n

1.235

=

E

s

200000 MPa

⋅

:=

d

0.76 m

=

l

7.5 m

⋅

:=

E

cd

E

cm

1.2

:=

l

o

15 m

⋅

:=

E

cd

2.417

10

4

×

MPa

=

λ

l

o

h

12

⋅

:=

λ

64.952

=

φ

t

2

:=

φ

ef

φ

t

M

oEqp

M

oEd

⋅

:=

φ

ef

1.6

=

ξ

eff.lim

0.8

0.0035

0.0035

f

yd1

E

s

+

⋅

:=

ξ

eff.lim

0.533

=

μ

eff.lim

ξ

eff.lim

1

0.5

ξ

eff.lim

⋅

−

(

)

⋅

:=

μ

eff.lim

0.391

=

A

smin

max 0.1

N

Ed

f

yd1

⋅

0.002 b

⋅ h

⋅

,

⎛

⎜

⎝

⎞

⎟

⎠

:=

A

smin

17.14 cm

2

=

1. Dane

Dane materiałowe

Dane geometryczne

Obciążenia

f

ck

20 MPa

⋅

:=

f

cd

0.85f

ck

1.4

:=

b

0.5 m

⋅

:=

N

Ed

6000 kN

⋅

:=

f

cd

12.143 MPa

=

h

0.8 m

⋅

:=

M

oEd

150 kNm

⋅

:=

f

yd1

350 MPa

⋅

:=

a

2

0.04 m

⋅

:=

M

oEqp

120 kNm

⋅

:=

f

yd2

350 MPa

⋅

:=

a

1

0.04 m

⋅

:=

n

N

Ed

b h

⋅ f

cd

⋅

:=

E

cm

29000 MPa

⋅

:=

d

h

a

1

−

:=

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

=

cm

m 10

2

−

⋅

:=

MPa

10

6

Pa

⋅

:=

kPa

1000 Pa

⋅

:=

kN

1000 kg

⋅

m

sec

2

⋅

:=

kNm

kN m

⋅

:=

min

, , ,

(

)