Armare grinda principala
1. Date generale:
N fyk N
fyk ≔ 355 ―― fyd ≔ ―― = 308.696 ―― fctm ≔ 2.6 MPa
mm 2 1.15 mm 2
fck T ≔ 10.5 m
N N
fck ≔ 25 ―― fcd ≔ ―― = 16.667 ――
mm 2 1.5 mm 2 L ≔ 7.5 m
hGP ≔ 1200 mm λ ≔ 0.8 hf ≔ 8 cm H≔4 m
bGP ≔ 600 mm μlim ≔ 0.372
2. Determinarea latimii efective in camp a grinzii principale:
b1 ≔ 0.5 ⋅ ⎛⎝T − bGP⎞⎠ = 4.95 m
bGP bGP
L0 ≔ T − ―― − ―― = 9.9 m
2 2
beff1.2 ≔ 0.2 ⋅ b1 + 0.1 ⋅ L0 = 1.98 m
beff ≔ min ⎛⎝beff1.2 ⋅ 2 + bGP , b1 ⋅ 2 + bGP , L0⎞⎠ = 4.56 m
beff.GP
beff.GP ≔ 4.6 m ――― >5=1
bGP
Med.camp ≔ 2226.32 kN ⋅ m
ϕsl ≔ 25 mm
cnom ≔ 35 mm ϕ
sl
d ≔ hGP − cnom − ― = 1.153 m
2
Med.camp
μ ≔ ―――― = 0.168
bGP ⋅ d 2 ⋅ fcd
η≔1
beff.GP hf ⎛ hf ⎞
μ < η ⋅ ――― ⋅ ―⋅ ⎜1 − 0.5 ⋅ ―⎟ = 1 Axa neutra se afla in placa
bGP d ⎝ d⎠
3.1 Armarea in camp:
Med.camp ≔ 2226.32 kN ⋅ m
beff.GP ≔ 4.5 m
d = 1.153 m
Med.camp
μ ≔ ――――― = 0.022 μ < μlim = 1
beff.GP ⋅ d 2 ⋅ fcd
ω ≔ 1 − ‾‾‾‾‾‾
1 − 2 ⋅ μ = 0.023
fcd
As.camp ≔ ω ⋅ beff.GP ⋅ d ⋅ ―― = 63.293 cm 2 Aleg 8 ϕ 32 cu Asl.camp ≔ 64.32 cm 2
fyd
3.2 Armarea reazamului marginal:
Med.r1 ≔ 701.61 kN ⋅ m
Med.r1
μ ≔ ―――― = 0.053 μ < μlim = 1
bGP ⋅ d 2 ⋅ fcd
fctm
As.min1 ≔ 0.26 ⋅ ―― ⋅ bGP ⋅ d = 0.001 m 2
ω ≔ 1 − ‾‾‾‾‾‾
1 − 2 ⋅ μ = 0.054 fyk
fcd
As.r1 ≔ ω ⋅ bGP ⋅ d ⋅ ―― = 20.271 cm 2 0.013 ⋅ bGP ⋅ d = 0.009 m 2
fyd
Aleg 2 ϕ 36 cu Asl.r1 ≔ 20.40 cm 2
3.3 Armarea reazamului central:
Med.r2 ≔ 2837.86 kN ⋅ m
ϕsl ≔ 36 mm
cnom ≔ 35 mm
28
d ≔ hGP − cnom − ― mm = 1.151 m
2
Med.r2
μ ≔ ―――― = 0.214 μ < μlim = 1
bGP ⋅ d 2 ⋅ fcd
ω ≔ 1 − ‾‾‾‾‾‾
1 − 2 ⋅ μ = 0.244
fcd
As.r2 ≔ ω ⋅ bGP ⋅ d ⋅ ―― = 90.967 cm 2 Aleg 9 ϕ 36 Asl.r2 ≔ 91.80 cm 2
fyd
fctm
As.min2 ≔ 0.26 ⋅ ―― ⋅ bGP ⋅ d = 13.151 cm 2 0.013 ⋅ bGP ⋅ d = 0.009 m 2
fyk
4 A l f t t i t
4. Armarea la forta taietoare:
4.1 Reazemul marginal:
Ved.A ≔ 1263.44 kN Asl.r1 ≔ 20.40 cm 2
0.18 N fyk N
CRdc ≔ ―――― = ⎛⎝1.2 ⋅ 10 5 ⎞⎠ Pa fywd ≔ ―― = 308.696 ――
1.5 mm 2
1.15 mm 2
k1 ≔ 0.15 fck ≔ 25
N
σcp ≔ 0 ――
mm 2
‾‾‾‾‾‾‾‾
200 mm
k≔1+ ―――= 1.417 k≤2=1
d
Asl.r1
ρl ≔ ――― = 0.003 ρl < 0.02 = 1
bGP ⋅ d
⎛ ―⎞
1
⎜ 3⎟
Vrd.c ≔ ⎝⎛⎝k1 ⋅ σcp⎞⎠ + CRdc ⋅ 1 ⋅ k ⋅ ⎛⎝100 ⋅ ρl ⋅ fck⎞⎠ ⎠ ⋅ bGP ⋅ d = 228.654 kN
⎛ fck ⎞
ν1 ≔ 0.6 ⋅ ⎜1 − ―― ⎟ = 0.525
⎝ 200 ⎠
⎛⎝1 ⋅ bGP ⋅ 0.9 ⋅ d ⋅ ν1 ⋅ fcd⎞⎠
VRd.max ≔ ―――――――= ⎛⎝1.875 ⋅ 10 3 ⎞⎠ kN
2.5 + 0.4
⎛ ⎛⎝⎛⎝Ved.A − Vrd.c⎞⎠ ⋅ 1.5⎞⎠ ⎞
ctθ1 ≔ 2.5 − ⎜――――――― ⎟ = 1.557
⎝ VRd.max − Vrd.c ⎠
ctθ1
s ≔ 0.9 ⋅ d ⋅ fywd ⋅ 226 ⋅ mm 2 ⋅ ――= 89.084 mm
Ved.A
Aleg etr.: ϕ12 /200:
⎛⎝226 ⋅ mm 2 ⎞⎠
VRd.s ≔ ――――⋅ 0.9 ⋅ d ⋅ fywd ⋅ ctθ1 = 562.761 kN
200 mm
VRd.s ≤ VRd.max = 1 ar ≔ 0.5 ⋅ 0.929 m ⋅ ctθ1 = 0.723 m
5.2 Reazemul central:
Ved.B ≔ 1665.24 kN Asl.r2 ≔ 91.80 cm 2
d 1 151
d = 1.151 m
0.18 N fyk N
CRdc ≔ ―――― = ⎛⎝1.2 ⋅ 10 5 ⎞⎠ Pa fywd ≔ ―― = 308.696 ――
1.5 mm 2
1.15 mm 2
k1 ≔ 0.15 fck ≔ 25
N
σcp ≔ 0 ――
mm 2
‾‾‾‾‾‾‾‾
200 mm
k≔1+ ―――= 1.417 k≤2=1
d
Asl.r2
ρl ≔ ――― = 0.013 ρl < 0.02 = 1
bGP ⋅ d
⎛ ―⎞
1
⎜ 3⎟
Vrd.c ≔ ⎝⎛⎝k1 ⋅ σcp⎞⎠ + CRdc ⋅ 1 ⋅ k ⋅ ⎛⎝100 ⋅ ρl ⋅ fck⎞⎠ ⎠ ⋅ bGP ⋅ d = 377.499 kN
⎛ fck ⎞
ν1 ≔ 0.6 ⋅ ⎜1 − ―― ⎟ = 0.525
⎝ 200 ⎠
⎛⎝1 ⋅ bGP ⋅ 0.9 ⋅ d ⋅ ν1 ⋅ fcd⎞⎠
VRd.max ≔ ―――――――= ⎛⎝1.875 ⋅ 10 3 ⎞⎠ kN
2.5 + 0.4
⎛ ⎛⎝⎛⎝Ved.B − Vrd.c⎞⎠ ⋅ 1.5⎞⎠ ⎞
ctθ2 ≔ 2.5 − ⎜――――――― ⎟ = 1.21
⎝ V Rd.max − V rd.c ⎠
ctθ2
s ≔ 0.9 ⋅ d ⋅ fywd ⋅ 226 ⋅ mm 2 ⋅ ――= 52.53 mm
Ved.B
Aleg etr.: ϕ12 /150:
⎛⎝226 ⋅ mm 2 ⎞⎠
VRd.s ≔ ――――⋅ 0.9 ⋅ d ⋅ fywd ⋅ ctθ2 = 583.169 kN
150 mm
VRd.s ≤ VRd.max = 1
ar1 ≔ 0.5 ⋅ 0.85 m ⋅ ctθ2 = 0.514 m
5. Lungimea de ancorare Marginal
α1 ≔ 1 α2 ≔ 1 α3 ≔ 1 α4 ≔ 1 α5 ≔ 0.7
d = 1.151 m
z ≔ 0.9 ⋅ d = 1.036 m
ctθ1 = 1.557 ctα ≔ 0
a1 ≔ 0.5 ⋅ z ⋅ ⎛⎝ctθ1 − ctα⎞⎠ = 0.807 m
Ved.A ⋅ a1
Fsd.1 ≔ ―――= 983.834 kN
z
Fsd.1 N
σsd.1 ≔ ――= 482.272 ――
Asl.r1 mm 2
Lungimea de ancorare la baza
ϕ1 ≔ 36 mm fbd ≔ 2.70 MPa
σsd.1
lbd.rqd1 ≔ 0.25 ⋅ ϕ1 ⋅ ―― = ⎛⎝1.608 ⋅ 10 3 ⎞⎠ mm
fbd
lb.min1 ≔ max ⎛⎝0.3 ⋅ lbd.rqd1 , 10 ⋅ ϕ1 , 100 mm⎞⎠ = 0.482 m
lbd.1 ≔ α1 ⋅ α2 ⋅ α3 ⋅ α4 ⋅ α5 ⋅ lbd.rqd1 = 1.125 m
lbd.1 ≥ lb.min1 = 1
Central
ϕ2 ≔ 36 mm
⎛ cnom − ϕ2 ⎞
α1 ≔ 1 α2 ≔ 1 − 0.15 ⋅ ⎜―――⎟ = 1.004
⎝ ϕ2 ⎠
k ≔ 0.1
12 ⋅ 226 − 0.25 ⋅ 1315.1
λ ≔ ―――――――= 1.812
1315.1
α3 ≔ 1 − k ⋅ λ = 0.819
α4 ≔ 0.7
α5 ≔ 1 − 0.04 ⋅ 0 = 1
d = 1.151 m
0 9 d 1 036
z ≔ 0.9 ⋅ d = 1.036 m
ctθ2 = 1.21 ctα ≔ 0
a2 ≔ 0.5 ⋅ z ⋅ ⎛⎝ctθ2 − ctα⎞⎠ = 0.627 m
Ved.B ⋅ a2
Fsd.2 ≔ ―――= ⎛⎝1.008 ⋅ 10 3 ⎞⎠ kN
z
Fsd.2 N
σsd.2 ≔ ――= 109.782 ――
Asl.r2 mm 2
Lungimea de ancorare la baza
ϕ2 ≔ 36 mm fbd ≔ 2.70 MPa
σsd.2
lbd.rqd2 ≔ 0.25 ⋅ ϕ2 ⋅ ―― = 365.942 mm
fbd
lb.min2 ≔ max ⎛⎝0.3 ⋅ lbd.rqd2 , 10 ⋅ ϕ2 , 100 mm⎞⎠ = 0.36 m
lbd.2 ≔ α1 ⋅ α2 ⋅ α3 ⋅ α4 ⋅ α5 ⋅ lbd.rqd2 lbd.2 ≔ 0.74 m
lbd.2 ≥ lb.min2 = 1
A t l
Armare stalp
Nmax ≔ 3192.37 kN Maf ≔ 701.612 kN ⋅ m
Mmax ≔ 2837.86 kN ⋅ m Naf ≔ 1565.14 kN
θy1 ≔ 0.071 rad
θz1 ≔ 0.027 rad
1. Determinarea coeficientului de zvelte:
bst ≔ 80 cm hst ≔ 80 cm
k1 ≔ 0.1 Nodul inferior incastrare perfecta
Nodul superior:
hst 3
I ≔ bst ⋅ ―― = ⎛⎝3.413 ⋅ 10 10⎞⎠ mm 4
12
E ≔ 18 GPa
θy1 ((E ⋅ I))
k2 ≔ ――⋅ ―― = 3.843
Mmax H
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
⎛ k1 ⎞ ⎛ k2 ⎞
l0 ≔ 0.5 ⋅ H ⋅ ⎜1 + ―――⎟ + ⎜1 + ―――⎟ = 3.508 m
⎝ 0.45 + k1 ⎠ ⎝ 0.45 + k2 ⎠
Nmax
n ≔ ―――― = 0.299
bst ⋅ hst ⋅ fcd
10.78
λlim ≔ ――= 19.705
‾‾
n
l0
λ ≔ ――― = 15.191
‾‾‾‾
bst 2
――
12
λ < λlim = 1 Deoarece coditia este verificata se pot neglija efectele de ordinul
2 dar se introduc excentricitati aditionale.
3 A l M iN
3. Armarea la M si N:
l0
e ≔ ―― = 0.877 cm
400
Nmax
Nmax ≔ 3192.37 kN Maf ≔ 701.612 kN ⋅ m ν ≔ ―――― = 0.299
bst ⋅ hst ⋅ fcd
Mmax ≔ 2837.86 kN ⋅ m Naf ≔ 1565.14 kN
Maf
Maf ≔ 701.612 kN ⋅ m + Nmax ⋅ e = 729.611 kN ⋅ m μ ≔ ―――― = 0.086
bst ⋅ hst 2 ⋅ fcd
Mmax ≔ 2837.86 kN ⋅ m + Naf ⋅ e = ⎛⎝2.852 ⋅ 10 3 ⎞⎠ kN ⋅ m
Mmax
Naf ≔ 1565.14 kN μ ≔ ―――― = 0.334
bst ⋅ hst 2 ⋅ fcd
Naf
ν ≔ ―――― = 0.147
bst ⋅ hst ⋅ fcd
ϕsl ≔ 25 mm
ϕsl
d1 ≔ cnom + ― + 8 mm = 55.5 mm
2
d1
―= 0.069
bst
Din figura 3.6 b) "Proiectearea structurilor din beton" Z. Kiss T. Onet pagina 60 se
determina coeficientul total de armare care pentru valorea μ sub 0.1 rezulta o armare
constructiva. In consecinda se va realiza armarea din combinatia de valori cu efort
axial maxim.
Nmax
As.min ≔ 0.1 ⋅ ――= 10.341 cm 2
fyd
As.min > 0.002 ⋅ bst 2 = 0
0.008 ⋅ bst 2 = 51.2 cm 2
Aleg: 9 ϕ28 : Asl ≔ 55.44 cm 2 Se alege aceasta arie de armatura pentru a
satisface distanta minima dintre bare si
procentul minim pentru clasa DCM.
Asl
――
2
ωtot ≔ ―――― = 0.08 Pe curba cu ωtot =0.09 si cu valoarea ν =0.224
fcd
bst ⋅ hst ⋅ ―― se determina valoarea μ =0.11
fyd
μ ≔ 0.11
Nrd ≔ bst ⋅ hst ⋅ fcd + fyd ⋅ Asl = ⎛⎝1.238 ⋅ 10 4 ⎞⎠ kN
Mrd ≔ bst ⋅ hst 2 ⋅ fcd ⋅ μ = 938.667 kN ⋅ m
3. Armarea la forta taietoare:
Ved ≔ 1665.24 kN
0.18 N fyk N
CRdc ≔ ―――― = ⎛⎝1.2 ⋅ 10 5 ⎞⎠ Pa fywd ≔ ―― = 308.696 ――
1.5 mm 2
1.15 mm 2
k1 ≔ 0.15 fck = 25
Nmax N
σcp ≔ ――― = 4.988 ―― η≔1
bst ⋅ hst mm 2
‾‾‾‾‾‾‾‾
200 mm
k≔1+ ―――= 1.5 k≤2=1
bst
Asl
ρl ≔ ――― = 0.009 ρl < 0.02 = 1
bst ⋅ hst
⎛ ―⎞
1
⎜ 3⎟
Vrd.c ≔ ⎝⎛⎝k1 ⋅ σcp⎞⎠ + CRdc ⋅ 1 ⋅ k ⋅ ⎛⎝100 ⋅ ρl ⋅ fck⎞⎠ ⎠ ⋅ bGP ⋅ d = 863.207 kN
⎛ fck ⎞
ν1 ≔ 0.6 ⋅ ⎜1 − ―― ⎟ = 0.525
⎝ 200 ⎠
⎛⎝1 ⋅ bGP ⋅ 0.9 ⋅ d ⋅ ν1 ⋅ fcd⎞⎠
VRd.max ≔ ―――――――= ⎛⎝1.875 ⋅ 10 3 ⎞⎠ kN
2.5 + 0.4
⎛ ⎛⎝⎛⎝Ved − Vrd.c⎞⎠ ⋅ 1.5⎞⎠ ⎞
ctθ ≔ 2.5 − ⎜―――――― ⎟ = 1.311
⎝ VRd.max − Vrd.c ⎠
ctθ
s ≔ 0.9 ⋅ d ⋅ fywd ⋅ 226 ⋅ mm 2 ⋅ ―― = 56.912 mm
Ved
Aleg etr.: ϕ12 /200:
⎛⎝226 ⋅ mm 2 ⎞⎠
VRd.s ≔ ――――⋅ 0.9 ⋅ d ⋅ fywd ⋅ ctθ = 473.861 kN
200 mm
VRd.s ≤ VRd.max = 1
⎛ ⎛⎝H − hGP⎞⎠ ⎞
lcr ≔ max ⎜hst , 450 mm , ―――― ⎟ = 80 cm Pentru clasa de ductilitate DCM
⎝ 6 ⎠
dbl ≔ 22 mm Diametrul barelor longitudinale
b0 ≔ hst − cnom ⋅ 2 − 8 mm ⋅ 2 = 71.4 cm Sectiunea utila a stalpului
⎛ b0 ⎞
scl.max ≔ min ⎜8 ⋅ dbl , ―, 175 mm⎟ = 17.5 cm Distanta maxima dintre etrieri in
⎝ 2 ⎠ zona critica
