Mathcad - fundatii izolate

FUNDATII IZOLATEFUNDATII IZOLATEFUNDATII IZOLATEFUNDATII IZOLATE

Date generale: n 4:= a 0:=

stalp: • bs 0.4:= armati cu: 8ϕϕϕϕ16 PC52 longitudinal

ls 0.5:= etrieri 10ϕϕϕϕ 100

pereti zidarie: • BCA 30 cm

Hp 3.0:=

cota terenului sistematizat:• CTA 0.4 a+( ) 0.4=:=

stratificatia terenului:•

0.00 ... -0.50: Umplutura

-0.50 ...-3.50: Argila prafoasa, cafenie,plastic consistenta

γγγγk1 18.2:= w1 22%:= Ip1 21%:= Ic1 0.70:=

ϕϕϕϕk1 12deg:= e1 0.69:= E1 12000:=

ck1 45 0.15n+( ) 45.6=:=

-3.50 ... -10.00: Nisip argilos ,galben cafeniu

γγγγk2 19:= w2 18%:= Ip2 11%:= ϕϕϕϕk2 24deg:=

e2 0.65:= E2 14000:= ck2 5 0.2n+( ) 5.8=:=

Incarcarile de calcul pentru fiecare stalp se vor determina considerand:

Incarcarea Distribuita Permanenta:•

Incarcarea Distribuita Variabila :•

pk 15 0.1n+( ) 15.4=:=

qk 8 0.1n+( ) 8.4=:=

Stalpul S1•

SLU - GEO+STR , CP3 : A1/A2+M2+R3 , conf SR EN 1997•SLEN conf SR EN 1997 (calculul tasarii prin insumare pe strate elementare)•

γγγγGf 1.0:= γγγγQf 0:= γγγγR 1.40:= γγγγc 1.25:=

γγγγGn 1.35:= γγγγQn 1.5:= γγγγϕϕϕϕ 1.25:=

1. ) PREDIMENSIONARE

Stabilirea Df :•

Df Hing 10...20( )cm+≥

Df. 0.90 0.2+ 1.1=:=

Stabilirea lui B si L ( bloc de fundare ) :•

peff pacc≤ AafS1 5 3⋅ 15=:=

pconv 408:= obtinut prin interpolare din tab de mai jos

pd pk γγγγGn⋅ 20.79=:= pacc pconv 0.4081

PakPa⋅=:=

qd qk γγγγQn⋅ 12.6=:=

NS1 AafS1 pd qd+( )⋅ 500.85=:=

Gf 0.25 NS1⋅ 125.213=:=

VdpS1 NS1 Gf+ 626.063=:=

ls

bs

1.25=L1

B1

ls

bs

= 1.25= L1 B1 1.25⋅=

VdpS1

pconv

1.534= B1 L1⋅VdpS1

pconv

= 2.181=

B1

VdpS1

1.25 pconv⋅1.108=:= L1 B1 1.25⋅ 1.385=:=

B , L multiplu de 5 cm

B , L > 40 cm

Bs1 1.50:=

alegem : Ls1 2.0:=

Stabilirea lui lc , bc ( cuzinet ) :•

lc1 0.5Ls1 1=:= alegem beton clasa C 12/15

bc1 0.5Bs1 0.75=:=

lcs1 1.10:=

alegembcs1 0.80:=

Stabilirea lui Hb :•

tan αααα( ) tan ααααadm( )≥

tan ααααadm( ) 1.52:=

clasa C 8/10

I.) tan αααα( )Hbloc

Ls1 lcs1−( )2

=

Hb1. 1.52Ls1 lcs1−( )

2⋅ 0.684=:=

II.) tan αααα( )Hbloc

Bs1 bcs1−( )2

=

Hb1.. 1.52Bs1 bcs1−( )

2⋅ 0.532=:=

Hb1 max Hb1. Hb1.., ( ) 0.684=:=

Hbloc-- multiplu de 5 cm

-- > 40 cmalegem : Hbloc1 0.70:=

Stabilirea lui hc :•

tan ββββ( ) tan ββββadm( )≥ tan ββββadm( ) 1.00:=

I.) tan ββββ( )hc

lcs1 ls−( )2

=

hc1. 1.00lcs1 ls−( )

2⋅ 0.3=:=

II.) tan ββββ( )hc

bcs1 bs−( )2

=

hc1.. 1.00bcs1 bs−( )

2⋅ 0.2=:=

hc max hc1. hc1.., ( ) 0.3=:=

hcz1-- multiplu de 5 cm

-- > 30 cmalegem : hcz1 0.30:=

VERIFICAREA LA CAPACITATE PORTANTA

( SLU - GEO )

C. P. 3•

Eforturile sectionale :

NS1 500.85=

Mx1 0.2 NS1⋅ 100.17=:= Tafx1 0.1 NS1⋅ 50.085=:=

My1 0.1 NS1⋅ 50.085=:= Tafy1 0.05 NS1⋅ 25.043=:=

hgr CTA 0.2+ 0.6=:=bgr 0.30:=

Adancimea de fundarehgr 2 bgr⋅ 0.6=:=

Df. 0.20 hcz1+ Hbloc1+ 1.2=:=Tr 5.0:=

Hz 3.0:=γγγγbet 25:=

bz 0.3:=

γγγγmed 22:= γγγγBCA 8:=Htot1 Df. CTA+ 1.6=:=

Vd Ned Npd+ Gfd+=

Npd1 γγγγGn hgr bgr⋅ Tr⋅ γγγγbet⋅ Hz bz⋅ Tr⋅ γγγγBCA⋅+( )⋅ 78.975=:=

Gfd1 γγγγGn Bs1 Ls1⋅ Htot1⋅ γγγγmed⋅( )⋅ 142.56=:=

Vd1 NS1 Npd1+ Gfd1+ 722.385=:=

e1

bs

2

bgr

2+ 0.35=:=

Mxfed1 Mx1 Tafx1 Hbloc1 hcz1+( )⋅+ Npd1 e1⋅+ 177.896=:=

Tyfed1 Tafy1 25.043=:=

Myfed1 My1 Tafy1 Hbloc1 hcz1+( )⋅+ 75.128=:=

Txfed1 Tafx1 50.085=:=

Hd1 Tyfed12Txfed1

2+

55.997=:=

Calculul presiunii active la talpa fundatiei•

pef

Vd

A1

=

pcr

Rd

A1

=

≤ A1 B1 L1⋅=

eB1

Myfed1

Vd1

0.104=:= B1 Bs1 2eB1− 1.292=:=

eL1

Mxfed1

Vd1

0.246=:= L1 Ls1 2eL1− 1.507=:=

eB1

Bs1

6<

A1 B1 L1⋅ 1.948=:=

eL1

Ls1

6<eB1

Bs1

2eL1

Ls1

2

+ 0.02=

E "VERIFICA"eB1

Bs1

2eL1

Ls1

2

+

1

9<if

"NU VERIFICA" otherwise

:=E "VERIFICA"=

pef1

Vd1

A1

370.899=:=

Calculul capacitatii portante•

-- se face in conditii drenate

αααα 0:= inclinarea bazei fundatiei

c1

ck1

γγγγc36.48=:=

ϕϕϕϕ1 9.56deg:=

Nq 2.374:=

Nc 8.155:=

Nγγγγ 0.463:=

mB

2B1

L1

+

1B1

L1

+

1.538=:=bq 1 αααα tan ϕϕϕϕ1( )2⋅−

1=:=

bγγγγ bq 1=:=

bc bq

1 bq−( )Nc tan ϕϕϕϕ1( )⋅( )

− 1=:=mL

2L1

B1

+

1L1

B1

+

1.462=:=

sq 1B1

L1

sin ϕϕϕϕ1( )⋅+ 1.142=:=

θθθθ 63.43deg:=

m mL cos θθθθ( )2⋅ mB sin θθθθ( )

2⋅+ 1.523=:=sγγγγ 1 0.3

B1

L1

⋅− 0.743=:=

sc

sq Nq⋅ 1−( )Nq 1−( )

1.246=:=Hd1 55.997=

iq 1Hd1

Vd1 A1 c1⋅1

tan ϕϕϕϕ1( )⋅+

−

m

0.895=:=

iγγγγ 1Hd1

Vd1 A1 c1⋅1

tan ϕϕϕϕ1( )⋅+

−

m 1+

0.831=:=

ic iq

1 iq−( )Nc tan ϕϕϕϕ1( )⋅( )

− 0.882=:=

q1 Df. γγγγk1⋅ 21.84=:=

pacc

Rd

A1

=

pacc1 c1 Nc⋅ bc⋅ sc⋅ ic⋅ q1 Nq⋅ bq⋅ sq⋅ iq⋅+ 0.5 γγγγk1⋅ B1⋅ Nγγγγ⋅ bγγγγ⋅ sγγγγ⋅ iγγγγ⋅+ 383.109=:=

pacc1 383.109=

pef1 370.899=

V "VERIFICA" pacc1 pef1≥if


:=V "VERIFICA"=

VERIFICAREA LA TASARE

( SLEN )

conditia SED SCD≤

SEDtasarea pamantului sub fundatie

SCDtasarea admisibila

Calculul luiSED folosind metoda insumarii pe straturi elementare

-- se folosesc valorile caracteristice γγγγG 1:= γγγγQ 1:=

Vd Ned Npd+ Gfd+=

Npk1 γγγγG hgr bgr⋅ Tr⋅ γγγγbet⋅ Hz bz⋅ Tr⋅ γγγγBCA⋅+( )⋅ 58.5=:=

Gfk1 γγγγG Bs1 Ls1⋅ Htot1⋅ γγγγmed⋅( )⋅ 105.6=:=

NSk1 AafS1 pk qk+( )⋅ 357=:=

Vk1 NSk1 Npk1+ Gfk1+ 521.1=:=

pef.1

Vk1

Bs1 Ls1⋅173.7=:=

pneta1 pef.1 γγγγk1 Df.⋅− 151.86=:=

straturile elementare :•

h1. 0.4 Bs1⋅ 0.6=:=

alegem h1 0.40m:=h1.. 1.0:=

h1 min h1. h1.., ( )=

εεεεσσσσ

ΕΕΕΕ= S

σσσσ

ΕΕΕΕH⋅= σσσσz 0.2 σσσσgz⋅≤

σσσσz αααα0 pneta1⋅= σσσσz efortul− transmis de structura in pamant

σσσσgz γγγγ Df⋅ + ΣΣΣΣhi γγγγi⋅+= σσσσgz presiunea− geologica

SED ββββ 100⋅ ΣΣΣΣ⋅ σσσσzi.med

hi

ΕΕΕΕ⋅

⋅= ββββ 0.8:=

din calcule rezulta : SED1 0.325cm:=

SCD1 5.0cm:= pentru pamanturi necoezive

S1 "VERIFICA" SED1 SCD1≤if


:=S1 "VERIFICA"=

?2= 19

?1= 18.2

Df= 1.2

Df*?1= 21.84

E1= 12000

E2= 14000

Sed1= 0.00325

Date generale

TASAREA S1

m m - - - kPa kPa kPa kPa kPa

1 0.4 0.4 0.267 1.333 0.914 138.8 145.33 29.12 25.48 5.096

2 0.4 0.8 0.533 1.333 0.708 107.5169 123.16 36.4 32.76 6.552

3 0.4 1.2 0.8 1.333 0.477 72.43722 89.977 43.68 40.04 8.008

4 0.4 1.6 1.067 1.333 0.359 54.51774 63.477 50.96 47.32 9.464

5 0.4 2 1.333 1.333 0.261 39.63546 47.077 58.24 54.6 10.92

6 0.3 2.3 1.533 1.333 0.211 32.04246 35.839 63.7 60.97 12.194

7 0.4 2.7 1.8 1.333 0.167 25.36062 28.702 71.3 67.5 13.5

8 0.4 3.1 2.067 1.333 0.132 20.04552 22.703 78.9 75.1 15.02

9 0.4 3.5 2.333 1.333 0.114 17.31204 18.679 86.5 82.7 16.54

10 0.4 3.9 2.6 1.333 0.095 14.4267 15.869 94.1 90.3 18.06

0.2*τgz med Nr strat

hi z z/B L/B α0 sup/inf τz sup/inf τz med τgz sup/inf τgz med

ARMAREA FUNDATIEI S1

Ned.c1 NS1 Npd1+ lcs1 bcs1⋅ hcz1⋅ γγγγbet⋅( ) γγγγGn⋅+ 588.735=:=

Mxed.c1 Mx1 Tafx1 hcz1⋅+ Npd1 e1⋅+ 142.837=:=

eL.

Mxed.c1

Ned.c1

0.243=:= pc1.2

Ned.c1

lc bc⋅1 6−

eL

lc

⋅+

⋅=

pc1

Ned.c1

lcs1 bcs1⋅1 6

eL.

lcs1

⋅+

⋅ 1.554 103

×=:=

pc2

Ned.c1

lcs1 bcs1⋅1 6

eL.

lcs1

⋅−

⋅ 216.335−=:=

pc2 0< admitem pc2=0

elcs1

60.183=:= c1

lcs1

2e− 0.367=:=

p12

3

Ned.c1

bcs1 c1⋅⋅ 1.338 10

3×=:= p0.1 975:=

Armatura As3•

cnom 0.05:=x1 1.345:=

Ta1 pc2

x1

2⋅ bcs1⋅ 116.388−=:=

alegem PC 52 cu : fyd345

1.15300=:=

armam constructiv cu 5 ϕϕϕϕ 10 ladistanta de 25 cm

A1.S3

Ta1

fyd

0.388=:=

A1.S3ef 3.14 104−

×:=ρρρρ3

A1.S3ef

bcs1 hcz1 cnom−( )⋅1.57 10

3−×=:=

Armatura As1•lcy1

lcs1 ls−( )2

0.3=:=

Mx.x1 bcs1 p0.1 lcy1⋅lcy1

2⋅ p1 p0.1−( )

lcy1

2⋅

2

3⋅ lcy1⋅+

⋅ 43.813=:=

cnom 0.05= ϕϕϕϕsl 0.014:= avem clasa de beton C12/15

fcd12

1.58=:=

dx hcz1 cnom−ϕϕϕϕsl

2− 0.243=:=

μμμμx μμμμlim<μμμμx

Mx.x1

bcs1 dx2⋅ fcd⋅ 1000⋅

0.116=:= μμμμlim 0.403:=

fyk 345MPa:= fctm 1.6MPa:=ωωωω 1 1 2μμμμx−− 0.124=:=

Asmin 0.26fctm

fyk

⋅ bcs1⋅ 104

⋅ dx⋅ 2.344=:=Asx ωωωω bcs1⋅ 10

4⋅ dx⋅

fcd

fyd

⋅ 6.406=:=

alegem ϕϕϕϕ 5 14 cu : As1ef 7.70cm2:=

distanta dintre bare : d = 17.5 cmA1.S1 7.70 10

4−⋅:=

ρρρρl

A1.S1

bcs1 dx⋅3.961 10

3−×=:= ρρρρmin 0.075% 7.5 10

4−×=:=

ρρρρ1 ρρρρmin>

Armatura As2•

pmed1

p1 0+( )2

669.017=:= lcx1

bcs1 bs−( )2

0.2=:=

My.y1 lcs1 pmed1 lcx1⋅lcx1

2⋅

⋅ 14.718=:=

ϕϕϕϕs2 0.010:= cnom 0.05=

dy hcz1 cnom−ϕϕϕϕs2

2− ϕϕϕϕsl− 0.231=:=

μμμμy

My.y1

lcs1 dy2⋅ fcd⋅ 10

3⋅

0.031=:= μμμμlim. 0.403:= μμμμy μμμμlim<

ωωωω 1 1 2μμμμy−− 0.032=:=Asmin. 0.26

fctm

fyk

⋅ lcs1⋅ dy⋅ 3.064 104−

×=:=

Asy ωωωω lcs1⋅ dy⋅fcd

fyd

⋅ 2.158 104−

×=:=


distanta dintre bare : d = 25 cm

A1.S2 3.92 104−

⋅:=

ρρρρ2

A1.S2

lcs1 dy⋅1.543 10

3−×=:= ρρρρmin. 0.075% 7.5 10

4−×=:=

ρρρρ2 ρρρρmin<

Lungimi de ancoraj + lungimea ciocurilor•

As1

Lcioc1 15 ϕϕϕϕsl⋅ 0.21=:=

As2

Lcioc2 15 ϕϕϕϕs2⋅ 0.15=:=

As3

αααα1 1:= αααα3 1:= αααα5 1:=lbrqd 94:=

αααα2 0.7:= αααα4 0.7:=

la αααα1 αααα2⋅ αααα3⋅ αααα4⋅ αααα5⋅ lbrqd⋅ 46.06=:=

Lanc3. la 25+ 71.06=:=

Lanc3 75cm:=

ϕϕϕϕ Armatura din stalp 8 16

lbrqd. 105cm:=

la. αααα1 αααα2⋅ αααα3⋅ αααα4⋅ αααα5⋅ lbrqd.⋅ 0.514m=:=

LancS1. la. 250mm+ 0.764m=:=

LancS1 80cm:=

Stalpii S2+S3•

1.) PREDIMENSIONARE S2

Stabilirea Df :•

Df Hing 10...20( )cm+≥

Df2 1.1:=

Stabilirea lui L2 , B2 ,H2 :•

peff pacc≤ AafS2 6 2.5⋅ 15=:=

pconv2 408:= obtinut prin interpolare

pacc2 pconv2 408=:=pd. 20.79:=

qd. 12.6:=

NS2 AafS2 pd. qd.+( )⋅ 500.85=:=L2

B2

ls

bs

= r=

Gf2 0.25 NS2⋅ 125.213=:=ls

bs

1.25=VdpS2 NS2 Gf2+ 626.063=:=

B2.

VdpS2

1.25 pconv2⋅1.108=:= L2. 1.25 B2.⋅ 1.385=:=

alegem : B2 1.25:=

L2 1.75:=

pef2

VdpS2

B2 L2⋅286.2=:=

luam clasa de beton C12/15 si rezulta ca H/L = min 0.29

H2. 0.29 L2⋅ 0.507=:=

H2 0.75:=

1'.) PREDIMENSIONARE S3

Stabilirea Df :•

Df Hing 10...20( )cm+≥

Df3 1.1:=

Stabilirea lui L3 , B3 ,H3 :•

peff pacc≤ AafS3 6 5⋅ 30=:=

pconv3 408:= obtinut prin interpolare

pacc3 pconv3 408=:=pd.. 20.79:=

qd3.. 12.6:=

L3

B3

ls

bs

= r=NS3 AafS3 pd. qd.+( )⋅ 1.002 10

3×=:=

Gf3 0.25 NS3⋅ 250.425=:= ls

bs

1.25=

VdpS3 NS3 Gf3+ 1.252 103

×=:=

B3.

VdpS3

1.25 pconv3⋅1.567=:= L3. 1.25 B3.⋅ 1.959=:=

alegem : B3 1.60:=

L3 2.40:=

pef3

VdpS3

B3 L3⋅326.074=:=

luam clasa de beton C12/15 si rezulta ca H/L = min 0.30

H3. 0.30 L3⋅ 0.72=:=

alegem : H3 0.75:= !!! din acest motiv modificam si inaltimea fundatiei de sub stalpul S2 , rezultand

H2 = 0.7 m

2.) VERIFICARE LA CAPACITATE PORTANTA S2+S3

consideram ca presiunea se transmite numai

pe talpa fundatiei

Kgr

Kst

10≤

Ed1efectul actiunilor la talpa fundatiei din axul 1,

de forma unei forte verticale 1). Ed1

A1.2

Rd1

A1.2

≤

Ed2efectul actiunilor la talpa fundatiei din axul 2,

de forma unei forte verticale

A1.2aria efectiva la talpa fundatiei din axul 12).

Ed2

A1.3

Rd2

A1.3

≤

A1.3aria efectiva la talpa fundatiei din axul 2

Eforturile sectionale :

NS2 500.85=

Mx2 0.2 NS2⋅ 100.17=:= Tafx2 0.1 NS2⋅ 50.085=:=

My2 0.1 NS2⋅ 50.085=:= Tafy2 0.05 NS2⋅ 25.043=:=

NS3 1.002 103

×=

Mx3 0.2 NS3⋅ 200.34=:= Tafx3 0.1 NS3⋅ 100.17=:=

My3 0.1 NS3⋅ 100.17=:= Tafy3 0.05 NS3⋅ 50.085=:=

Gfd2. L2 B2⋅ H2⋅ γγγγbet⋅ 41.016=:=

Gfd3. L3 B3⋅ H3⋅ γγγγbet⋅ 72=:=

Grinda de fundare :•bs

20.2=

lgr. 5 B2

B3

2+

− 0.2+ 3.15=:=

hgr

bgr

1.5.....3= bgr = minim latimea stalpului ls. 0.5:=

bgr. 0.50:=

hgr. 0.75:= luam inaltimea grinzii egala cu inaltimea fundatiei

lgr. 3.15:=

Ggr bgr. hgr.⋅ lgr.⋅ γγγγbet⋅ 29.531=:=

Gfd2 Gfd2.

Ggr

2+ 55.781=:=

Gfd3 Gfd3.

Ggr

2+ 86.766=:=

Myfed2 My2 Tafy2 H2⋅+ 68.867=:=

Myfed3 My3 Tafy3 H3⋅+ 137.734=:=

ΣΣΣΣM1 0= obtinem E2

ΣΣΣΣM2 0= obtinem E1

ΣΣΣΣM1 0=

Ed2

Myfed3 NS3 4.575⋅+ Gfd3 4.575⋅+ Myfed2− NS2 0.425⋅−( )4.575

1.057 103

×=:=

ΣΣΣΣM2 0=

Ed1

Myfed2 NS2 5⋅+ Gfd2 4.575⋅+ Myfed3−( )4.575

588.105=:=

VERIFICARE : NS2 NS3+ Gfd2+ Gfd3+ Ed1− Ed2− 0=

Axul 1 :•

Mxfed2 Mx2 Tafx2 H2⋅+ 137.734=:=

eL2

Mxfed2

Ed1

0.234=:=

B1.2 B2 1.25=:= L1.2 L2 2eL2− 1.282=:=

A1.2 B1.2 L1.2⋅ 1.602=:=

Axul 2 :•

Mxfed3 Mx3 Tafx3 H3⋅+ 275.468=:=

eL3

Mxfed3

Ed2

0.261=:=

B1.3 B3 1.6=:= L1.3 L3 2eL3− 1.879=:=

A1.3 B1.3 L1.3⋅ 3.006=:=

Hd2 Tafx22Tafy2

2+ 55.997=:=

Hd3 Tafx32Tafy3

2+ 111.993=:=

presiunile

efective :pef2.

Ed1

A1.2

367.106=:= pef3.

Ed2

A1.3

351.623=:=

Calculul capacitatii portante S2•


ϕϕϕϕ1 9.56 deg⋅=αααα 0:= inclinarea bazei fundatiei

c1 36.48:=

Nq2 Nq 2.374=:=

Nc2 Nc 8.155=:=

Nγγγγ2 Nγγγγ 0.463=:=

mB2

2B1.2

L1.2

+

1B1.2

L1.2

+

1.506=:=bq2 1 αααα tan ϕϕϕϕ1( )2⋅−

1=:=

bγγγγ2 bq 1=:=

bc2 bq


− 1=:=

mL2

2L1.2

B1.2

+

1L1.2

B1.2

+

1.494=:=

sq2 1B1.2

L1.2

sin ϕϕϕϕ1( )⋅+ 1.162=:=

θθθθ 63.43deg:=

m mL2 cos θθθθ( )2⋅ mB2 sin θθθθ( )

2⋅+ 1.504=:=sγγγγ2 1 0.3

B1.2

L1.2

⋅− 0.707=:=

sc2

sq2 Nq⋅ 1−( )Nq 1−( )

1.28=:=

iq2 0.91:=

iγγγγ2 0.854:=

ic2 0.84:=

q2 20.02:=

pacc

Rd

A1

=

pacc2 c1 Nc⋅ bc⋅ sc⋅ ic⋅ q2 Nq⋅ bq⋅ sq⋅ iq⋅+ 0.5 γγγγk1⋅ B1.2⋅ Nγγγγ⋅ bγγγγ⋅ sγγγγ⋅ iγγγγ⋅+ 378.584=:=

pacc2 378.584=

pef2. 367.106=

V2 "VERIFICA" pacc2 pef2.≥if


:=V2 "VERIFICA"=

Calculul capacitatii portante S3•


ϕϕϕϕ1 9.56 deg⋅=αααα 0:= inclinarea bazei fundatiei

c1 36.48=

Nq3 Nq 2.374=:=

Nc3 Nc 8.155=:=

Nγγγγ3 Nγγγγ 0.463=:=

mB3

2B1.3

L1.3

+

1B1.3

L1.3

+

1.54=:=bq3 1 αααα tan ϕϕϕϕ1( )2⋅−

1=:=

bγγγγ3 bq 1=:=

bc3 bq


− 1=:=

mL3

2L1.3

B1.3

+

1L1.3

B1.3

+

1.46=:=

sq3 1B1.3

L1.3

sin ϕϕϕϕ1( )⋅+ 1.141=:=

θθθθ 63.43deg:=

sγγγγ3 1 0.3B1.3

L1.3

⋅− 0.745=:=

m mL3 cos θθθθ( )2⋅ mB3 sin θθθθ( )

2⋅+ 1.524=:=

sc3

sq3 Nq⋅ 1−( )Nq 1−( )

1.244=:=

iq3 0.902:=

iγγγγ3 0.843:=

ic3 0.831:= pacc

Rd

A1

=

q3 20.02:=

pacc3. c1 Nc3⋅ bc3⋅ sc3⋅ ic3⋅ q3 Nq3⋅ bq3⋅ sq3⋅ iq3⋅+ 0.5 γγγγk1⋅ B1.3⋅ Nγγγγ3⋅ bγγγγ3⋅ sγγγγ3⋅ iγγγγ3⋅+:=

pacc3. 360.796=

pef3. 351.623=

V3 "VERIFICA" pacc3. pef3.≥if


:=V3 "VERIFICA"=

ARMARE FUNDATIEI S2

p1.2

Ed1

L2 B2⋅1 6−

eL2

L2

+

⋅= eL2

L2

6≤

eL2.

L2

60.292=:=

p1.2

Ed1

L2 B2⋅1 6

eL2.

L2

+

⋅ 537.696=:=

p2.2

Ed1

L2 B2⋅1 6

eL2.

L2

−

⋅ 0=:= p02 350.51:=

Armatura As1•

lcy2 0.6:=cnom 0.05= ϕϕϕϕs2 0.016:=

lcx2 0.425:=

Mx.x2 B2 p02 lcy2⋅lcy2

2⋅ p1.2 p02−( )

lcy2

2⋅

2

3⋅ lcy2⋅+

⋅ 106.943=:=

fcd12

1.58=:=

dx2 H2 cnom−ϕϕϕϕs2

2− 0.692=:=

μμμμx μμμμlim<μμμμx2

Mx.x2

B2 dx22⋅ 10

3⋅ fcd⋅

0.022=:= μμμμlim 0.403:=

ωωωω2 1 1 2μμμμx2−− 0.023=:=Asmin 0.26

fctm

fyk

⋅ B2⋅ dx2⋅ 104

⋅ 10.43=:=

Asx2 ωωωω2 B2⋅ dx2⋅ 104

⋅fcd

fyd

⋅ 5.21=:=

alegem ϕϕϕϕ 6 16 cu : d = 23 cm

As1ef2 12.06cm2:=

A1.S2 12.06 104−

⋅:=

ρρρρ2

A1.S2

B2 dx⋅3.97 10

3−×=:= ρρρρmin 0.075% 7.5 10

4−×=:=

ρρρρ2 ρρρρmin>

Armatura As2•

pmed2

p1.2 p2.2+( )2

268.848=:=

My.y2 L2 pmed2 lcx2⋅lcx2

2⋅

⋅ 42.491=:=

L2 1.75=

dy2 H2 cnom−ϕϕϕϕs2

2− ϕϕϕϕs2− 0.676=:=

μμμμy2

My.y2

L2 dy22⋅ 10

3⋅ fcd⋅

6.642 103−

×=:= μμμμlim 0.403:=

μμμμx μμμμlim<

ωωωω2 1 1 2μμμμy2−− 6.664 103−

×=:=

Asmin 0.26fctm

fyk

⋅ L2⋅ dy2⋅ 104

⋅ 14.265=:=Asy2 ωωωω2 L2⋅ dy2⋅ 10

4⋅

fcd

fyd

⋅ 2.102=:=

A2.S2 16.94 104−

×:= alegem ϕϕϕϕ 11 14 cu : d = 16.5 cm

As1ef 16.94cm2:=

ρρρρy2

A2.S2

L2 dy2⋅1.432 10

3−×=:= ρρρρmin. 7.5 10

4−×= ρρρρ2 ρρρρmin>

ARMARE FUNDATIEI S3

p1.2

Ed2

L2 B2⋅1 6−

eL2

L2

+

⋅= eL2

L2

6≤

eL3.

B3

60.267=:=

p1.3

Ed2

L3 B3⋅1 6

eL3.

L3

+

⋅ 458.764=:=

p2.3

Ed2

L3 B3⋅1 6

eL3.

L3

−

⋅ 91.753=:= p03 172.13:=

Armatura As1•

lcy3 0.95:=cnom 0.05= ϕϕϕϕs3 0.016:=

lcx3 0.6:=

Mx.x3 B3 p03 lcy3⋅lcy3

2⋅ p1.3 p03−( )

lcy3

2⋅

2

3⋅ lcy3⋅+

⋅ 262.244=:=

dx3 H3 cnom−ϕϕϕϕs3

2− 0.692=:=

μμμμx3

Mx.x3

B3 dx32⋅ 10

3⋅ fcd⋅

0.043=:= μμμμlim 0.403:=

μμμμx μμμμlim<

ωωωω3 1 1 2μμμμx3−− 0.044=:=

Asmin 0.26fctm

fyk

⋅ B3⋅ dx3⋅ 104

⋅ =:=Asx3 ωωωω3 B3⋅ dx3⋅ 10

4⋅

fcd

fyd

⋅ 12.915=:=

A1.S3 14.07 104−

×:= alegem ϕϕϕϕ 7 16 cu : d = 25 cm

As1ef 14.07cm2:=

ρρρρx3

A1.S3

B3 dx3⋅1.271 10

3−×=:= ρρρρmin. 7.5 10


Armatura As2•

pmed3

p1.3 p2.3+( )2

275.258=:=

My.y3 L3 pmed3 lcx3⋅lcx3

2⋅

⋅ 118.912=:=

dy3 H3 cnom−ϕϕϕϕs3

2− ϕϕϕϕs3− 0.676=:=

μμμμy3

My.y3

L3 dy32⋅ 10

3⋅ fcd⋅

0.014=:=μμμμlim 0.403:= μμμμy μμμμlim<

ωωωω3. 1 1 2μμμμy3−− 0.014=:=

Asmin 0.26fctm

fyk

⋅ L3⋅ dy3⋅ 104

⋅ 19.563=:=Asy3 ωωωω3. L3⋅ dy3⋅ 10

4⋅

fcd

fyd

⋅ 5.904=:=


distanta dintre bare : d = 23 cm

A2.S3 22.11 104−

×:=

ρρρρy3

A2.S3

L3 dy3⋅1.363 10

3−×=:= ρρρρmin. 7.5 10


ϕϕϕϕLUNGIMEA CIOCURILOR = 15

GRINDA DE ECHILIBRARE

Hgr 0.75:= Bgr 0.5:=

Vmax 500.85:= Mmax 281.37:=

armatura longitudinala :•

ϕϕϕϕsg 0.022:= cnom 0.05=

dg Hgr cnom−ϕϕϕϕsg

2− 0.689=:=

μμμμg

Mmax

Bgr dg2⋅ 10

3⋅ fcd⋅

0.148=:=μμμμlim 0.403:= μμμμy μμμμlim<

ωωωωg 1 1 2μμμμg−− 0.161=:=

Asmin 0.26fctm

fyk

⋅ Bgr⋅ dg⋅ 104

⋅ 4.154=:=Asg ωωωωg Bgr⋅ dg⋅ 10

4⋅

fcd

fyd

⋅ 14.806=:=

alegem ϕϕϕϕ 4 22 cu : As1efg 15.20cm2:=

Al.gr 15.20 104−

×:=

ρρρρg

Al.gr

Bgr dg⋅4.412 10

3−×=:= ρρρρmin. 7.5 10


armatura transversala :•

In fundatia S2 :

cRd.c 0.12:= ρρρρl ρρρρg 4.412 103−

×=:= fck 8:=

ηηηη 1:= k 1200

dg 103

⋅+ 1.539=:=

VRd.c. cRd.c ηηηη⋅ k⋅ 100 ρρρρl⋅ fck⋅( )1

3⋅ dg⋅ Bgr⋅ 10

6⋅ 9.686 10

4×=:=

VRd.c

VRd.c.

100096.856=:=

ctgθθθθ 1.75:= z 0.9 dg⋅ 0.62=:=

Vmax VRd.c> fywd fyd 300=:=

Asw

s

Vmax

z fywd⋅ ctgθθθθ⋅≥

impunem s 0.10:= Asw

Vmax 106

⋅ s⋅

z fywd⋅ ctgθθθθ⋅ 103

⋅153.846=:=

ϕϕϕϕalegem Etrier 2 10 la distanta de 10 cm cu : Asw. 1.57 cm2⋅:=

In camp si in fundatia S3

Ved 31.474:=

VRd.c 96.856=

Ved VRd.c< ϕϕϕϕalegem constructiv Etrier 10 la distanta de 25 cm

Mathcad - fundatii izolate

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