e
s2
0.187 m
=
e
s2
0.5 h
⋅
e
−
a
2
−
:=
e
s1
0.533 m
=
e
s1
0.5 h
⋅
e
+
a
1
−
:=
e
0.173 m
=
e
η e
o
⋅
:=
η
2.767
:=
λ > λ
lim
stąd istnieje konieczność uwzględnienia efektów II rzędu
λ
lim
10.497
=
λ
lim
20 A
⋅ B
⋅ C
⋅
n
:=
C
0.7
:=
B
1.1
:=
A
0.758
=
A
1
1
0.2
φ
ef
⋅
+
:=
λ
64.952
=
λ
l
o
h
12
⋅
:=
2.2 Wstępne przyjęcie mimośrodów II rzędu
e
o
0.063 m
=
e
o
e
e
e
a
+
:=
e
a
0.038 m
=
e
a
max
l
o
400
h
30
,
0.02 m
⋅
,
⎛
⎜
⎝
⎞
⎟
⎠
:=
e
e
0.025 m
=
e
e
M
oEd
N
Ed
:=
2.1 Wyznaczenie mimośrodów I rzędu
2. Wymiarowanie:
A
s2
58.784 cm
2
=
A
s2
max A
s21
(
)
:=
A
s21
58.784
8.571
⎛
⎜
⎝
⎞
⎟
⎠
cm
2
=
A
s21
N
Ed
e
s1
⋅
x
eff
d
0.5 x
eff
⋅
−
(
)
⋅
b
⋅ f
cd
⋅
−
f
yd2
d
a
2
−
(
)
⋅
0.5A
smin
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎦
:=
ξ
eff
0.854
=
ξ
eff
x
eff
d
:=
x
eff
0.649 m
=
x
eff
if
a
2
2
2 N
Ed
⋅
e
s2
⋅
b f
cd
⋅
+
a
2
+
d
<
a
2
2
2 N
Ed
⋅
e
s2
⋅
b f
cd
⋅
+
a
2
+
,
d
,
⎛
⎜
⎝
⎞
⎟
⎠
:=
Ponieważ A
s1
<0 mamy do czynienia z przypadkiem małego mimośrodu
A
s1
28.653
−
cm
2
=
A
s1
ξ
eff
b
⋅ d
⋅ f
cd
⋅
A
s2
f
yd2
⋅
+
N
Ed
−
f
yd1
:=
ξ
eff
0.533
=
ξ
eff
1
1
2
μ
eff
⋅
−
−
:=
μ
eff
0.391
=
μ
eff
N
Ed
e
s1
⋅
f
yd2
A
s2
⋅
d
a
2
−
(
)
⋅
−
f
cd
b
⋅ d
2
⋅
:=
A
s2
72.46 cm
2
=
A
s2
max A
s21
(
)
:=
A
s21
72.462
8.571
⎛
⎜
⎝
⎞
⎟
⎠
cm
2
=
A
s21
N
Ed
e
s1
⋅
μ
eff.lim
f
cd
⋅
b
⋅ d
2
⋅
−
f
yd2
d
a
2
−
(
)
⋅
0.5A
smin
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎦
:=
η
2.767
=
η
rz
2.766
=
η
rz
1
1
N
Ed
N
B
−
:=
N
B
9.398
10
3
×
kN
=
N
B
π
2
EJ
⋅
l
o
2
:=
EJ
214.243 m
4
MPa
=
EJ
K
c
E
cd
⋅
J
c
⋅
E
s
J
s
⋅
+
:=
K
c
0.077
=
K
c
k
1
k
2
⋅
1
φ
ef
+
:=
k
2
0.2
=
k
2
min n
λ
170
⋅
0.20
,
⎛⎜
⎝
⎞⎟
⎠
:=
k
1
1
=
k
1
f
ck
20MPa
:=
J
c
0.021 m
4
=
J
c
b h
3
⋅
12
:=
J
s
8.729
10
4
−
×
m
4
=
J
s
A
s2
0.5 h
⋅
a
2
−
(
)
2
⋅
A
s1
0.5 h
⋅
a
1
−
(
)
2
⋅
+
⎡
⎣
⎤
⎦
:=
2.3 Korekta mimośrodów II rzędu (metoda nominalnej sztywności)
A
s2
58.784 cm
2
=
A
s1
8.571 cm
2
=
A
s1
max A
s11
(
)
:=
A
s11
8.571
8.571
⎛
⎜
⎝
⎞
⎟
⎠
cm
2
=
A
s11
if x
eff
d
<
0.5A
smin
,
N
Ed
e
s2
⋅
b h
⋅
0.5 h
⋅
a
2
−
(
)
⋅
f
cd
⋅
−
f
yd2
d
a
2
−
(
)
⋅
,
⎡
⎢
⎣
⎤
⎥
⎦
0.5A
smin
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎦
:=
Sprawdzenie poprawności założenia małego mimośrodu:
ξ
eff
x
eff
d
:=
ξ
eff
0.854
=
Ostateczny przekrój zbrojenia:
A
s2
58.784 cm
2
=
A
s1
8.571 cm
2
=
ρ
s
A
s2
A
s1
+
(
)
100
⋅
%
⋅
b h
⋅
:=
ρ
s
1.684 %
=
η
rz
2.766
=
η
2.767
=
