## 3 Results
Substituting in expression (12) the values included in it, we obtain  
\[\sigma_{F}=\frac{F_{t}}{bm}\;\frac{1}{\cos\alpha}\;\left(\frac{\epsilon\gamma _{h}\cos\alpha_{\alpha}}{\gamma_{S}^{2}}-\frac{\sin\alpha_{\alpha}}{\gamma_{S} }\right)\;K_{T} \tag{23}\]  
Denoted by  
\(Y_{F}=\frac{1}{\cos\alpha}\left(\frac{\epsilon\gamma_{h}\cos\alpha_{\alpha} }{\gamma_{S}^{2}}-\frac{\sin\alpha_{\alpha}}{\gamma_{S}}\right)K_{T}\)--tooth shape coefficient, and introducing design load coefficients \(K_{F\beta}\) and \(K_{FV}\), we obtain a formula for the verification calculation of straight wheel teeth for bending fatigue resistance  
\[\sigma_{F}=\frac{F_{t}Y_{F}}{bm}\;K_{F\beta}K_{FV}\leq[\sigma_{F}] \tag{24}\]  
Substituting in the form (14) \(F_{t}=\frac{2T}{d}=\frac{2T}{mz}\), we obtain  
\[\sigma_{F}=\frac{2 TY_{F}}{bm2}\;K_{F\beta}K_{FV}\leq[\sigma_{F}] \tag{25}\]  
To obtain the formula for the design calculation it is necessary to substitute in expression (15) \(b=\psi_{bd}d_{1}=\psi_{bd}mz_{1}\) and solving it with respect to the tooth modulus \(m\)  
\[F_{t}^{\prime}=F_{n}\cos\alpha_{\alpha}=F_{t}\cos\alpha_{\alpha}/\cos\alpha\]0  
Denoting by \(K_{m}=\sqrt[3]{2K_{FV}}=1.4\)--auxiliary coefficient  
We finally obtain  
\[F_{t}^{\prime}=F_{n}\cos\alpha_{\alpha}=F_{t}\cos\alpha_{\alpha}/\cos\alpha\]1  
where \(T_{1}\)--torque on the gear shaft, N\(\cdot\cdot\cdot\)m; \(z_{1}\)--number of gear teeth;  
\(\big{[}\sigma_{F_{1}}\big{]}\)--permissible bending stress for gear material, MPa;  
\(\psi_{bd}\)--gear width coefficient relative to the diameter of the dividing circle (taken from the data in Table 1). Tooth shape coefficient \(Y_{F}\) (\(Y_{F_{1}}>Y_{F_{2}}\)). To ensure approximately equal operability of the gear and wheel teeth, the gear is made stronger than the wheel. Gear and wheel teeth have equal resistance to bending fatigue under the condition  
\[F_{t}^{\prime}=F_{n}\cos\alpha_{\alpha}=F_{t}\cos\alpha_{\alpha}/\cos\alpha\]2  
III. Next, we calculate the contact fatigue resistance of the active surfaces of the teeth of helical gears. The derivation of the formula for the check calculation is based on the replacement of helical wheels by equivalent spur wheels [4; 5; 6; 7]. The calculation scheme for the contact fatigue resistance calculation of the active surfaces of the teeth of helical gear wheel teeth is shown in \[F_{t}^{\prime}=F_{n}\cos\alpha_{\alpha}=F_{t}\cos\alpha_{\alpha}/\cos\alpha\]3 Figure 6 also shows the forces acting on the tooth of helical wheel and the formation of an equivalent spur wheel. The calculation of helical gear teeth in the simplest form can be reduced to the calculation of spur wheels (see paragraphs I and II of this article), the strengths of which are mutually equivalent to each other [4; 5; 6; 7]. In this case, the initial formula for determining contact stresses is as follows  
\[F_{t}^{\prime}=F_{n}\cos\alpha_{\alpha}=F_{t}\cos\alpha_{\alpha}/\cos\alpha\]4  
The reduced radius of curvature of tooth profiles of equivalent spur wheels is determined with regard to Formula (2) by the formula  
\[F_{t}^{\prime}=F_{n}\cos\alpha_{\alpha}=F_{t}\cos\alpha_{\alpha}/\cos\alpha\]5  
Taking into account Expressions (5) and (19), we obtain Formula (18) in the form  
\[F_{t}^{\prime}=F_{n}\cos\alpha_{\alpha}=F_{t}\cos\alpha_{\alpha}/\cos\alpha\]6  
Denoted by:  
\(Z_{H}=\sqrt{\frac{2(\text{cos}\,\beta)^{2}}{\text{sin}\,2\alpha_{n}}}=1.76 \text{cos}\,\beta\)--coefficient that takes into account the shape of the mating tooth surfaces;  
\(Z_{M}=\sqrt{\frac{E_{red}}{\pi\ (1-\mu^{2})}}\)-coefficient, taking into account mechanical properties of materials of contacting wheels (for steel wheels \(Z_{M}=\)275 MPa\({}^{1/2}\));  
\(Z_{\epsilon}=\sqrt{\frac{1}{\varepsilon_{a}K_{\epsilon}}}\)-coefficient taking into account the total length of contact lines depending on  
\(\varepsilon_{a}=\left(1.88-3.2\left(\frac{1}{z_{1}}\pm\frac{1}{z_{2}}\right) \right)\text{cos}\,\beta\) and \(K_{\epsilon}=0.9\ldots 0.95\).