9.2.6.3 State II (cracked)

State II (cracked)

In contrast to the cross-section properties in the uncracked state (state I), the cross-section properties in state II (cracked sections) are quite difficult to determine manually. Determining the strain distribution (general case: ε₀ + (1/r)_y ⋅ y + (1/r)_z ⋅ z) for a particular action constellation with the stress-strain relations defined in the standards for nonlinear methods already represents a problem. For further information, refer to the corresponding literature [7].

To determine the stresses and strains for crack formation, we can normally make simplified assumptions (linear elastic material rules). We can justify this approach by the fact that the ratio of stress to strain for concrete behaves nearly linearly up to a stress of σ_c ≅ 0.4 ⋅ f_c. For reinforcing steel, we can roughly assume this fact until yielding is reached anyway. Thus, if we have a structural component with a crack moment in the characteristic load level, we can calculate stresses and strains with sufficient accuracy using these simplified approaches.

Without an acting axial force, the solution for a triangular compression zone leads to a quadratic equation (with axial force: cubic equation) when calculating the neutral axis depth x (height of compression zone). Due to the assumed linearity of stresses and strains, the neutral axis depth is decoupled from the applied moment.

$\rho = \frac{A_{s1}}{b · d} = \frac{6.22 {\mathrm{cm}}^2}{100 · 13.5 {\mathrm{cm}}^2} = 0.004607$

$$\begin{gathered} \xi = -{\alpha }_e · \rho · \left(1 + \frac{A_{s2}}{A_{s1}}\right) + \sqrt{{\alpha }_e · \rho · {\left(1 + \frac{A_{s2}}{A_{s1}}\right)}^{\text{'}\text{'}} + 2 · {\alpha }_e · \rho · \left(1 + \frac{A_{s2} · d_2}{A_{s1} · d}\right) } = \\ = - 26.33 · 0.004607 · \left(1 + 1\right) + \\ +\sqrt{{\left[26.33 · 0.004607 · \left(1 + 1\right)\right]}^{\text{'}\text{'}} + 2 · 26.33 · 0.004607 · \left(1 + \frac{1 · 2.5}{1 · 13.5}\right) } = \\ = 0.3459 \end{gathered}$$

$x_{II} = \xi · d = 0.346 · 13.5 = 4.67 {\mathrm{cm}}^4$

$$\begin{gathered} \kappa = 4 · {\xi }^3 + 12 · {\alpha }_e · \rho · {\left(1 - \xi \right)}^2 + 12 · {\alpha }_e ·\rho · \frac{A_{s2}}{A_{s1}} · {\left(\xi - \frac{d_2}{2}\right)}^2 = \\ = 4 · 0.{346}^3 + 12 · 26.33 · 0.00460 · {\left(1 - 0.346\right)}^2 + 12 · 26.33 · 0.00460 · 1 ·{\left(0.346 - \frac{2.5}{13.5}\right)}^2 = \\ = 0.826 \end{gathered}$$

$I_{c,II} = \kappa · b · \frac{d^3}{12} = 0.826 · 100 · \frac{13.5^3}{12} = 16 935 {\mathrm{cm}}^4$

${\sigma }_{cr,II} = \frac{M}{I_{y,II}} · x = \frac{1210}{16935} · 4.67 · 10 = 3.34 \mathrm{N}/{\mathrm{mm}}^2$

${\sigma }_{sr1,II} = {\alpha }_e · \frac{M}{I_{y,II}} · \left(d - x\right) = 26.33 · \frac{1210}{16935} · \left(13.5 - 4.7\right) · 10 = 166.12 \mathrm{N}/{\mathrm{mm}}^2$

${\sigma }_{sr2,II} = {\sigma }_{sr1,II} · \frac{x - d_2}{d - x} = 166.12 · \frac{4.67 - 2.5}{13.5 - 4.67} = 40.82 \mathrm{N}/{\mathrm{mm}}^2$

${\epsilon }_{sr1,II} = \frac{{\sigma }_{sr1,II}}{E_s} = \frac{166.012}{200 000} · 1 000 = 0.8306 \text{‰}$

A simplified calculation of stresses and strains, as was done for the crack moment, cannot be applied without due consideration. The stresses and strains for the effective moment M = 17.64 kNm required for the calculation of the curvatures and stiffnesses are determined in a comparative calculation using the exact stress-strain curves for concrete and reinforcing steel according to EN 1992-1-1, Figure 3.2 or 3.3.

The accurate calculation of stresses in the cracked state is performed by means of a third party application used for exact stress integration, leading to the following results for M = 17.64 kNm:

• σ_s1,II = 242.27 N/mm²

• σ_s2,II = −59.07 N/mm²

• ε_s1,II = 1.211 ‰

• ε_s2,II = −0.638 ‰
• ε_c,II = −0.6378 ‰