2.7.10.7 Calculation of distribution coefficient

Calculation of distribution coefficient

The maximum stress in the uncracked state is:

$$\begin{gathered} {\sigma }_{max,{\varphi }_1} = \frac{n_{{\varphi }_1} + n_{sh,{\varphi }_1}}{A_{{\varphi }_1,l}} + \frac{m_{{\varphi }_1} - n_{{\varphi }_1} \left({\chi }_{l,{\varphi }_1} - {\displaystyle \frac{h}{2}}\right) + m_{sh,l,{\varphi }_1}}{I_{{\varphi }_1,l}} · \left(h - {\chi }_{l,{\varphi }_1}\right) = \\ = \frac{-{100}_{ }{·}_{ }{{10}^3}_{ }{+}_{ }101.5_{ }{·}_{ }{10}^3}{0.206} + {\frac{{30}_{ }{·}_{ }{{10}^3}_{ }{-}_{ }\left(-{100}_{ }{·}_{ }{10}^3\right) {\left(0.{101}_{ }{-}_{ }{\displaystyle \frac{0.2}{2}}\right)}_{ }{+}_{ }4.778 ·{10}^3}{6.816 · {10}^{-4}}}_{ }{·}_{ }\left(0.2_{ }{-}_{ }0.1\right) \\ = 5.1\text{ Mpa} \end{gathered}$$

We assume a long-term loading:

${\beta }_{{\phi }_1}= 0.5$

Taking Tension Stiffening into account, the distribution coefficient is calculated according to the following equation:

• for σ_max,φ1 > f_ctm:

${\zeta }_d = 1 - {\beta }_{{\varphi }_1} · {\left(\frac{f_{ctm}}{{\sigma }_{max,{\varphi }_1}}\right)}^2$

• for σ_max,φ1 ≤ f_ctm:

${\zeta }_{sh,c,{\varphi }_1} = 0$

In the example, the maximum tension stress in the concrete is larger than the concrete tensile strength.

${\sigma }_{max,{\varphi }_1} > f_{ctm}$

$5.1 > 2.9$

Thus, the distribution coefficient is:

${\zeta }_{{\varphi }_1} = 1- {\beta }_{{\varphi }_1} · {\left(\frac{f_{ctm}}{{\sigma }_{max,{\varphi }_1}}\right)}^2 = 1 - 0.5 · {\left(\frac{2.9}{5.1}\right)}^2 = 0.835$