Breviar Calcul Acoperis 45 Grade
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Transcript of Breviar Calcul Acoperis 45 Grade
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