actioneaza de la dreapta la stanga
Momentul capabil pentru 2Φ16:
As.eff.1 4.02cm2
:= η 1:= λ 0.8:=
xAs.eff.1 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.044m=:=
zreal d1λ x⋅( )
2− 375.522 mm⋅=:=
Mcap.1 As.eff.1 fyd⋅ zreal⋅ 65.635 kN m⋅⋅=:=
Momentul capabil pentru 2ϕ16 si 1ϕ18 :
As.eff.2 5.65cm2
:= η 1:= λ 0.8:=
xAs.eff.2 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.061m=:=
zreal d1λ x⋅( )
2− 368.435 mm⋅=:=
Mcap.2 As.eff.2 fyd⋅ zreal⋅ 90.507 kN m⋅⋅=:=
Momentul capabil petntru 5 ϕ18 :
As.eff.3 12.8cm2
:= η 1:= λ 0.8:=
xAs.eff.3 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.139m=:=
zreal d1λ x⋅( )
2− 337.348 mm⋅=:=
Mcap.3 As.eff.3 fyd⋅ zreal⋅ 187.741 kN m⋅⋅=:=
ΣMRb Mcap.1 Mcap.2+ 156.141 kN m⋅⋅=:= moment capabil grinda
ΣMRc 2 Mcap.3⋅ 375.483 kN m⋅⋅=:= moment capabil stalpi care intra in nod
ΣMRc
ΣMRb
2.405=
Mdb.i γRb. MRb.i⋅ min 1ΣMRc
ΣMRb
,
⋅=
γRb 1.20:= factor de suprarezistenta
MRb.1 As.eff.1 d1⋅ fyd⋅ 68.69 kN m⋅⋅=:=
MRb.2 As.eff.2 d1⋅ fyd⋅ 96.541 kN m⋅⋅=:=
Mdb.1 γRb MRb.1⋅ 1⋅ 82.427 kN m⋅⋅=:=
Mdb.2 γRb MRb.2⋅ 1⋅ 115.85 kN m⋅⋅=:=
L1 6m:= bst.marginal 45cm:= bst.central 55cm:= γb.a 25kN
m3
:= hpl 13cm:=
Ppardoseala 1.4kN
m2
:=lcl L1
bst.marginal
2
bst.central
2+
− 5.5m=:=
Agr 3.9m 5.5⋅ m 21.45m2
=:=
U1 2kN
m2
:= incarcare utila din plansee
U2 0.8kN
m2
:= incarcare utila din pereti despartitori
qU1 U2+( )Agr
lcl
10.92kN
m⋅=:=
gPpardoseala hpl γb.a⋅+( ) Agr⋅
lcl
bgr hgr⋅ γb.a⋅+ 21.51kN
m⋅=:=
ψ2 0.3:=
VEd.st
Mdb.1 Mdb.2+
lcl
g ψ2 q⋅+( ) lcl⋅
2+
104.212 kN⋅=:=
VEdc 88.4kN:=
VEd max VEd.st VEdc, ( ) 104.212 kN⋅=:=
Aseff As.eff.1 As.eff.2+ 9.67 cm2
⋅=:=
ρ1
Aseff
bgr d1⋅0.008=:= σcp 0:= k1 0.15:=
k 1200mm
d1
+ 1.713=:=
γc 1.5:= bgr 0.3m= d1 0.393m=crdc
0.18
γc
0.12=:=
VRdc crdc k⋅ 100ρ1 fck⋅( )1
3⋅ k1 σcp⋅+
bgr⋅ d1⋅ 1⋅ MPa
2
366.349 kN⋅=:=
VEd VRdc> 1= este necesar calculul etrierilor
νmin 0.035 k
3
2⋅ fck⋅ 1⋅ MPa
1
2392.48
kN
m2
⋅=:=
VRdc.min νmin k1 σcp⋅+( ) bgr⋅ d1⋅ 46.273 kN⋅=:=
VRdc VRdc.min> 1=
aleg ctgθ 1.75:= tgθ1
ctgθ0.571=:= z 0.9 d1⋅ 0.354m=:= αcw 1:= v1 0.6:=
VRd.max
αcw bgr⋅ z⋅ v1⋅ fcd⋅
tgθ ctgθ+457.089 kN⋅=:=
VRds VEd 104.212 kN⋅=:= VRd min VRds VRd.max, ( ) 104.212 kN⋅=:=
Asw 101mm2
:= pentru etrieri ϕ8
sAsw z⋅ fyd⋅ ctgθ⋅
VRds
0.261m=:=
aleg s 250mm:=
smax 0.75 d1⋅ 0.295m=:=-distanta maxima intre etrieri:
slmax min 7 14⋅ mm 150mm, hgr
4,
9.8 cm⋅=:=-pentru zona critica:
lcr 1.5 hgr⋅ 0.675m=:=
scr 10cm:=
Coeficient de armare transversala:
ρw
Asw
s bgr⋅0.001=:=
ρw.min 0.08fck 1⋅ MPa
0.5
fyk
⋅ 0.001=:= ρw ρw.min> 1=
Verificarea ariei efective maxime:
Asw fyd⋅
bgr s⋅0.5αcw v1⋅ fcd⋅< 1=
xlcl
2
VRdc
lcl
2⋅
VEd
− 0.999m=:=
-pe distanta 0-1.0m aleg etrieri ϕ8/10cm
-pe distanta 1.0-4.55m aleg etrieri ϕ8/25cm
-pe distanta 4.55-5.55m aleg etrieri ϕ8/10cm
Deschiderea B-C:
-seismul actioneaza de la stanga la dreapta
Momentul capabil pentru 2Φ16:
As.eff.1 4.02cm2
:= η 1:= λ 0.8:=
xAs.eff.1 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.044m=:=
zreal d1λ x⋅( )
2− 375.522 mm⋅=:=
Mcap.1 As.eff.1 fyd⋅ zreal⋅ 65.635 kN m⋅⋅=:=
Momentul capabil pentru 2Φ16 si 1 Φ18 :
As.eff.2 5.65cm2
:= η 1:= λ 0.8:=
xAs.eff.2 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.061m=:=
zreal d1λ x⋅( )
2− 368.435 mm⋅=:=
Mcap.2 As.eff.2 fyd⋅ zreal⋅ 90.507 kN m⋅⋅=:=
Momentul capabil pentru 5Φ18:
As.eff.3 12.8cm2
:= η 1:= λ 0.8:=
xAs.eff.3 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.139m=:=
zreal d1λ x⋅( )
2− 337.348 mm⋅=:=
Mcap.3 As.eff.3 fyd⋅ zreal⋅ 187.741 kN m⋅⋅=:=
ΣMRb Mcap.1 Mcap.2+ 156.141 kN m⋅⋅=:= moment capabil grinda
ΣMRc 2 Mcap.3⋅ 375.483 kN m⋅⋅=:= moment capabil stalpi care intra in nod
ΣMRc
ΣMRb
2.405=
Mdb.i γRb. MRb.i⋅ min 1ΣMRc
ΣMRb
,
⋅=
γRb 1.20:= factor de suprarezistenta
MRb.1 As.eff.1 d1⋅ fyd⋅ 68.69 kN m⋅⋅=:=
MRb.2 As.eff.2 d1⋅ fyd⋅ 96.541 kN m⋅⋅=:=
Mdb.1 γRb MRb.1⋅ 1⋅ 82.427 kN m⋅⋅=:=
Mdb.2 γRb MRb.2⋅ 1⋅ 115.85 kN m⋅⋅=:=Ppardoseala 1.4
kN
m2
:=bst.central 55cm:= hpl 13cm:= γb.a 25
kN
m3
:=L2 3.9m:=
lcl L2 bst.central( )− 3.35m=:=
Agr 3.9m 3.35⋅ m 13.065m2
=:=
U1 2kN
m2
:= incarcare utila din plansee
U2 0.8kN
m2
:= incarcare utila din pereti despartitori
qU1 U2+( )Agr
lcl
10.92kN
m⋅=:=
gPpardoseala hpl γb.a⋅+( ) Agr⋅
lcl
bgr hgr⋅ γb.a⋅+ 21.51kN
m⋅=:=
ψ2 0.3:=
VEd.st
Mdb.1 Mdb.2+
lcl
g ψ2 q⋅+( ) lcl⋅
2+
100.704 kN⋅=:=
-seismul actioneaza de la dreapta la stanga
Momentul capabil pentru 2Φ16:
As.eff.1 4.02cm2
:= η 1:= λ 0.8:=
xAs.eff.1 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.044m=:=
zreal d1λ x⋅( )
2− 375.522 mm⋅=:=
Mcap.1 As.eff.1 fyd⋅ zreal⋅ 65.635 kN m⋅⋅=:=
Momentul capabil petntru 2Φ16 si 1 Φ18 :
As.eff.2 5.65cm2
:= η 1:= λ 0.8:=
xAs.eff.2 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.061m=:=
zreal d1λ x⋅( )
2− 368.435 mm⋅=:=
Mcap.2 As.eff.2 fyd⋅ zreal⋅ 90.507 kN m⋅⋅=:=
Momentul capabil petntru 5Φ18:
As.eff.3 12.8cm2
:= η 1:= λ 0.8:=
xAs.eff.3 fyd⋅( )λ η⋅ bgr⋅ fcd⋅
0.139m=:=
zreal d1λ x⋅( )
2− 337.348 mm⋅=:=
Mcap.3 As.eff.3 fyd⋅ zreal⋅ 187.741 kN m⋅⋅=:=
ΣMRb Mcap.1 Mcap.2+ 156.141 kN m⋅⋅=:= moment capabil grinda
ΣMRc 2 Mcap.3⋅ 375.483 kN m⋅⋅=:= moment capabil stalpi care intra in nod
ΣMRc
ΣMRb
2.405=
Mdb.i γRb. MRb.i⋅ min 1ΣMRc
ΣMRb
,
⋅=
γRb 1.20:= factor de suprarezistenta
MRb.1 As.eff.1 d1⋅ fyd⋅ 68.69 kN m⋅⋅=:= a /???
MRb.2 As.eff.2 d1⋅ fyd⋅ 96.541 kN m⋅⋅=:=
Mdb.1 γRb MRb.1⋅ 1⋅ 82.427 kN m⋅⋅=:=
Mdb.2 γRb MRb.2⋅ 1⋅ 115.85 kN m⋅⋅=:=
γb.a 25kN
m3
:= Ppardoseala 1.4kN
m2
:=L2 3.9m:= bst.central 55cm:= hpl 13cm:=
lcl L2 bst.central( )− 3.35m=:=
Agr 3.9m 3.35⋅ m 13.065m2
=:=
U1 2kN
m2
:= incarcare utila din plansee
U2 0.8kN
m2
:= incarcare utila din pereti despartitori
qU1 U2+( )Agr
lcl
10.92kN
m⋅=:=
gPpardoseala hpl γb.a⋅+( ) Agr⋅
lcl
bgr hgr⋅ γb.a⋅+ 21.51kN
m⋅=:=
ψ2 0.3:=
VEd.st
Mdb.1 Mdb.2+
lcl
g ψ2 q⋅+( ) lcl⋅
2+
100.704 kN⋅=:=
VEdc 63.2kN:=
VEd max VEd.st VEdc, ( ) 100.704 kN⋅=:=
Aseff As.eff.1 As.eff.2+ 9.67 cm2
⋅=:=
ρ1
Aseff
bgr d1⋅0.008=:= σcp 0:= k1 0.15:=
k 1200mm
d1
+ 1.713=:=
γc 1.5:= bgr 0.3m= d1 0.393m=crdc
0.18
γc
0.12=:=
VRdc crdc k⋅ 100ρ1 fck⋅( )1
3⋅ k1 σcp⋅+
bgr⋅ d1⋅ 1⋅ MPa
2
366.349 kN⋅=:=
VEd VRdc> 1= este necesar calculul etrierilor
νmin 0.035 k
3
2⋅ fck⋅ 1⋅ MPa
1
2392.48
kN
m2
⋅=:=
VRdc.min νmin k1 σcp⋅+( ) bgr⋅ d1⋅ 46.273 kN⋅=:=
VRdc VRdc.min> 1=
aleg ctgθ 1.75:= tgθ1
ctgθ0.571=:= z 0.9 d1⋅ 0.354m=:= αcw 1:= v1 0.6:=
VRd.max
αcw bgr⋅ z⋅ v1⋅ fcd⋅
tgθ ctgθ+457.089 kN⋅=:=
VRds VEd 100.704 kN⋅=:= VRd min VRds VRd.max, ( ) 100.704 kN⋅=:=
Asw 101mm2
:= pentru etrieri ϕ8
sAsw z⋅ fyd⋅ ctgθ⋅
VRds
0.27m=:=
aleg s 250mm:=
smax 0.75 d1⋅ 0.295m=:=-distanta maxima intre etrieri:
slmax min 7 14⋅ mm 150mm, hgr
4,
0.098m=:=-pentru zona critica:
lcr 1.5 hgr⋅ 0.675m=:=
scr 10cm:=
Coeficient de armare transversala:
ρw
Asw
s bgr⋅0.001=:=
ρw.min 0.08fck 1⋅ MPa
0.5
fyk
⋅ 0.001=:= ρw ρw.min> 1=
Verificarea ariei efective maxime:
Asw fyd⋅
bgr s⋅0.5αcw v1⋅ fcd⋅< 1=
xlcl
2
VRdc
lcl
2⋅
VEd
− 0.571m=:=
-pe distanta 0-0.6m aleg etrieri ϕ8/10cm
-pe distanta 0.6-4.95m aleg etrieri ϕ8/25cm
-pe distanta 4.95-5.55m aleg etrieri ϕ8/10cm
Deschiderea C-D:
Datoria simetriei deschiderea C-D se calculeaza ca si deschiderea A-B astfel rezulta:
-pe distanta 0-1.0m aleg etrieri ϕ8/10cm
-pe distanta 1.0-4.55m aleg etrieri ϕ8/25cm
-pe distanta 4.55-5.55m aleg etrieri ϕ8/10cm
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