## 2 Materials and Methods
\begin{array}{c}q=\frac{Q}{T}h^{5/2}\widetilde{K}_{1}^{+}=\Big{(}\frac{2}{3} \widetilde{d}_{1}\Big{)}^{3/2}\Big{(}2\widetilde{L}_{3}-\widetilde{d}\Big{)} \widetilde{d}_{1}=d_{1}/h\widetilde{L}_{3}=L_{3}/h\widetilde{d}=d/h,\\ \widetilde{K}_{1}^{-}=0,\widetilde{K}_{2}^{+}=\frac{\sqrt{T}}{4}\Big{[} \big{(}\widetilde{d}_{1}-\overline{1}\big{)}^{2}+\widetilde{d}_{1}^{2}-2 \Big{]}\widetilde{J}=l/h,\\ \widetilde{K}_{2}^{-}=\frac{\sqrt{T}}{4}\Big{[}2\big{(}\widetilde{d}_{1}- \overline{1}\big{)}^{2}-2\widetilde{d}_{1}\overline{1}+3\overline{1}^{2}-2 \Big{]},\end{array}  \tag{29}\]  
where sign \"\(+\)\" corresponds to tip \(A\) and \"\(-\)\" to tip \(B\). In order to determine the ultimate value of force \(Q\)' causing the crack propagation, we use the energy fracture criterion [15; 16; 17]  
\[\gamma^{*}=\frac{\pi}{8\widetilde{b}^{2}E}\Big{(}K_{1}^{2}+K_{2}^{2}\Big{)}, \tag{30}\]  
where \(E\) is [PERSON]'s modulus, \(\gamma^{*}-\) the effective density of surface energy of the crack for beam material. Using (28) dependence (30) can be rewritten as  
\[Q^{*\pm}=\frac{Q}{\widetilde{b}^{2}\sqrt{E\gamma^{*}h^{3}}}=\frac{8\sqrt{2}}{ 3\sqrt{\pi}}\frac{1}{\sqrt{\big{(}\widetilde{K}_{1}^{\pm}\big{)}^{2}+\big{(} \widetilde{K}_{2}^{\pm}\big{)}^{2}}}. \tag{31}\]  
Case 2. Crack with non-contacting faces. Now we consider the case when crack faces are not contacting. This means that the crack is completely located in the tensile zone and its faces are unloaded. In this case, \(d_{2}>-\frac{1}{3}d_{1}\) must be fulfilled. The part of the axis containing the crack is denoted by \(L_{1}\). On \(L_{1}\), the following boundary conditions take place  
\[\sigma_{yy}^{\pm}=0,\sigma_{phy}^{\pm}=0,x\in L_{1}, \tag{32}\]  
and they also may be written in form (9). Repeating the appropriate transformations we obtain dependence (11). The boundary condition (32) can be rewritten as  
\[\Big{(}\sigma_{yy}-i\alpha_{xy}\Big{)}^{+}+\Big{(}\sigma_{yy}-i\alpha_{xy} \Big{)}^{-}=0,x\in L_{1}.\]  
Taking into account (3) and (11), we come to the linear conjunction problem  
\[\Big{[}\Phi(x)-\frac{1}{2}D(x)\Big{]}^{+}+\Big{[}\Phi(x)-\frac{1}{2}D(x)\Big{]} ^{-}=0,x\in L_{1}.\]  
Solving this problem and taking into account the behavior of function \(\Phi(z)\) for large values of \(|z|\) (5) we have  
\[\Phi(z)=\frac{1}{2}D(z)+\sum_{j=0}^{3}\alpha_{j}z^{j}/X(z), \tag{33}\]  
where  
\[ \begin{array}{c}\alpha_{3}=2 ia,\alpha_{2}=2b-ia(d_{1}-d_{2}),\alpha_{1}=-ia \Big{[}2h^{2}+\frac{(d_{1}-d_{2})^{2}}{4}\Big{]}-b(d_{1}+d_{2}),\\ \alpha_{0}=ia(d_{1}+d_{2})\Big{[}h^{2}-\frac{1}{8}(d_{1}-d_{2})^{2}\Big{]}-b \frac{(d_{1}-d_{2})^{2}}{4}.\end{array}  \tag{34}\]  
In this case, we find the stress intensity factor using Formulas (26) and (33)  
\[K_{1}^{\pm}=q\widetilde{K}_{j}^{\pm}, \tag{35}\]where  
\[\widetilde{K}_{1}^{+}=\frac{\sqrt{l}l}{2}\big{(}2\widetilde{L}_{3}-\widetilde{d} \big{)}\big{(}2\widetilde{d}_{1}-\widetilde{l}\big{)}\widetilde{K}_{1}^{-}= \frac{\sqrt{l}l}{2}\big{(}2\widetilde{L}_{3}-\widetilde{d}\big{)}\big{(}2 \widetilde{d}_{1}-3\widetilde{l}\big{)}.\]  
The coefficients \(q\) and \(\widetilde{K}_{2}^{\pm}\) are expressed by Formulas (29) at \(\widetilde{l}<\frac{2}{3}\widetilde{d}_{1}\). The ultimate value of force we obtain from formula  
\[Q^{\pm}=\frac{8\sqrt{2}}{3\sqrt{\pi}}\frac{1}{\sqrt{\left(\widetilde{K}_{1}^{ \pm}\right)^{2}+\left(\widetilde{K}_{2}^{\pm}\right)^{2}}}. \tag{36}\]  
Case 3. Crack with perfectly contacting faces. Now we consider the case when crack faces were in perfect contact \((d_{1}<0)\). The part of the \(Ox-\) axis where the contact take place is denoted by \(L\). In this case, we have the following boundary conditions  
\[\sigma_{yy}^{+}=\sigma_{yy}^{-}\sigma_{xy}^{\pm}=0,v^{+}-v^{-}=0,x\in L. \tag{37}\]  
Similarly to the previous case, we write boundary conditions at \(L\) in form (6) and come to the linear conjunction problem (9). Solution of this problem has the form (11).