bp
0.3m
:=
beffprzeslo
2.12m
:=
fcd
21.43MPa
:=
hp
0.6m
:=
beffpodpora
1.08m
:=
hf
0.1m
:=
fyk
500MPa
:=
dprzeslo
54.9cm
:=
fyd
434.78MPa
:=
ξeff.lim
0.509
:=
dpodpora
53.8cm
:=
fctm
2.9MPa
:=
ZBROJENIE NA ZGINANIE
1.Zbrojenie dolne w przęśle AB
MEd
144.54kN m
⋅
:=
Mf
fcd beffprzeslo
⋅
hf
⋅
dprzeslo 0.5 hf
⋅
−
(
)
⋅
2.267 MN m
⋅
⋅
=
:=
MEd Mf
<
1
=
PRZEKRÓJ POZORNIE TEOWY
Sc.eff
MEd
fcd bp
⋅
dpodpora
(
)
2
⋅
0.078
=
:=
ξeff
1
1
2 Sc.eff
⋅
−
(
)
1
2
−
0.081
=
:=
ξeff ξeff.lim
<
1
=
xeff
ξeff dprzeslo
⋅
4.444 cm
⋅
=
:=
As1.AB
fcd bp
⋅
xeff
⋅
(
)
fyd
6.572 cm
2
⋅
=
:=
2. Zbrojenie dolne w przęśle BB'
MEd
137.14kN m
⋅
:=
Mf
fcd beffprzeslo
⋅
hf
⋅
dprzeslo 0.5 hf
⋅
−
(
)
⋅
2.267 MN m
⋅
⋅
=
:=
MEd Mf
<
1
=
PRZEKRÓJ POZORNIE TEOWY
Sc.eff
MEd
fcd bp
⋅
dprzeslo
(
)
2
⋅
=
:=
ξeff
1
1
2 Sc.eff
⋅
−
(
)
1
2
−
0.073
=
:=
ξeff ξeff.lim
<
1
=
xeff
ξeff dprzeslo
⋅
4.034 cm
⋅
=
:=
As1.BB
fcd bp
⋅
xeff
⋅
(
)
fyd
5.965 cm
2
⋅
=
:=
3.Zbrojenie górne na krawędzi podpory:
MEd.B
244.38kN m
⋅
:=
MEd.A
178.99kN m
⋅
:=
Sc.eff
MEd.B
fcd bp
⋅
dpodpora
(
)
2
⋅
=
:=
Sc.eff.A
MEd.A
fcd bp
⋅
dpodpora
(
)
2
⋅
0.096
=
:=
ξeff
1
1
2 Sc.eff
⋅
−
(
)
1
2
−
0.141
=
:=
ξeff.A
1
1
2 Sc.eff.A
⋅
−
(
)
1
2
−
0.101
=
:=
ξeff ξeff.lim
<
1
=
ξeff.A ξeff.lim
<
1
=
xeff.A
dpodpora ξeff.A
⋅
5.451 cm
⋅
=
:=
xeff
dpodpora ξeff
⋅
7.603 cm
⋅
=
:=
As1.A
fcd bp
⋅
xeff.A
⋅
(
)
fyd
8.06 cm
2
⋅
=
:=
As1.B
fcd bp
⋅
xeff
⋅
(
)
fyd
11.242 cm
2
⋅
=
:=
WARUNEK ZBROJENIA MINIMALNEGO I MAKSYMALNEGO
Zbrojenie przęsłowe dolne
As.min
5.22cm
2
:=
As1.AB 6.572 cm
2
⋅
=
As1.AB As.min
>
1
=
As1.AB.prz
As1.AB 6.572 cm
2
⋅
=
:=
As1.BB 5.965 cm
2
⋅
=
As1.BB As.min
>
1
=
As1.BB.prz
As1.BB 5.965 cm
2
⋅
=
:=
Zbrojenie krawędziowe górne
As.min
5.6cm
2
:=
As1.B 11.242 cm
2
⋅
=
As1.A 8.06 cm
2
⋅
=
As1.B As.min
>
1
=
As1.A As.min
>
1
=
As1.B.pod
As1.B 11.242 cm
2
⋅
=
:=
As1.A.pod
As1.A 8.06 cm
2
⋅
=
:=
ROZSTAW MINIMALNY PRĘTÓW (w świetle)
smin
max 1 20
⋅
mm 16mm
5mm
+
,
20mm
,
(
)
21 mm
⋅
=
:=
DŁUGOŚĆ ZAKOTWIENIA (pręty dolne)
η1
1
:=
η2
1
:=
fctk.05
2MPa
:=
fctd
fctk.05
1.4
1.429 MPa
⋅
=
:=
fbd
2.25
η1
⋅
η2
⋅
fctd
⋅
3.214 MPa
⋅
=
:=
σsd
fyd 434.78 MPa
⋅
=
:=
lbd.rqd
20mm
σsd
⋅
4 fbd
⋅
0.676 m
=
:=
α1
1
:=
cd
min 0.5 3
⋅
cm 35mm
,
(
)
0.015 m
=
:=
α2
min 1 1
0.15
cd 3 20
⋅
mm
−
(
)
20mm
⋅
−
,
1
=
:=
λ
20mm
(
)
4
2
π
⋅
0.25
20mm
(
)
4
2
⋅
π
⋅
−
20mm
(
)
4
2
π
⋅
0.75
=
:=
K
0.05
:=
α3
min 1 max 1
K
λ
⋅
−
0.7
,
(
)
,
(
)
0.963
=
:=
α2 α3
⋅
0.7
≥
1
=
lb.min
max 0.3 lbd.rqd
⋅
10 20
⋅
mm
,
100mm
,
(
)
202.897 mm
⋅
=
:=
lbd
1
α2
⋅
α3
⋅
lbd.rqd
⋅
650.962 mm
⋅
=
:=
lbd lb.min
>
1
=
Przyjmuję
lbd
655mm
:=
DŁUGOŚĆ ZAKOTWIENIA (pręty górne)
η1
0.7
:=
η2
1
:=
fctk.05
2MPa
:=
fctd
fctk.05
1.4
1.429 MPa
⋅
=
:=
fbd
2.25
η1
⋅
η2
⋅
fctd
⋅
2.25 MPa
⋅
=
:=
σsd
fyd 434.78 MPa
⋅
=
:=
lbd.rqd
16mm
σsd
⋅
4 fbd
⋅
0.773 m
=
:=
α1
1
:=
cd
min 0.5 3
⋅
cm 30mm
,
(
)
0.015 m
=
:=
α2
min 1 1
0.15
cd 16mm
−
(
)
16mm
⋅
−
,
1
=
:=
λ
16mm
(
)
4
2
π
⋅
0.25
16mm
(
)
4
2
⋅
π
⋅
−
16mm
(
)
4
2
π
⋅
0.75
=
:=
K
0.05
:=
α3
min 1 max 1
K
λ
⋅
−
0.7
,
(
)
,
(
)
0.963
=
:=
α2 α3
⋅
0.7
≥
1
=
lb.min
max 0.3 lbd.rqd
⋅
10 16
⋅
mm
,
100mm
,
(
)
231.883 mm
⋅
=
:=
lbd
1
α2
⋅
α3
⋅
lbd.rqd
⋅
743.957 mm
⋅
=
:=
lbd lb.min
>
1
=
Przyjmuję
lbd
750mm
:=
DŁUGOŚĆ ZAKŁADU (pręty górne)
α1 1
=
α2 1
=
α3
1
:=
ρ1
1.5
:=
α6
min 1.5 max 1
ρ1
25
,
,
1
=
:=
l0.min
max 0.3 lbd.rqd
⋅
α6
⋅
15 16
⋅
mm
,
200mm
,
(
)
240 mm
⋅
=
:=
l0
α1 α2
⋅
α3
⋅
α6
⋅
lbd.rqd
⋅
772.942 mm
⋅
=
:=
l0 l0.min
>
1
=
Przyjmuję połączenie na zakład
l0
800mm
:=
ŚCINANIE
Przyjmuje się strzemiona czterocięte
ϕs
8mm
:=
α
90deg
:=
bw
bp 0.3 m
=
:=
Asw1
4
π
⋅
0.5
ϕs
⋅
(
)
2
⋅
2.011 cm
2
⋅
=
:=
smax
min 0.75 min dpodpora dprzeslo
,
(
)
⋅
600.mm
,
(
)
40.35 cm
⋅
=
:=
cnom
35cm
:=
Podpora A
d
dpodpora 53.8 cm
⋅
=
:=
z
0.9 d
⋅
48.42 cm
⋅
=
:=
As1.A
10.05cm
2
:=
k
min 1
200
d
mm
+
2
,
1.61
=
:=
CRd.c
0.18
γc
0.129
=
:=
ρ
min 0.02
As1.A
bw d
⋅
,
6.227
10
3
−
×
=
:=
νmin
0.035 k
3
2
⋅
fck MPa
⋅
(
)
1
2
⋅
0.392 MPa
⋅
=
:=
VRd.c.1
νmin cnom
⋅
d
⋅
73.723 kN
⋅
=
:=
VRd.c.2
CRd.c k
⋅
100
ρ fck
⋅
MPa
2
⋅
1
3
⋅
bw
⋅
d
⋅
88.632 kN
⋅
=
:=
VRd.c
max VRd.c.1 VRd.c.2
,
(
)
88.632 kN
⋅
=
:=
Vp.A
VEd 149.32 kN
⋅
=
:=
VEd VRd.c
>
1
=
odcinek 2 rodzaju
αcw
1
:=
θ
45deg
:=
Asw
4
π ϕs
2
⋅
4
⋅
2.011
10
4
−
×
m
2
=
:=
s
Asw fyd
⋅
z
⋅
cot
θ
( )
⋅
VEd
28.347 cm
⋅
=
:=
ν1
0.6 1
fck
250MPa
−
⋅
0.528
=
:=
VRd.s
Asw
s
z
⋅
fyd
⋅
cot
θ
( )
⋅
149.32 kN
⋅
=
:=
VRd.max
αcw bw
⋅
z
⋅
ν1
⋅
fcd
⋅
cot
θ
( )
tan
θ
( )
+
821.811 kN
⋅
=
:=
VRd.max VEd
>
1
=
ok
przyjmuję rozstaw strzemion czterociętych #8co 28cm
ρw
Asw
s bw
⋅
sin
α
( )
⋅
0.236 %
⋅
=
:=
ρw.min
0.08
fck
MPa
fyk
MPa
⋅
0.088 %
⋅
=
:=
ρw ρw.min
>
1
=
s
smax
<
1
=
ok
Podpora B
d
dpodpora 53.8 cm
⋅
=
:=
z
0.9 d
⋅
48.42 cm
⋅
=
:=
As1.B
12.06cm
2
:=
k
min 1
200
d
mm
+
2
,
1.61
=
:=
CRd.c
0.18
γc
0.129
=
:=
ρ
min 0.02
As1.B
bw d
⋅
,
=
:=
νmin
0.035 k
3
2
⋅
fck MPa
⋅
(
)
1
2
⋅
0.392 MPa
⋅
=
:=
VRd.c.1
νmin cnom
⋅
d
⋅
73.723 kN
⋅
=
:=
VRd.c.2
CRd.c k
⋅
100
ρ fck
⋅
MPa
2
⋅
1
3
⋅
bw
⋅
d
⋅
94.186 kN
⋅
=
:=
VRd.c
max VRd.c.1 VRd.c.2
,
(
)
94.186 kN
⋅
=
:=
Vp.B
VEd 161.85 kN
⋅
=
:=
VEd VRd.c
>
1
=
odcinek 2 rodzaju
αcw
1
:=
θ
45deg
:=
Asw
4
π ϕs
2
⋅
4
⋅
2.011
10
4
−
×
m
2
=
:=
s
Asw fyd
⋅
z
⋅
cot
θ
( )
⋅
VEd
26.152 cm
⋅
=
:=
ν1
0.6 1
fck
250MPa
−
⋅
0.528
=
:=
VRd.s
Asw
s
z
⋅
fyd
⋅
cot
θ
( )
⋅
161.85 kN
⋅
=
:=
VRd.max
αcw bw
⋅
z
⋅
ν1
⋅
fcd
⋅
cot
θ
( )
tan
θ
( )
+
821.811 kN
⋅
=
:=
VRd.max VEd
>
1
=
ok
przyjmuję rozstaw strzemion czterociętych #8co 28cm
ρw
Asw
s bw
⋅
sin
α
( )
⋅
0.256 %
⋅
=
:=
ρw.min
0.08
fck
MPa
fyk
MPa
⋅
0.088 %
⋅
=
:=
ρw ρw.min
>
1
=
s
smax
<
1
=
ok
Przeslo AB
d
dprzeslo 54.9 cm
⋅
=
:=
z
0.9 d
⋅
49.41 cm
⋅
=
:=
As1
8.04cm
2
:=
k
min 1
200
d
mm
+
2
,
1.604
=
:=
CRd.c
0.18
γc
0.129
=
:=
ρ
min 0.02
As1.B
bw d
⋅
,
=
:=
νmin
0.035 k
3
2
⋅
fck MPa
⋅
(
)
1
2
⋅
0.389 MPa
⋅
=
:=
VRd.c
νmin cnom
⋅
d
⋅
74.8 kN
=
:=
Vp.AB
VEd 13.01 kN
⋅
=
:=
VEd VRd.c
>
0
=
odcinek 1 rodzaju
VRd.I.rodz
0.5 bw
⋅
d
⋅
ν1
⋅
fcd
⋅
931.794 kN
⋅
=
:=
VRd.I.rodz VEd
>
1
=
ok
Przyjmuję rozstaw strzemion czterociętych #8 co 28cm
s
28cm
:=
ρw
Asw1
s bw
⋅
sin
α
( )
⋅
0.239 %
⋅
=
:=
ρw.min
0.08
fck
MPa
fyk
MPa
⋅
0.088 %
⋅
=
:=
ρw ρw.min
>
1
=
s
smax
<
1
=
ok
Przeslo
BB'
d
dprzeslo 54.9 cm
⋅
=
:=
z
0.9 d
⋅
49.41 cm
⋅
=
:=
As1
6.03cm
2
:=
k
min 1
200
d
mm
+
2
,
1.604
=
:=
CRd.c
0.18
γc
0.129
=
:=
ρ
min 0.02
As1.B
bw d
⋅
,
7.322
10
3
−
×
=
:=
νmin
0.035 k
3
2
⋅
fck MPa
⋅
(
)
1
2
⋅
0.389 MPa
⋅
=
:=
VRd.c
νmin cnom
⋅
d
⋅
74.8 kN
⋅
=
:=
Vp.AB
VEd 10.74 kN
⋅
=
:=
VEd VRd.c
>
0
=
odcinek 1 rodzaju
VRd.I.rodz
0.5 bw
⋅
d
⋅
ν1
⋅
fcd
⋅
931.794 kN
⋅
=
:=
VRd.I.rodz VEd
>
1
=
ok
Przyjmuję rozstaw strzemion czterociętych #8 co 28cm
s
28cm
:=
ρw
Asw1
s bw
⋅
sin
α
( )
⋅
0.239 %
⋅
=
:=
ρw.min
0.08
fck
MPa
fyk
MPa
⋅
0.088 %
⋅
=
:=
ρw ρw.min
>
1
=
s
smax
<
1
=
ok
Ś
CINANIE MIĘDZY PÓŁKĄ A ŚRODNIKIEM
Półka ściskana
1.
xeff.1
0
:=
2.
Mp.AB.down
144.54kN m
⋅
:=
półka ściskana:
MEd.2
Mp.AB.down
2
72.27 kN m
⋅
⋅
=
:=
Scc.eff.2
MEd.2
beff.przeslo dprzeslo
2
⋅
fcd
⋅
5.278
10
3
−
×
=
:=
xeff.2
1
1
2 Scc.eff.2
⋅
−
−
(
)
dprzeslo
⋅
0.291 cm
⋅
=
:=
3.
Mp.AB.down
140.66kN m
⋅
:=
półka ściskana:
MEd.3
Mp.AB.down
2
70.33 kN m
⋅
⋅
=
:=
Scc.eff.3
MEd.3
beff.przeslo dprzeslo
2
⋅
fcd
⋅
5.136
10
3
−
×
=
:=
xeff.3
1
1
2 Scc.eff.3
⋅
−
−
(
)
dprzeslo
⋅
0.283 cm
⋅
=
:=
4.
xeff.4
0
:=
Fd1
0.5xeff.1 beff.przeslo bw
−
(
)
⋅
fcd
⋅
0
=
:=
Fd2
0.5xeff.2 beff.przeslo bw
−
(
)
⋅
fcd
⋅
56.655 kN
⋅
=
:=
Fd3
0.5xeff.3 beff.przeslo bw
−
(
)
⋅
fcd
⋅
55.131 kN
⋅
=
:=
Fd4
0.5xeff.4 beff.przeslo bw
−
(
)
⋅
fcd
⋅
0
=
:=
∆x12
0.99m
:=
∆Fd12
Fd2 Fd1
−
56.655 kN
⋅
=
:=
∆x23
2.17m
:=
∆Fd23
Fd3 Fd2
−
1.525
−
kN
⋅
=
:=
∆x34
0.92m
:=
∆Fd34
Fd4 Fd3
−
55.131
−
kN
⋅
=
:=
Naprężenia w styku
VEd.12
∆Fd12
hf ∆x12
⋅
572.278
kN
m
2
⋅
=
:=
VEd.23
∆Fd23
hf ∆x23
⋅
7.027
−
kN
m
2
⋅
=
:=
VEd.34
∆Fd34
hf ∆x34
⋅
599.247
−
kN
m
2
⋅
=
:=
VEd.max
599.247
kN
m
2
:=
Sprawdzenie czy konieczne jest dodatkowe zbrojenie w styku
571.429 kPa
VEd.12
<
1
=
0.4 fctd
⋅
571.429 kPa
⋅
=
571.429 kPa
VEd.23
>
1
=
0.4 fctd
⋅
571.429 kPa
⋅
=
571.429 kPa
VEd.34
<
1
=
0.4 fctd
⋅
571.429 kPa
⋅
=
Sprawdzenie krzyżulców betonowych ściskanych z uwagi na zmiażdżenie
5.658
10
3
×
kN
m
2
VEd.max
>
1
=
ν1 fcd
⋅
sin
θ
( )
⋅
cos
θ
( )
⋅
5.658
10
3
×
kN
m
2
⋅
=
krzyżulce betonowe nie ulegną zniszczeniu
Zbrojenie zszywające
Asfr
1.01cm
2
:=
sf
30cm
:=
VRd.r
Asfr fyd
⋅
cot
θ
( )
⋅
sf hf
⋅
1.464
10
3
×
kN
m
2
⋅
=
:=
∆V12
VEd.12
VRd.r
−
:=
∆V34
VEd.34
VRd.r
−
:=
∆V12 891.482
kN
m
2
⋅
=
∆V34 864.513
kN
m
2
⋅
=
Asfr.dod
1.01cm
2
:=
dla 2ϕ8 co 30cm
VRd.dod
Asfr.dod fyd
⋅
cot
θ
( )
⋅
sf hf
⋅
1.464
10
3
×
kN
m
2
⋅
=
:=
VRd.dod ∆V12
>
1
=
VRd.dod.34
Asfr.dod fyd
⋅
cot
θ
( )
⋅
sf hf
⋅
1.464
10
3
×
kN
m
2
⋅
=
:=
VRd.dod.34 ∆V34
>
1
=
półka rozciągana:
As1.polka
2.01cm
2
:=
∆Fd
As1.polka fyd
⋅
87.391 kN
⋅
=
:=
∆x
1.25
2
m
0.625 m
=
:=
νEd
∆Fd
hf ∆x
⋅
:=
νEd 0.527MPa
=
zbrojenie
νEd 0.4 fctd
⋅
≤
1
=
pomijam obliczenia
krzyżulce
νEd ν1 fcd
⋅
sin
θ
( )
⋅
cos
θ
( )
⋅
≤
1
=
Przyjmuję zagęszczenie zbrojenia rozdzielczego wkładkami #8
co 30cm
srozdz
30cm
:=
swkladki
30cm
:=
STRZEMIONA PRZY ŻEBRACH:
Rzebro.max
76.67kN
:=
hż
30cm
:=
zasieg_strzemion
min
hp
2
hż
2
hp
3
+
,
0.3 m
=
:=
nn
0.5
Rzebro.max
fyd Asw1
⋅
0.439
=
:=
przyjeto 1 strzemię #6 po kazdej stronie zebra na dystansie
30cm od osi połączenia żebra z podciągiem
STAN GRANICZNY UŻYTKOWALNOŚCI- RYSY (metoda uproszczona):
Ecm
32GPa
:=
Es
200GPa
:=
fct.eff
2.9MPa
:=
bp 30 cm
⋅
=
hp 60 cm
⋅
=
Ec.eff
Ecm
1
2.1
+
10.323 GPa
⋅
=
:=
Mp.AB.down 144.54 kN m
⋅
⋅
=
As1.prz.AB 6.572 cm
2
⋅
=
αe.lt
Es
Ec.eff
19.375
=
:=
MEqp
Mp.AB.down
116.25kN
0.8 177.21
⋅
kN
+
1.35 116.25
⋅
kN
1.5 177.21
⋅
kN
+
⋅
88.217 kN m
⋅
⋅
=
:=
xII
αe.lt
As1.prz.AB
bp
⋅
1
−
1
2bp dprzeslo
⋅
αe.lt As1.prz.AB
⋅
+
+
⋅
17.756 cm
⋅
=
:=
ξ
0.8
:=
dlatego, że ρ 1.25%
=
σs1.przyb
MEqp
ξ dprzeslo
⋅
As1.prz.AB
⋅
305.646 MPa
⋅
=
:=
-szacunkowy zasięg strefy ściskanej
σs1
MEqp
As1.prz.AB dprzeslo
xII
3
−
⋅
274.064 MPa
⋅
=
:=
-dokładny zasięg strefy ściskanej
ϕ's.max
18mm
:=
-interpolacja z tabl. 7.2N
kc.
0.4
:=
hc
0.5 hp
⋅
0.3 m
=
:=
STAN GRANICZNY UŻYTKOWALNOŚCI- UGIĘCIA (metoda uproszczona):
ρ
As1.prz.AB
bp dprzeslo
⋅
0.399 %
⋅
=
:=
ρ0
fck
MPa
10
3
−
⋅
0.548 %
⋅
=
:=
ρ'
0%
:=
lp
6.5m
:=
K
1.5
:=
-Tab. 7.4N- wewnętrzne przęsła belek oraz płyt
jednokierunkowo lub dwukierunkowo zbrojonych
L_d
K 11
1.5
fck
MPa
⋅
ρ0
ρ
⋅
+
3.2
fck
MPa
⋅
ρ0
ρ
1
−
3
2
⋅
+
⋅
ρ
ρ0
≤
if
K 11
1.5
fck
MPa
⋅
ρ0
ρ
ρ'
−
⋅
+
1
12
fck
MPa
ρ'
ρ0
⋅
⋅
+
⋅
ρ
ρ0
>
if
39.4
=
:=
lp
dprzeslo
11.84
=
< L_d 39.4
=
warunek spełniony
As1.prov
25.12cm
2
:=
As1.req
23cm
2
:=
lp
dprzeslo
500
fyk
MPa
⋅
As1.prov
As1.req
⋅
12.931
=
>
lp
dprzeslo
11.84
=
dopuszczalne ugięcie
podciągu nie będzie
przekroczone
OBWIEDNIA NOŚNOŚCI ZBROJENIA ROZCIĄGANEGO I SIŁ W TYM ZBROJENIU
Wpływ ścinania na wzrost siły w zbrojeniu podłużnym
ramię sił wew. w przęsłach AB i BC:
zprzeslo
0.9 dprzeslo
⋅
49.41 cm
⋅
=
:=
al
0.5 zprzeslo
⋅
cot
θ
( )
sin
α
( )
−
(
)
⋅
24.846 cm
⋅
=
:=
przyjmuję rozsunięcie wykresu w przęsłach o
25cm
ramie sił wew. nad podporami A i B
zpodpora
0.9 dpodpora
⋅
48.42 cm
⋅
=
:=
al
0.5 zpodpora
⋅
cot
θ
( )
sin
α
( )
−
(
)
⋅
24.348 cm
⋅
=
:=
przyjmuję rozsunięcie wykresu w przęsłach o
25cm
w przęśle
As1.AB.prz
As1.BB.prz
fyd
⋅
349.563
262.172
kN
⋅
=
As1.A.pod fyd
⋅
436.954 kN
=
w podporze
As1.B.pod fyd
⋅
524.345 kN
=
Przęsła :
AB -
144.54kN m
⋅
zprzeslo
258.107 kN
=
140.66kN m
⋅
zprzeslo
251.179 kN
=
BB -
137.14kN m
⋅
zprzeslo
244.893 kN
=
Podpora :
A-
139.95kN m
⋅
zpodpora
289.033 kN
=
223.28kN m
⋅
zpodpora
461.132 kN
=
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