SITUATIE EXISTENTA

CARACTERISTICILE CONSTRUCTIEI Ionut CARP Locatie: Constanta

Invelitoare: Tigla ceramica

Panta: 2 pante

α : 45 °

1. Incarcarea din zapada Sk= μi∙Ce∙Ct∙sok → Sk= [kN/m²]

unde: μi= μi=

Ce= (expunere partiala) Ce=

Ct= Ct=

sok= valoarea caracteristica a incarcarii din zapada pe sol in amplasament sok= [kN/m²]

2. Incarcarea din vant → [kPa]

unde: presiunea de referinta a vantului [kN/m²]

factorul de expunere la inaltimea z deasupra terenului

coeficientul aerodinamic de presiune

F

F

H I

H J I G

45° G G

H I

F

[m] F

[m]

[m]

3. Incarcarea utila qu= [kN/m²]

4. Incarcarea permanenta

Material Incarcare Coef Incarc. De calc.

Tigla

Sipca

Hidroizol.

0.80

1.00

2.00

coeficient de expunere al amplasamentului

coeficient termic

1.00

0.045

0.001

1.35

1.35

1.35

0.1323

0.0608

0.0014

0.4

I. EVALUAREA INCARCARILOR

coeficient de forma pentru incarcarea din zapada pe acoperis

0.64

qref= 0.5

Ce(z)= 0.665

Cp= 0.7

W(Z)= qref∙ce(z)∙Cp W(z)=

qref=

Ce(z)=

Cp=

0.2328

0.098

0.1944

11.25

11.25

15.34

5.625

SITUATIE EXISTENTA

II. CALCULUL SIPCILOR

1.Evaluarea incarcarilor

a) Incarcarea permanenta aferenta unei sipci 50 x 50 [mm]

Ip= 0.098+0.045+0.001 → Ip= [kN/m²]

qp= Ip∙1.35∙c → qp= [kN/m²]

c= [m]

qp∙sinα → [kN/m²]

qp∙cosα → [kN/m²]

b) Incarcarea din vant aferenta unei sipci

qv= w∙1.05∙c → qv= [kN/m²]

0 [kN/m²]

qv= [kN/m²]

c) Incarcarea din zapada aferenta unei sipci

qz= Sk∙1.5∙c → qz= [kN/m²]

qz∙sinα → [kN/m²]

qz∙cosα → [kN/m²]

d) Incarcarea utila aferenta unei sipci

Iu= [daN]

qu= Iu∙1.5 → qu= [kN/m²]

qu∙sinα → [kN/m²]

qu∙cosα → [kN/m²]

2. Ipoteze de incarcarea) q1= qp+qz

[kN/m²]

[kN/m²]

b) q2= qp+(1/2)qz+qv

[kN/m²]

[kN/m²]

c') Forta uniform distribuita q3= qp

[kN/m²]

[kN/m²]

qpy=

qpx=

qpy=

0.0579

0.0357

0.0855

0.144

0.35

0.068

qpx=

qvx=

qvy= 0.0855

0.336

qzx=

qzy=

qzx=

qzy=

0.2859

0.1765

q1x= 0.3438

q1y= 0.2123

q2x=

100

1.5

qux=

quy=

qux=

quy=

1.2764

0.788

q2y=

0.2008

0.2095

q3x=

q3y=

0.0579

0.0357

SITUATIE EXISTENTA

c'') Forta concentrata q3= qu

[kN/m²]

[kN/m²]

3. Calculul eforturilor in sipci

d= distanta interax intre sipci d= [m]

a) → → [kN∙m]

→ [kN∙m]

b) → → [kN∙m]

→ [kN∙m]

c) → → [kN∙m]

→ [kN∙m]

4. Verificarea rezistentei la capacitate portanta

→ → < 1

a) [kN∙m]

[kN∙m]

b) [kN∙m]

[kN∙m]

Rincc∙Wx∙mT∙m1 → [kN∙m]

[kN∙m]

mui∙mdi∙ →

1

→ →

Ri= [N/mm²]

γi=

→ → [mm³]

q3x=

q3y=

1.2764

0.788

0.35

M1= M1x= M1x= 0.015

M1y= M1y= 0.0093

M2= M2x= M2x= 0.0088

M2y= M2y= 0.0092

M3= M3x=

M3y=

M3x=

M3y=

0.0558

0.0345

M3x= 0.0558

Mefx= Mxmax= M3x= 0.0558

Mefy= Mycores= M3y= 0.0345

0.3217

Wx= Wx= Wx= 20833

mdi= 0.9805

16.8

1.1

Rincc= 14.975

Mrx=

Rincc=

mui=

mdi= mdi=

Mefy= Mymax= M3y= 0.0345

Mefx= Mxcors=

Mrx= 0.2808

Mry= 0.2808

𝑞1 ∙ 𝑑²

8

𝑞1𝑥 ∙ 𝑑²

8

𝑞1𝑦 ∙ 𝑑²

8

𝑞2 ∙ 𝑑²

8

𝑞2𝑥 ∙ 𝑑²

8

𝑞2𝑦 ∙ 𝑑²

8

𝑞3 ∙ 𝑑²

8

𝑞3𝑥 ∙ 𝑑²

8

𝑞3𝑦 ∙ 𝑑²

8

𝑀𝑒𝑓𝑥

𝑀𝑟𝑥 +

𝑀𝑒𝑓𝑦

𝑀𝑟𝑦 ≤ 1

𝑅𝑖

γ𝑖

𝑞𝑝 ∙ 0.55 + 𝑞𝑢 ∙ 1

𝑞𝑝 + 𝑞𝑢

0.068 ∙ 0.55 + 1.5 ∙ 1

0.068 + 1.5

𝑏 ∙ ℎ²

50 ∙ 50²

6

0.0558

0.2808 +

0.0345

0.2808 ≤ 1

SITUATIE EXISTENTA

1

5. Verificarea de rezistenta la incovoiere staticafmaxf ≤fadm → < se verifica la incovoiere statica

→ [mm]

a) Sageata din incarcarea permanenta aferenta unei sipci

fpx= fpstx(1+kdef) → fpx= [mm]

fpy= fpsty(1+kdef) → fpy= [mm]

→ [mm]

→ [mm]

Ix= Iy= → Ix=

E=

b) Sageata din incarcarea data de vant aferenta unei sipci

0

fvx= 0

fvy= fvsty(1+kdef) → fvy= [mm]

→ [mm]

c) Sageata din incarcarea data de zapada aferenta unei sipci

fzx= fzstx(1+kdef) → fzx= [mm]

fzy= fzsty(1+kdef) → fzy= [mm]

→ [mm]

→ [mm]

d) Sageata din incarcarea utila aferenta unei sipci

0

fux= fustx(1+kdef) → fux= [mm]

fuy= fusty(1+kdef) → fuy= [mm]

mT= 0.9

fadm= fadm= fadm= 0.3333

kdef= 0.8

m1=

kdef=

fvsty= fvst

y= 0.0116

0.0116

0.0038

0.011

0.0068

11000

fpstx=

fpsty=

fpstx=

fpsty=

520833 [mm⁴]

0.0061

0.0004

0.0002

0.0406 0.3333

0.0271

0.0167

0.0406

0.0251

kdef=

kdef= 0.5

fzstx= fzst

x=

fzsty= fzst

y=

6

6

𝐿

150

50

150

5 ∙ 𝑞𝑝𝑦 ∙ 𝑑4

384 ∙ 𝐸 ∙ 𝐼𝑦

𝑏 ∙ ℎ³

12

5 ∙ 𝑞𝑣𝑦 ∙ 𝑑4

384 ∙ 𝐸 ∙ 𝐼𝑦

5 ∙ 𝑞𝑧𝑥 ∙ 𝑑4

384 ∙ 𝐸 ∙ 𝐼𝑦

5 ∙ 𝑞𝑧𝑦 ∙ 𝑑4

384 ∙ 𝐸 ∙ 𝐼𝑦

1 𝑞 𝑑3

SITUATIE EXISTENTA

→ [mm]

→ [mm]

III. CALCULUL CAPRIORILOR

1.Evaluarea incarcarilor

a) Incarcarea permanenta aferenta unui caprior x [mm]

Ip= Ips+Ipc → Ip= [kN/m]

G= 0.1∙0.15∙2∙6 → G= [kN/m]

Ip∙1.35∙d∙cosα → [kN/m]

d= [m]

b) Incarcarea din vant aferenta unui caprior

Qv= w∙1.05∙d → Qv= [kN/m²]

c) Incarcarea din zapada aferenta unui caprior

Qz= Sk∙1.5∙d∙cosα → Qz= [kN/m]

d) Incarcarea utila aferenta unui caprior

Qu= Iu∙1.5∙cosα → Qu= [kN/m]

2. Ipoteze de incarcarea) Q1=Qp+Qz

[kN/m²]

b) Q2= Qp+(1/2)Qz+Qv

[kN/m²]

c') Forta uniform distribuita Q3= Qp

[kN/m]

c'') Forta concentrata Q3= Qu

[kN/m]

3. Calculul eforturilor in capriori

l= [m]

a) → → [kN∙m]

b) → → [kN∙m]

0.18

0.324

fustx= fust

x=

fusty= fust

y=

0.0004

0.0002

Q3= 0.1137

Q1= 0.3632

Q2= 0.3594

0.2495

0.788

Qp= Qp=

0.4948

0.1137

0.1209

3.97

0.7156

M2= M2= M2= 0.708

Q3= 0.788

M1= M1=

100 150

M1=

1 ∙ 𝑞𝑢𝑥 ∙ 𝑑3

48 ∙ 𝐸 ∙ 𝐼𝑦

1 ∙ 𝑞𝑢𝑦 ∙ 𝑑3

48 ∙ 𝐸 ∙ 𝐼𝑦

𝑄1 ∙ 𝑙²

8

0.4301 ∙ 3.5²

8

𝑄2 ∙ 𝑙²

8

0.3947 ∙ 3.5²

8

SITUATIE EXISTENTA

c) + → + →

4. Verificarea rezistentei la capacitate portanta

≤ Me → < se verifica la capacitate portanta

=max(M1;M2;M3) → Mmaxcap= max(0.6586;0.6044;2.054) → Mmaxcap= [kN∙m]

Rincc∙W∙mT∙m1 → [kN∙m]

mui∙mdi∙ →

1

W= → W= [mm²]

5. Verificarea la moment incovoietor

a) Sageata din incarcarea permanenta aferenta unui caprior

fpc= fpci(1+kdef) → fpx= [mm]

→ [mm]

E=

I=

b) Sageata din incarcarea data de vant aferenta unui caprior

0

fvc= fvci(1+kdef) → fvx= [mm]

→ [mm]

c) Sageata din incarcarea data de zapada aferenta unui caprior

fzc= fzci(1+kdef) → fzx= [mm]

→ [mm]

d) Sageata din incarcarea utila aferenta unui caprior

0

fuc= fuci(1+kdef) → fux= [mm]

→ [mm]

Calculul sagetii pe ipoteze de incarcarefmaxf ≤fadm → < se verifica sageata

0.0056

fvci= fvc

i= 0.0056

kdef=

0.024

fzci= fzc

i= 0.0166

0.45

0.9805mdi=

kdef= 0.8

fpci= 0.0059

4.219E+09

0.0106

kdef=

kdef=

0.0366

fuci= fuc

i= 0.0366

0 19.85

fpci=

11000

Me=

Rincc=

mui=

Rincc= 14.975

Mmaxcap

M3= M3=

Mmaxcap

375000

Me= 5.0539

2.6274 5.0539

2.6274

M3= 2.6274

5 ∙ 𝑄𝑝 ∙ 𝑙4

384 ∙ 𝐸 ∙ 𝐼𝑦

8

8

𝑞3 ∙ 𝑙²

8

𝑄3 ∙ 𝑙

4

1.27 ∙ 3.5²

8

0.1149 ∙ 3.5

4

𝑅𝑖

γ𝑖

𝑏 ∙ ℎ²

6

5 ∙ 𝑄𝑣 ∙ 𝑙4

384 ∙ 𝐸 ∙ 𝐼𝑦

5 ∙ 𝑄𝑧 ∙ 𝑙4

384 ∙ 𝐸 ∙ 𝐼𝑦

5 ∙ 𝑄𝑢 ∙ 𝑙4

384 ∙ 𝐸 ∙ 𝐼𝑦

SITUATIE EXISTENTA

→ [mm]

max(f1;f2;f3)

IV. CALCULUL PANEI CENTRALE

1. Evaluarea incarcarilor

a) Incarcarea permanenta aferenta unei pane x [mm]

→ [kN/m]

v·γ → [kN/m]

0

[kN/m]

dp= [m]

b) Incarcarea din vant aferenta unei pane

→ [kN/m]

→ [kN/m]

qvp·cosα → [kN/m]

c) Incarcarea din zapada aferenta unei pane

qzp=Sk∙1.05∙dp → qzp= [kN/m]

0

[kN/m]

d) Incarcarea utila aferenta unei pane

Iu∙1.5 → [kN/m]

0 [kN/m]

[kN/m]

2. Ipoteze de incarcarea) q1=qp+qz

qpx+qzx → 0 [kN/m²]

qpy+qzy → [kN/m²]

b) q2= qp+(1/2)qz+qv

qpx+(1/2)qzx+qvx → [kN/m²]

qpy+(1/2)qzy+qvy → [kN/m²]

c') Forta uniform distribuita q3= qp

qpx= 0 [kN/m²]

fadm= fadm= fadm= 19.85

fmaxf=

qpy= qpp=

1.544

1.544

qvp= qvp= 1.8692

qvpx= qvp·sinα qvpx= 1.5905

150 200

qpana=

qpp= qpp=

qpana= 0.18

qup= qup= 1.5

qvpy= qvpy= 0.9819

2.8125

1.89

qzpx=

qzpy= 1.89qzp=

3.4709

q3x=

q1x= q1x=

q1y= q1y= 3.434

q2x= q2x=

qpx=

qupx=

qupy= qup= 1.5

1.5905

q2y= q2y=

𝐼𝑝𝑐 ∙ 1.35

𝑐𝑜𝑠α∙ 𝑑𝑝 + 𝑞𝑝𝑎𝑛𝑎 ∙ 1.35

𝑊 ∙ 1.5

𝑐𝑜𝑠α∙ 𝑑𝑝

𝐿

200

3500

200

SITUATIE EXISTENTA

qpy= [kN/m²]

c'') Forta concentrata q3= qu

qux= 0 [kN/m²]

quy= [kN/m²]

3. Calculul eforturilor in pana centralalp= [m]

a) 0 [kN∙m]

→ [kN∙m]

b) → [kN∙m]

→ [kN∙m]

c) 0 [kN∙m]

+ → [kN∙m]

4. Verificarea rezistentei la capacitate portanta

a) [kN∙m]

[kN∙m]

b) [kN∙m]

[kN∙m]

Rincc∙Wx∙mT∙m1 → [kN∙m]

[kN∙m]

mui∙mdi∙ →

1

→

Ri= [N/mm²]

γi=

→ → [mm³]

→ → [mm³]

q3y=

q3y= 1.544

q3x=

1.522

1.717

M2x=

M2y=

M2x=

M2y=

0.7952

1.7355

M3x=

M3y= M3y=

1.5

2.00

M1x=

M1y= M1y=

Mefx= Mxmax= M2x= 0.7952

Mefy= Mycores= M2y= 1.7355

Mefy= Mymax= M2y= 1.7355

Mefx= Mxcors= M2x= 0.7952

Mrx= Mrx=

mdi= 0.8406

16.8

1.1

Wx= Wx=

11.55

Mry= 8.666

Rincc= Rinc

c= 12.838

mui=

mdi=

Wx= 1.00E+06

Wy= Wy= Wy= 7.50E+05

𝑞1𝑦 ∙ 𝑙𝑝²

8

𝑞2𝑥 ∙ 𝑙𝑝²

8

𝑞2𝑦 ∙ 𝑙𝑝²

8

𝑞3𝑦 ∙ 𝑙𝑝²

8

𝑄3𝑦 ∙ 𝑙𝑝

4

𝑀𝑒𝑓𝑥

𝑀𝑟𝑥 +

𝑀𝑒𝑓𝑦

𝑀𝑟𝑦 ≤ 1

𝑅𝑖

γ𝑖

𝑞𝑝 ∙ 0.55 + 𝑞𝑣 + 𝑞𝑧/2

𝑞𝑝 + 𝑞𝑣 + 𝑞𝑧/2

𝑏 ∙ ℎ²

6

150 ∙ 200²

6

ℎ ∙ 𝑏²

6

200 ∙ 150²

6

SITUATIE EXISTENTA

1

5. Verificarea de rezistenta la incovoiere staticafmaxf ≤fadm → <

→ [mm]

a) Sageata din incarcarea permanenta aferenta unei pane

fpx= fpstx(1+kdef) → fpx= [mm]

fpy= fpsty(1+kdef) → fpy= [mm]

→ [mm]

→ [mm]

Ix= → Ix= [mm⁴] Iy= → Iy= [mm⁴]

E=

b) Sageata din incarcarea data de vant aferenta unei pane

0

fvx= fvstx(1+kdef) → fvx= [mm]

fvy= fvsty(1+kdef) → fvy= [mm]

→ [mm]

→ [mm]

c) Sageata din incarcarea data de zapada aferenta unei pane

fzx= fzstx(1+kdef) → fzx= [mm]

fzy= fzsty(1+kdef) → fzy= [mm]

→ [mm]

→ [mm]

d) Sageata din incarcarea utila aferenta unei pane

0

10

kdef= 0.8

mT= 0.9

m1=

0

0.9358

fpstx= fpst

x= 0

fpsty= fpst

y= 0.520

0.9358 10

fadm= fadm= fadm=

0

fzsty= fzst

y= 0.2386

kdef=

11000

kdef=

0.3306

fvsty= fvst

y= 0.33

kdef= 0.5

0

5.63E+07

fvstx= fvst

x= 0.30

0.3012

1.00E+08

0.358

fzstx= fzst

x=

se verifica la incovoiere statica

𝐿𝑝

200

2000

200

5 ∙ 𝑞𝑝𝑦 ∙ 𝑑4

384 ∙ 𝐸 ∙ 𝐼𝑦

𝑏 ∙ ℎ³

12

5 ∙ 𝑞𝑣𝑦 ∙ 𝑑4

384 ∙ 𝐸 ∙ 𝐼𝑦

5 ∙ 𝑞𝑧𝑥 ∙ 𝑑4

384 ∙ 𝐸 ∙ 𝐼𝑦

5 ∙ 𝑞𝑧𝑦 ∙ 𝑑4

384 ∙ 𝐸 ∙ 𝐼𝑦

5 ∙ 𝑞𝑝𝑥 ∙ 𝑑4

384 ∙ 𝐸 ∙ 𝐼𝑥

ℎ ∙ 𝑏³

12

5 ∙ 𝑞𝑣𝑥 ∙ 𝑑4

384 ∙ 𝐸 ∙ 𝐼𝑥

SITUATIE EXISTENTA

fux= fustx(1+kdef) → fux= [mm]

fuy= fusty(1+kdef) → fuy= [mm]

→ [mm]

fusty= → [mm]

Calculul sagetii pe ipoteze de incarcarefmaxf ≤fadm → < se verifica

→ [mm]

max(f1;f2;f3)

1. f1= → f1= [mm]

f1x= fpx+fzx → f1x= 0 [mm]

f1y= fpy+fzy → f1y= [mm]

2. f2= → f2= [mm]

f2x= fpx+fvx+fzx/2 → f2x= [mm]

f2y= fpy+fvy+fzy/2 → f2x= [mm]

3. f3= → f3= [mm]

f3x= fpx+fux → f3x= 0 [mm]

f3y= fpy+fuy → f3y= [mm]0.9359

0.9359

1.4764 10

fadm= fadm= fadm= 10

0

0.0002

fustx= fust

x= 0

fusty= 0.0002

fmaxf=

1.2937

1.2937

0.3012

1.4453

1.4764

1 ∙ 𝑞𝑢𝑥 ∙ 𝑑3

48 ∙ 𝐸 ∙ 𝐼𝑦

1 ∙ 𝑞𝑢𝑦 ∙ 𝑑3

48 ∙ 𝐸 ∙ 𝐼𝑦

𝐿

200

2000

200

(𝑓1𝑥2 + 𝑓1𝑦

2 )

(𝑓2𝑥2 + 𝑓2𝑦

2 )

(𝑓3𝑥2 + 𝑓3𝑦

2 )

SITUATIE EXISTENTA

V. CALCULUL POPILOR

1. Evaluarea incarcarilor

a) Incarcarea permanenta aferenta unui pop d= [m]

→ Npp= [kN/m]

v·γ → [kN/m]

Asect= → Asect= Acalc= 0.3*0.3 → Acalc=

t= [m]

hp= [m]

b) Incarcarea din vant aferenta unui pop

→ [kN/m]

c) Incarcarea din zapada aferenta unui pop

Nzp= Sk∙1.05∙dp∙t → Nzp= [kN/m]

d) Incarcarea utila aferenta unui pop

Iu∙1.5 → [kN/m]

2. Ipoteze de incarcare

a) N1=Np+Nz → [kN/m]

b) N2= Np+(1/2)Nz+Nv → [kN/m]

3. Verificarea la compresiune cu flambaj

Cr= Rc''∙Acalc∙mTc∙ϕ

max(N1;N2) [kN/m]

Nmax ≤ Cr

Nmax= Nmax= 27.94

1.5Nu= Nu=

N1= 27.94

N2= 25.463

Nvp= Nvp= 2.6384

3.835

5.625

[m²]

10.231

0.16

0.0201 [m²]

17.709

0.09

Npp=

qpana= qpana= 0.18

𝐼𝑝𝑐 ∙ 1.35

𝑐𝑜𝑠α∙ 𝑑𝑝 ∙ 𝑡 + 𝑞𝑝 ∙ 𝑡 ∙ 1.35 + 𝐴𝑠 ∙ ℎ𝑝 ∙ γ ∙ 1.35

𝑊 ∙ 1.5

𝑐𝑜𝑠α∙ 𝑑𝑝

π ∙ 𝑑²

4

SITUATIE EXISTENTA

→

→

ϕ= f(λ) → ϕ= → ϕ=

λ= → λ= > 75

If= [m]

i= 0.25∙dpop → i=

mTc=

Cr=

→ <

VI. CALCULUL TALPII

N < Qr → <

unde: N=

N= [kN/m]

Qr=

Qr= Ac∙Rc''∙mTc∙mr → Qr=

Ac=

Ac= → Ac= [m²]

coeficient de reazemmr=

mr= 1.6

182.31

27.94 182.31 TRUE

27.94

capacitatea portanta a elementelor din lemn masiv cu sectiunea simpla, solicitatala compresiune

perpendicular pe directia fibrelor

0.0192

aria de contact dintre cele doua elemente, aria popului la rezemarea pe talpa se considera ca popul se

imbina cu talpa cu cep cu dimensiunile 3x3 cm

Rc''= 6.5955

mTc= 0.9

0.9

102.57

Nmax ≤ Cr 27.94 102.57 TRUE

Verificarea la strivire a talpii se face cu relatia:

incarcarea verticala a popului

0.687

6595.5

Acalc= 0.09

5.00

0.04

125 TRUE

0.0192

Rc''= Rc''=

mdi= mdi=

𝑀𝑢𝑐 ∙ 𝑚𝑑𝑖 ∙ 𝑅𝑐

γ𝑐

𝑄𝑝 ∙ 0.55 + 𝑄𝑣 + 𝑄𝑧/2

𝑄𝑝 + 𝑄𝑣 + 𝑄𝑧/2

𝐼𝑓

𝑖

3 ∙ 100

λ²

π∙𝑑²

4 - 0.03·0.03