# Insights into Bending Stiffness Modeling of Elementary Flexure Hinges
## 4 Bending Stiffness Model of a Flexure Hinge
### Solution of the Model for an Elliptical Flexure Hinge  
An analytical expression for \(\widetilde{\theta}(\widetilde{x})\) can be found from solving the integral in Equation (14), according to the results presented by [PERSON] et al. [50], with some reworking. For the sake of conciseness, a compact form is here proposed:  
\[\widetilde{\theta}(\widetilde{x})=\frac{\rho^{3}M_{f}}{E\;a_{x}^{2}}\;f\big{(} \widetilde{x},\beta_{y}\big{)}+\mu_{Q}\;\rho^{3}g\big{(}\widetilde{x},\beta_{ y}\big{)}. \tag{37}\]  
where \(\mu_{Q}\) is defined, as analogously done for \(\mu_{F}\) in Equation (24) referring to \(F\), as \(\mu_{Q}=\frac{Q}{E\;I\;a_{x}}\). In Figures 9 and 10, functions \(f\big{(}\widetilde{x},\beta_{y}\big{)}\) and \(g\big{(}\widetilde{x},\beta_{y}\big{)}\) are represented for some discrete values of \(\beta_{y}\). As known in the literature [51], it is:  
\[\sigma(x)=-E\;\kappa(x)\;y(x), \tag{38}\]where \(\sigma(x)\) represents the axial stress in the beam. The negative sign is due to the convention chosen; namely, a positive curvature means a counterclockwise rotation of the beam with the upper fibers compressed. Conventionally, a compression tension is considered negative. Equation (38) can be updated with the simplification made on \(\kappa(x)\) and can be made dimensionless as follows:  
\[\overline{\sigma}(\widetilde{x})=-\frac{E}{\rho}\ \widetilde{\theta}^{\prime}( \widetilde{x})\ \widetilde{y}(\widetilde{x}). \tag{39}\]  
In the first place, it is assumed that the most stressed section of the flexure is in \(\widetilde{x}\ =\ 0\). When the maximum stress in the elastic domain is set in such a section, given by the yield stress \(\sigma_{sn}\) of the material, the maximum moment that can be applied to the free end of the beam can be found for positive values of \(y(x)\) as a function of \(\rho\), \(\beta_{y}\) and \(Q\):  
\[M_{f,\max}=-a_{x}Q\pm\frac{2}{3}a_{x}^{2}l\left(\frac{\beta_{y}}{\rho}\right)^ {2}\sigma_{sn}, \tag{40}\]  
where the positive sign in Equation (40) has to be chosen to have a counterclockwise rotation, whereas the negative sign has to be chosen to have a clockwise rotation. Such equation results from a combination of Equations (5), (12), (14), (21) and (24) with Equation (39). It is demonstrated here below that the hypothesis of having the maximum stress in \(\widetilde{x}=0\). From this relation, Equation (40) is more valid when small values of \(\mu_{Q}\) are considered in the module. First of all, a development in the Taylor series of the stress \(\overline{\sigma}(\widetilde{x})\) in Equation (39) up to the second order allows to look for stationary points. The maximum of \(\overline{\sigma}(\widetilde{x})\) is obtained in:  
\[\widetilde{x}_{s}=-\frac{\beta_{y}}{2}\frac{1}{1+\frac{M_{f}}{l\ \widetilde{a}_{x}^{2}\mu_{Q}\ E}}. \tag{41}\]  
Figure 10: Trend of function \(g\) in terms of \(\widetilde{x}\) and \(\beta_{y}\). The expression of the maximum moment \(M_{f,\max}\), assuming that a positive moment generates a counterclockwise rotation inducing a compressive tension on the fibers with positive \(y(\widetilde{x})\), evaluated when \(\overline{\sigma}(\widetilde{x}_{s})=-\sigma_{sn}\), is:  
\[M_{f,\max}=-a_{x}Q\pm\left\{\frac{1}{3}a_{x}^{2}l\Big{(}\frac{\beta_{y}}{\rho} \Big{)}^{2}\sigma_{sn}+\sqrt{\left[\frac{1}{3}a_{x}^{2}l\Big{(}\frac{\beta_{y} }{\rho}\Big{)}^{2}\sigma_{sn}\right]^{2}-\frac{\beta_{y}}{4}(a_{x}Q)^{2}} \right\}. \tag{42}\]  
The positive sign in Equation (42) has to be chosen to have a counterclockwise rotation, as already mentioned. The term in square brackets exists if \(|\mu_{Q}|\leq 3\frac{\beta_{y}^{3/2}}{\rho^{2}}\frac{\sigma_{sn}}{E}\). The constraint on \(|\mu_{Q}|\) imposes a limit to the possible values of \(Q\).