Mathcad - Armare Grinda La M-Tbeff

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⋅


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:=


Asmin 0.5fctm

fyk

⋅ bgr⋅ d1⋅ 3.065 cm2

⋅=:=


Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2

⋅=:=

Asmin Aseff.1.dr.sup≤ Asmax≤ 1=

M1.dr.inf.red 8.2kN m⋅:=



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=:=




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⋅:=



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=:=




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:


M2.st.sup 120.7kN 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:=


Asmin 0.50fctm

fyk

⋅ bgr⋅ d1⋅ 3.065 cm2

⋅=:=


Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2

⋅=:=





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=:=




o sectiune ''T''.

beff

bgr

5> 1=

μM2.st.inf.red

beff d12

⋅ fcd⋅

0.011=:= rezulta ω 0.011:=


fyd

⋅ 2.595 cm2

⋅=:= aleg 2ϕ16 Aseff.2.st.inf 4.02cm2

:=

Asmin Aseff.2.st.inf≤ Asmax≤ 1=



μlim 0.372:=


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:=


Asmin 0.5fctm

fyk

⋅ bgr⋅ d1⋅ 3.065 cm2

⋅=:=


Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2

⋅=:=





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=:=




o sectiune ''T''.

beff

bgr

5> 1=

μM2.dr.inf.red

beff d12

⋅ fcd⋅

0.011=:= rezulta ω 0.011:=


fyd

⋅ 2.595 cm2

⋅=:= aleg 2ϕ16 Aseff.2.dr.inf. 4.02cm2

:=

Asmin Aseff.2.dr.inf.≤ Asmax≤ 1=

Moment in campul 2:




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=:=




o sectiune ''T''.

beff

bgr

5> 1=

μM2.cent

beff d12

⋅ fcd⋅

0.004=:= rezulta ω 0.0101:=


fyd

⋅ 2.383 cm2

⋅=:= aleg 2ϕ16 Aseff.2.cent 4.02cm2

:=

Asmin Aseff.2.cent≤ Asmax≤ 1=

Campul 3:


M3.st.sup 120.8kN m⋅:=


μlim 0.372:=

μM3.st.sup.red

bgr hgr2

⋅ fcd⋅


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:=


Asmin 0.5fctm

fyk

⋅ bgr⋅ d1⋅ 3.065 cm2

⋅=:=


Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2

⋅=:=





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=:=




o sectiune ''T''.

beff

bgr

5> 1=

μM3.st.inf.red

beff d12

⋅ fcd⋅

0.002=:= rezulta ω 0.0101:=


fyd

⋅ 2.383 cm2

⋅=:= aleg 2ϕ16 Aseff3.st.inf 4.02cm2

:=

Asmin Aseff3.st.inf≤ Asmax≤ 1=



μlim 0.372:=


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:=


Asmin 0.5fctm

fyk

⋅ bgr⋅ d1⋅ 3.065 cm2

⋅=:=


Asmax 0.04 bgr⋅ d1⋅ 47.16 cm2

⋅=:=





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=:=




o sectiune ''T''.

beff

bgr

5> 1=

μM3.dr.inf.red

beff d12

⋅ fcd⋅

0.005=:= rezulta ω 0.0101:=


fyd

⋅ 2.383 cm2

⋅=:= aleg 2ϕ16 Aseff3.dr.inf 4.02cm2

:=

Asmin Aseff3.dr.inf≤ Asmax≤ 1=

Moment in campul 3:




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=:=




o sectiune ''T''.

beff

bgr

5> 1=

μM3.cent

beff d12

⋅ fcd⋅

0.015=:= rezulta ω 0.0151:=


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:=


0.071m=:=

zreal d1λ x⋅( )

2− 364.478 mm⋅=:=


Momentul capabil petntru 5Φ18:

As.eff.3 12.8cm2

:= η 1:= λ 0.8:=


0.139m=:=

zreal d1λ x⋅( )

2− 337.348 mm⋅=:=


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=


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:=


0.044m=:=

zreal d1λ x⋅( )

2− 375.522 mm⋅=:=


Momentul capabil pentru 2ϕ16 si 1ϕ18 :

As.eff.2 5.65cm2

:= η 1:= λ 0.8:=


0.061m=:=

zreal d1λ x⋅( )

2− 368.435 mm⋅=:=


Momentul capabil petntru 5 ϕ18 :

As.eff.3 12.8cm2

:= η 1:= λ 0.8:=


0.139m=:=

zreal d1λ x⋅( )

2− 337.348 mm⋅=:=




ΣMRc

ΣMRb

2.405=


ΣMRb

,

⋅=




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


U2 0.8kN

m2


qU1 U2+( )Agr

lcl

10.92kN

m⋅=:=


lcl


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


Deschiderea B-C:

-seismul actioneaza de la stanga la dreapta


As.eff.1 4.02cm2

:= η 1:= λ 0.8:=


0.044m=:=

zreal d1λ x⋅( )

2− 375.522 mm⋅=:=


Momentul capabil pentru 2Φ16 si 1 Φ18 :

As.eff.2 5.65cm2

:= η 1:= λ 0.8:=


0.061m=:=

zreal d1λ x⋅( )

2− 368.435 mm⋅=:=



As.eff.3 12.8cm2

:= η 1:= λ 0.8:=


0.139m=:=

zreal d1λ x⋅( )

2− 337.348 mm⋅=:=




ΣMRc

ΣMRb

2.405=


ΣMRb

,

⋅=




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


U2 0.8kN

m2


qU1 U2+( )Agr

lcl

10.92kN

m⋅=:=


lcl


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


As.eff.1 4.02cm2

:= η 1:= λ 0.8:=


0.044m=:=

zreal d1λ x⋅( )

2− 375.522 mm⋅=:=


Momentul capabil petntru 2Φ16 si 1 Φ18 :

As.eff.2 5.65cm2

:= η 1:= λ 0.8:=


0.061m=:=

zreal d1λ x⋅( )

2− 368.435 mm⋅=:=


Momentul capabil petntru 5Φ18:

As.eff.3 12.8cm2

:= η 1:= λ 0.8:=


0.139m=:=

zreal d1λ x⋅( )

2− 337.348 mm⋅=:=




ΣMRc

ΣMRb

2.405=


ΣMRb

,

⋅=


MRb.1 As.eff.1 d1⋅ fyd⋅ 68.69 kN m⋅⋅=:= a /???


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


U2 0.8kN

m2


qU1 U2+( )Agr

lcl

10.92kN

m⋅=:=


lcl


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=:=




Deschiderea C-D:

Datoria simetriei deschiderea C-D se calculeaza ca si deschiderea A-B astfel rezulta:




Mathcad - Armare Grinda La M-Tbeff

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Transcript of Mathcad - Armare Grinda La M-Tbeff