9.1.4 Curvature for Cracked Sections (State II)

Curvature for Cracked Sections (State II)

When characteristic loads are applied, concrete shows linear elastic behavior. The concrete stress distributed over the compression zone is assumed to be triangular.

The depth of the concrete compression area can be determined as follows:

$$\begin{gathered} x = \rho · {\alpha }_e · d · \left[-1 + \sqrt{1 + \frac{2}{\rho · {\alpha }_e}}\right] = \\ = 0.0026 · 20.0 ·17\text{ cm} · \left[-1 + \sqrt{1 + \frac{2}{0.0026 · 20.0}}\right] = 4.68 \text{cm} \end{gathered}$$

The tension stress in the reinforcement is determined with M_Ed = 18.50 kNm as follows:

${\sigma }_s = \frac{M}{A_s · \left(d - {\displaystyle \frac{x}{3}}\right)} = \frac{18.5 · {10}^{-3}}{4.45 · {10}^{-4} · \left(0.17 - {\displaystyle \frac{0.0468}{3}} \right)} = 269.60 \mathrm{N}/{\mathrm{mm}}^2$

The curvature in the final crack state is determined as follows:

${\left(\frac{1}{r}\right)}_{M,II} = \frac{{\epsilon }_s}{d - x} = \frac{1.346 · {10}^{-3}}{170 - 46.8} = 0.010931 {\mathrm{m}}^{-1}$

where

${\epsilon }_s = \frac{{\sigma }_s}{E_s} = \frac{269.26}{200 000} = 1.346 · {10}^{-3}$

In manual calculations, the curvature for cracked sections (state II) is determined by means of a table from [12] (see Figure 9.2).

${\omega }_1 = {\alpha }_e · \frac{A_s}{b · d} = 20.0 · \frac{4.45 {\mathrm{cm}}^2}{100 \mathrm{cm} · 17 \mathrm{cm}} = 0.052 \to \beta = 1.10$

${\left(\frac{1}{r}\right)}_{cs,II} = {\epsilon }_{cs\infty } · {\alpha }_e ·\frac{S_{II}}{I_{II}} = {\epsilon }_{cs\infty } · \beta ·\frac{1}{d} = 0.0005 · 1.10 · \frac{1}{0.17 \text{m}} = 0.00324 {\text{m}}^{-1}$

${\left(\frac{1}{r}\right)}_{tot,II} = {\left(\frac{1}{r}\right)}_{M,II} = {\left(\frac{1}{r}\right)}_{cs,II} = 0.01093 + 0.00324 = 0.01417 {\mathrm{m}}^{-1}$