1da93f57-fdb0-479e-ac12-c4a21cd5b96a 45 4 5 decimal

&[DATE] &[TIME] - &[FILENAME]

Dane wymiarowe:

Dane siłowe:

h 500 mm * :

b 200 mm * :

Założenie strefy ściskanej x.compression 156.67 mm * :

M.Ed 144.25 kN * m * :

a.1 30 mm * :

N.Ed 135.03 kN * - :

0.85 - NO
1.0 - SWE α.cc 1 :

Założenie odkształceń betonu ε.c 1.5 :

Dane materiałowe:

f.ck 35 MPa * :

γ.c 1.5 :

Pręty do rozdziału siły normalnej n.bars 4 :

A.s1 2 3.14 * cm 2^* :

E.s 200 GPa * :

Wykres naprężenie - odkształcenie

f.cd α.cc f.ck γ.c / * : MPa 23.3333

ε.c2 f.ck 55 MPa * ≤ 2 zdefiniuj inaczej if : 2

ε.cu2 f.ck 55 MPa * ≤ 3.5 zdefiniuj inaczej if : 3.5

n f.ck 55 MPa * ≤ 2 zdefiniuj inaczej if : 2

x σ.c x 0 ≤ 0 x ε.c2 ≤ f.cd ε.c2 / x * f.cd if if :

x σ.c MPa /

x σ.c.εc2 f.cd ε.c2 / x * :

x σ.c.fcd f.cd :

Wartość całki pod wykresem ε.c σ.c.int ε.c 0 ≤ 0 ε.c ε.c2 ≤ x σ.c.εc2 x 0 ε.c int x σ.c.εc2 x 0 ε.c2 int x σ.c.fcd x ε.c2 ε.c int + if if :

ε.c σ.c MPa 17.5

ε.c σ.c.int MPa 13.125

Wyprowadzenie położenia siły wypadkowej ε.c σ.c_x.int ε.c 0 ≤ 0 ε.c ε.c2 ≤ x σ.c.εc2 x * x 0 ε.c int MPa / x σ.c.εc2 x * x 0 ε.c2 int x σ.c.fcd x * x ε.c2 ε.c int + MPa / if if :

ε.c σ.c.int MPa / 13.12

ε.c σ.c_x.int 13.125

ε.c 1.5

Położenie siły wpadkowej (bezwymiarowe, względne) x x_b x σ.c_x.int x σ.c.int MPa / x * / :

x x_t 1 x x_b - :

Wspolczynnik przed naprezeniami itd x ω x σ.c.int x σ.c x * / :

ε.c ω 0.5

x N.c x ω x σ.c * x.compression * b * :

Siła w bryle naprężeń betonu ε.c N.c kN 274.1725

x.compression 0.1567 m *

b 0.2 m *

x N.c MPa / x σ.c MPa / x ω

d h a.1 - : 0.47 m *

x z d x x_t x.compression * - :

ε.c z mm 417.7767

x N.s M.Ed x z / N.Ed abs n.bars / + :

ε.c N.s kN 379.0377

ε.c z 0.4178 m *

ε.c N.s kN 379.0377

ε.c N.c kN 274.1725

x steel x N.s h 2 / a.1 - (* :

x concrete x N.c h 2 / x x_t x.compression * - (* :

ε.c steel kN m * 83.3883

ε.c concrete kN m * 54.2249

x M.Ed M.Ed kN m * / : 144.25

x 230.77 :

x sum.momentów x steel x concrete + x M.Ed - :

1. There are two equations: steel(x)+concrete(x)-MEd(x)=0 & N_s(x)+Ned(x)-N_c(x)=0 x - needed value

At the beginning I have defined value x_compression, however this value has to be different to get proper strains in section. How can I calculate x_compression (it was needed for calculation of first step - therefore has to be defined. First value, was just guess).

Probably x_compression has to be involved in function of strain (or other?) somehow, any ideas?

Finally, strains shall be x=1.5% and y=3.0%, while x_compression=156.67mm

Example:

A.s1 cm 2^6.28

E.s MPa 2105^*

x 1.5 :

x σ.s x ε.s E.s * :

x ε.s x N.s A.s1 E.s * / :

Where "x" stands for ε_c

x ε.s % 0.3018

y x ε.s : % 0.3018

x steel x concrete + x M.Ed 230.77 5000 230.77 1000 - 2 2 mat 3 1 sys

x sum.momentów #

acc 103 -^: a 0 : b 700 : xn b a + (2 / : 41 line

Dichotomy:a division or contrast between two things that are or are represented as being opposed or entirely different. 'ε': accuracy of the solution [a,b] the bracketed initial guesses over possible solution. "sanity": insert solution in the original system to check if solution found is valid ... here: answer is YES .

Dichotomy n 0 : xn steel xn concrete + xn M.Ed - abs acc > xn steel xn concrete + xn M.Ed - (a steel a concrete + a M.Ed - (* 0 ≤ b xn : a xn : if xn b a + (2 / : n n 1 + : 31 line while 31 line

.. sanity ..

xn 350

xn steel xn concrete + xn M.Ed - #

n 0

xn steel 1.0857 105^* J *

xn concrete 2.3929 108^* N *

x concrete 1.8979 108^* N *