## 2 NSC Formed by CRMs with Quadratic Polynomial Trajectory
Based on the integrated model in [26] and the purpose of cubic nonlinear restoring force, the trajectory of the roller center with respect to \(u\) is defined as the quadratic function:  
\[x(u)=au^{2} \tag{1}\]  
where \(x\) is the compression deformation length of the horizontal spring caused by \(u\), and \(a\) is the quadratic coefficient of the trajectory. The pressure of the horizontal spring keeps the roller in contact with the cam. So, the cam generates the force \(f_{\mathrm{c}}\) on the roller along the normal direction, which is equal to the force of the roller acting on the cam and opposite in direction. The horizontal component force \(f_{\mathrm{h}}\) of \(f_{\mathrm{c}}\) depends on the stiffness \(k_{\mathrm{n}}\) and total deformation length of the horizontal spring, which can be obtained as:  
\[f_{\mathrm{h}}(u)=k_{\mathrm{n}}\Big{(}x-a\Delta y^{2}-\delta_{\mathrm{n}} \Big{)} \tag{2}\]  
where \(\delta_{\mathrm{n}}\) is the pre-compressed length of the horizontal spring without the rate mass, and \(\Delta y\), the static vertical displacement caused by rated mass, is selected as the maximum working displacement. The angle between the horizontal force and the normal vector of the roller center trajectory is defined as \(\beta\). Its tangent value can be obtained as \(\mathrm{tan}\beta=2 au\) by differentiating \(x\) with respect to \(u\). Then, the vertical component force \(F_{\mathrm{n}}\) on the roller, namely the force exerted on the upper loading platform by the QCRM, can be gained as:  
\[F_{\mathrm{n}}(u)=f_{\mathrm{h}}\tan\beta=2a^{2}k_{\mathrm{n}}u^{3}-2 ak_{ \mathrm{n}}(a\Delta y^{2}+\delta_{\mathrm{n}})u \tag{3}\]  
Figure 3: Static analysis of cam-roller mechanism with a quadratic polynomial trajectory (QCRM). (a) Roller at static equilibrium position; (b) roller deviating from equilibrium position. When the horizontal spring has the largest compression length, \(\tan\beta=0\) which satisfies the requirement that there is no vertical component force at the equilibrium position. Nondimensionalizing Equation (3), the non-dimensional force-displacement relationship can be expressed as:  
\[\overline{F}_{\mathrm{n}}(\overline{u})=2\overline{a}^{2}\overline{u}^{3}-2 \overline{a}^{2}\overline{u}-2\overline{a}\overline{\delta}_{\mathrm{n}} \overline{u} \tag{4}\]  
where \(\overline{F}_{\mathrm{n}}=F_{\mathrm{n}}/\left(k_{n}\Delta y\right)\) is the non-dimensional vertical force on the roller, \(\overline{u}=u/\Delta y\) is the non-dimensional relative displacement, \(\overline{a}=a\Delta y\) is the non-dimensional quadratic coefficient, and \(\overline{\delta}_{\mathrm{n}}=\delta_{\mathrm{n}}/\Delta y\) is the non-dimensional pre-compressed length of the horizontal spring. Next, differentiating Equation (4) with respect to \(\overline{u}\), the non-dimensional stiffness \(\overline{K}_{\mathrm{n}}\) is also obtained as:  
\[\overline{K}_{\mathrm{n}}(\overline{u})=6\overline{a}^{2}\overline{u}^{2}-2 \overline{a}^{2}-2\overline{a}\overline{\delta}_{\mathrm{n}} \tag{5}\]  
The non-dimensional force-displacement characteristics and non-dimensional stiffness for different \(\overline{\delta}_{\mathrm{n}}\) and \(\overline{a}\) are plotted in Figures 4 and 5. At the static equilibrium position (\(\overline{u}=0\)), the restoring force provided by the QCRM is equal to zero and changes direction, and correspondingly the negative stiffness is symmetrical and reaches the minimum. Then, with the displacement moving away from the symmetric point, both the force and stiffness increase. From Figure 4, it can be found that for a fixed \(\overline{a}\), the change of \(\overline{\delta}_{\mathrm{n}}\) will shift the stiffness curves longitudinally, but not change their shape. Raising \(\overline{\delta}_{\mathrm{n}}\) can broaden the negative stiffness region and provide a lower negative stiffness. In Figure 5, \(\overline{a}\) is varied, but \(\overline{\delta}_{\mathrm{n}}\) is fixed.