# Simulation of Crack Propagation in Reinforced Concrete Elements
## 4 Simulation and Calculation Procedure  
With the preceding explanations, the CPM can be used to simulate the cracking of a reinforced concrete tensile specimen. In the following, the programme sequence will be briefly explained. A representation can be taken from Figure 12.  
In the first calculation step, the entire model or component is uncracked. A load increase follows until the concrete tensile strength is exceeded at the \"weakest\" point. Then, the stiffness of the cracked concrete element is reduced to zero (\((EA)_{\text{c1}}=0\)) (also see Figure 3). The model is recalculated with the adjusted stiffness value (cracked concrete element). The load is then increased further until the concrete tensile strength is exceeded at another point. The process is repeated until the desired load level \(N_{\text{s,end}}\) is reached. The iterative calculation procedure in the CPM will be illustrated below using an example. The calculations are based on the parameters listed in Table 2.  
Figure 11: Extension of the CPM to a three-bar model. \begin{table}
\begin{tabular}{c c c} \hline \hline \multicolumn{2}{c}{**Geometry**} \\ \hline Reinforcement ratio & \(\rho\) & 2\% \\ Diameter of reinforcing bar & \(\lozenge\) & 16 mm \\ Length of model & \(l\) & 2000 mm \\ Element length & \(l_{\text{e}}\) & 20 mm \\ \hline \multicolumn{2}{c}{**Material**} \\ \hline Concrete: C20/25 & \\ Mean value of compressive strength of concrete & \(f_{\text{cm}}\) & 28 N/mm\({}^{2}\) \\ Mean value of tensile strength of concrete & \(f_{\text{ct,m}}\) & 2.2 N/mm\({}^{2}\) \\ Mean value of modulus of elasticity of concrete & \(E_{\text{cm}}\) & 30,000 N/mm\({}^{2}\) \\ Reinforcement: B500 & \\ Modulus of elasticity of reinforcing steel & \(E_{\text{s}}\) & 200,000 N/mm\({}^{2}\) \\ \hline \multicolumn{2}{c}{**Bond–slip relationship**} \\ \hline Model Code 2010 (good bond conditions) & \\ Bond stress & \(\tau_{\text{b}}(s)\ =\ 13.23\cdot\left(s/1.0\right)^{0.4}\) \\ Reduction factor & \(\lambda(x)\ =\ 0.5\cdot x/16\leq 1.0\) \\ \hline \multicolumn{2}{c}{**Scattering**} \\ \hline Tensile strength of concrete & \\ Coefficient of variation & \(v_{\text{k}}\) & 10\% \\ Correlation length & \(L_{\text{x}}\) & 20 mm \\ \hline \hline \end{tabular}
\end{table} 
Table 2: CPM parameters for the case example. Figure 12: Iterative calculation procedure for the CPM. Figure 13 shows the strains determined with the CPM for different load levels (expressed by the maximum steel stress \(\sigma_{\rm s,max}\)). At the cracks, the concrete stress is zero. The first crack occurs with a steel stress of \(\sigma_{\rm s,max}=100\) N/mm\({}^{2}\) at the point of lowest concrete tensile strength, depending on the distribution of the specimen tensile strength (for comparison, without scattering of the concrete tensile strength \(v_{\rm k}=0\), cracking would not occur until \(\sigma_{\rm s,max}=125\) N/mm\({}^{2}\)). Outside the transmission length, the steel and concrete stresses between the cracks are identical until the completed crack pattern with \(\sigma_{\rm s,max}=114\) N/mm\({}^{2}\) is reached. Up to this loading level five cracks are formed, mainly at the low points of the strength distribution. Due to the non-linear bond law, a further increase in the load level leads to an increase in the bond forces, and thus also to a greater introduction of force into the concrete. This leads to further cracking occurring in the stabilized cracking stage (progressive stabilized cracking). These cracks predominantly occur centrally between the existing cracks, so that under certain circumstances, regions with higher strength values are also affected by crack formation, i.e., \"late\" cracks can emerge under relatively high stresses. Figure 13: CPM case example showing the stress distributions and crack sequences along the model axis for different stress levels. The individual crack spacing and crack widths for each stress level can then be read out from the CPM calculation results.