9.1.3 Zakřivení ve stavu I

Zakřivení ve stavu I

Výška tlakové oblasti betonu x_I

$x_I = \frac{b · h · {\displaystyle \frac{h}{2}} + {\alpha }_e · A_s · d - A_s · d}{{\alpha }_e · A_s - A_s + b· h} = \frac{100 · 20 · {\displaystyle \frac{20}{2}} + 20.0 · 4.45 ·17 - 4.45 · 17}{20 · 4.45 - 4.45 + 100 · 20} = 10.28\text{ cm}$

$\xi = x_I/h = 10.28/20 = 0.514$

$$\begin{gathered} \kappa = 1 + 12 · {\left(0.5 - \xi \right)}^2 + 12 · \left(\alpha - 1\right) · \rho · {\left(\frac{d}{h} - \xi \right)}^2 = \\ = 1 + 12 · {\left(0.5 - 0.514\right)}^2 + 12 · \left(20 - 1\right) · 0.002225 · {\left(\frac{17}{20} - 0.514\right)}^2 = 1.06 \end{gathered}$$

$I_{cI} = \kappa · b · \frac{h^3}{12} = 1.06 · 100 · \frac{{20}^3}{12} = 70 667 {\mathrm{cm}}^4 = 0.0007067{\text{ m}}^4$

$W_{c,I} = \frac{I_{c,I}}{h - x_I} = \frac{70 667}{20 - 10.28} = 7 270.2 {\mathrm{cm}}^3$

$S_I= A_s · \left(d - x_I\right) = 4.45 · \left(17 - 10.28\right) = 29.904 {\mathrm{cm}}^3 = 0.0000299 {\mathrm{m}}^3$

${\left(\frac{1}{r}\right)}_M = \frac{M_{Ed}}{E_{c,\text{eff}} · I_c} = \frac{0.01850 \mathrm{MNm}}{10 000 \mathrm{MN}/{\mathrm{m}}^2 · 0.0007067 {\mathrm{m}}^4} = 0.00262 {\mathrm{m}}^{-1}$

${\left(\frac{1}{r}\right)}_{cs} = {\epsilon }_{cs\infty } · {\alpha }_e · \frac{S_I}{I_c} = 0.0005 · 20.0 · \frac{0.0000299 {\mathrm{m}}^3}{0.0007067 {\mathrm{m}}^4} = 0.00042{\mathrm{m}}^{-1}$

${\left(\frac{1}{r}\right)}_{tot,I} = {\left(\frac{1}{r}\right)}_M + {\left(\frac{1}{r}\right)}_{cs} = 0.00262 + 0.00042 = 0.00304 {\mathrm{m}}^{-1}$