2. D-H methodology

In this section, the inverse kinematics is solved using the D-H methodology. The geometric description of the single leg and a UPS leg is shown in Fig. 4a. The d_1i and d_2i are the displacements of the linear actuators, and l₁ is the link length of the transmission link. $\theta _{1i}$ and $\theta _{2i}$ are the angle of rotational joints C and D, respectively.

Figure 4.

Frame assignment for (a) a single leg and (b) passive load-balancing UPS leg.

The frame assignment for the passive load-balancing UPS leg is shown in Fig. 4b. The first universal joint is assigned with frame 0 and frame 1, followed by the prismatic joint assigned with frame 2, followed by the second universal joint, which is assigned with frames 3 and 4. The final frame 5 is assigned to the mobile platform. There are five joints serially connected, starting from two rotational joints with the same origin, then a prismatic joint and two rotational joints with the same origin. The respective joint axes $x_{n}$ and $z_{n}$ , are assigned to the joints. The angle for the rotational joint is $\theta _{n}$ , and the displacement for the linear joint is $d_{n}$ . The D-H link and joint parameters are listed in Table I.

Table I.

D-H table for passive UPS leg of the 3-PPRS manipulator.

The following equation gives the basic link transformation matrix,

(1) \begin{equation}{ }^{i-1}{T_{i}}{} = \left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c} C\theta _{i} & -{S\theta _{i}}C\alpha _{i} & S\theta _{i}S\alpha _{i} & {a_{i}}C\theta _{i}\\ S\theta _{i} & C\theta _{i}C\alpha _{i} & -C\theta _{i}S\alpha _{i} & {a_{i}}S\theta _{i}\\ 0 & S\alpha _{i} & C\alpha _{i} & d_{i}\\ 0 & 0 & 0 & 1 \end{array}\right]\end{equation}

The D-H transformation matrices are derived by substituting the D-H parameters in the fundamental link transformation matrix. The D-H transformation matrices are as follows:

(2) \begin{equation}{ }^{0}{T_{3}}{}\,{=}\,{ }^{0}{T_{2}}{}\times { }^{2}{T_{3}}{}\,{=}\,\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c} C\theta _{1} & -{S\theta _{1}}C\theta _{2} & S\theta _{1}S\theta _{2} & -{S\theta _{1}}C\theta _{2}d_{3}\\ S\theta _{1} & C\theta _{1}C\theta _{2} & -C\theta _{1}S\theta _{2} & C\theta _{1}C\theta _{2}d_{3}\\ 0 & S\theta _{2} & C\theta _{2} & S\theta _{2}d_{3}\\ 0 & 0 & 0 & 1 \end{array}\right]\end{equation}

The stiffness analysis was carried out on the manipulator in its home position, with the mobile platform at zero orientation. In other words, without any tilt. As a result, the $\theta _{4};\, \theta _{5}$ ; and $\theta _{6}$ influences on end-effector orientation were considered to be zero.

(3) \begin{equation}{ }^{0}{T_{6}}{}\,{=}\,{ }^{0}{T_{3}}{}\times { }^{3}{T_{6}}{} = \left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c} S\theta _{1}S\theta _{2} & C\theta _{1} & -S\theta _{1}C\theta _{2} & -{S\theta _{1}}C\theta _{2}d_{3}\\ -C\theta _{1}S\theta _{2} & S\theta _{1} & C\theta _{1}C\theta _{2} & C\theta _{1}C\theta _{2}d_{3}\\ C\theta _{2} & 0 & S\theta _{2} & S\theta _{2}d_{3}\\ 0 & 0 & 0 & 1 \end{array}\right]\end{equation}

The directions of the joint axes, $P_{i-1}$ , derived using the equation $P_{i-1}={ }^{0}{R_{i-1}}{}\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]$ , are as follows:

The rotation matrix,

(4) \begin{equation}{ }^{0}{R_{0}}{}\,{=}\,\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]{;}\,P_{0}={ }^{0}{R_{0}}\widehat{u}={ }^{0}{R_{0}}{}\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right];\end{equation}

(5) \begin{equation}{ }^{0}{R_{1}}{}\,{=}\,\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c} -S\theta _{1} & 0 & C\theta _{1}\\ C\theta _{1} & 0 & S\theta _{1}\\ 0 & 1 & 0 \end{array}\right]{;}\,P_{1}={ }^{0}{R_{1}}{}\widehat{u}={ }^{0}{R_{1}}{}\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=\left[\begin{array}{c} C\theta _{1}\\ S\theta _{1}\\ 0 \end{array}\right];\end{equation}

(6) \begin{equation}{ }^{0}{R_{2}}{}\,{=}\,\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c} S\theta _{1}S\theta _{2} & C\theta _{1} & -{S\theta _{1}}C\theta _{2}\\ -C\theta _{1}S\theta _{2} & S\theta _{1} & C\theta _{1}C\theta _{2}\\ C\theta _{2} & 0 & S\theta _{2} \end{array}\right]{;}\,P_{2}={ }^{0}{R_{2}}{}\widehat{u}={ }^{0}{R_{2}}{}\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=\left[\begin{array}{c} -{S\theta _{1}}C\theta _{2}\\ C\theta _{1}C\theta _{2}\\ S\theta _{2} \end{array}\right];\end{equation}

(7) \begin{equation}{ }^{0}{R_{3}}{}\,{=}\,\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c} C\theta _{1} & -{S\theta _{1}}C\theta _{2} & S\theta _{1}S\theta _{2}\\ S\theta _{1} & C\theta _{1}C\theta _{2} & -C\theta _{1}S\theta _{2}\\ 0 & S\theta _{2} & C\theta _{2} \end{array}\right];\, P_{3}={ }^{0}{R_{3}}{}\widehat{u}={ }^{0}{R_{3}}{}\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=\left[\begin{array}{c} S\theta _{1}S\theta _{2}\\ -C\theta _{1}S\theta _{2}\\ C\theta _{2} \end{array}\right];\end{equation}

(8) \begin{equation}{ }^{0}{R_{4}}{}\,{=}\,\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c} -{S\theta _{1}}C\theta _{2} & S\theta _{1}S\theta _{2} & C\theta _{1}\\ C\theta _{1}C\theta _{2} & -C\theta _{1}S\theta _{2} & S\theta _{1}\\ S\theta _{2} & C\theta _{2} & 0 \end{array}\right]{;}\,P_{4}={ }^{0}{R_{4}}{}\widehat{u}={ }^{0}{R_{4}}{}\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=\left[\begin{array}{c} C\theta _{1}\\ S\theta _{1}\\ 0 \end{array}\right];\end{equation}

(9) \begin{equation}{ }^{0}{R_{5}}{}\,{=}\,\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c} S\theta _{1}S\theta _{2} & C\theta _{1} & -{S\theta _{1}}C\theta _{2}\\ -C\theta _{1}S\theta _{2} & S\theta _{1} & C\theta _{1}C\theta _{2}\\ C\theta _{2} & 0 & S\theta _{2} \end{array}\right]{;}\,P_{5}={ }^{0}{R_{5}}{}\widehat{u}={ }^{0}{R_{5}}{}\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=\left[\begin{array}{c} -{S\theta _{1}}C\theta _{2}\\ C\theta _{1}C\theta _{2}\\ S\theta _{2} \end{array}\right];\end{equation}

The position vectors ${ }^{i-1}{P_{6}}{}$ , derived by applying the equation, ${ }^{n}{P_{i}}{}={ }^{0}{T_{i}}{}O_{i}-{ }^{0}{T_{n}}{}O_{i}$ , are listed below

(10) \begin{equation}{ }^{0}{P_{6}}{}={ }^{0}{T_{6}}{}O_{6}-{ }^{0}{T_{0}}{}O_{6}=\left[\begin{array}{c} -{S\theta _{1}}C\theta _{2}d_{3}\\ C\theta _{1}C\theta _{2}d_{3}\\ S\theta _{2}d_{3}\\ 1 \end{array}\right]-\left[\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right]=\left[\begin{array}{c} -{S\theta _{1}}C\theta _{2}d_{3}\\ C\theta _{1}C\theta _{2}d_{3}\\ S\theta _{2}d_{3}\\ 0 \end{array}\right];\end{equation}

(11) \begin{equation}{ }^{1}{P_{6}}{}={ }^{0}{T_{6}}{}O_{6}-{ }^{0}{T_{1}}{}O_{6}=\left[\begin{array}{c} -{S\theta _{1}}C\theta _{2}d_{3}\\ C\theta _{1}C\theta _{2}d_{3}\\ S\theta _{2}d_{3}\\ 1 \end{array}\right]-\left[\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right]=\left[\begin{array}{c} -{S\theta _{1}}C\theta _{2}d_{3}\\ C\theta _{1}C\theta _{2}d_{3}\\ S\theta _{2}d_{3}\\ 0 \end{array}\right];\end{equation}

The Jacobian matrix does not require the position vector ${ }^{2}{P_{6}}{}$ . Because it corresponds to the prismatic joint, only the direction of the joint axis $P_{2}$ is required.

Also,

(12) \begin{equation}{ }^{3}{P_{6}}{}={ }^{4}{P_{6}}{}={ }^{5}{P_{6}}{}=\left[\begin{array}{c} 0\\ 0\\ 0 \end{array}\right];\end{equation}

3. Stiffness analysis using Jacobian matrix

The Jacobian matrix for a manipulator with non-planar links is obtained using the screw theory principle. The input/actuator velocities are transformed to the end-effector output velocity using a Jacobian velocity matrix. The Jacobian matrix can also be used to find singular configurations of the manipulator in which it gains one or more DOF.

The end-effector’s twist is related to the joint rate vector $\dot{\boldsymbol{\theta}}$ by

(13) \begin{equation}J\dot{\boldsymbol{\theta}}=t\end{equation}

(14) \begin{equation} \dot{\boldsymbol{\theta}}=\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c} \dot{d_{1i}} & {\dot{d_{2i}}} & \dot{\theta _{1i}} & \dot{\theta _{4i}} & \dot{\theta _{5i}} & \dot{\theta _{6i}} \end{array}\right]^{T};i \text{denotes leg }i, t=\left[\begin{array}{c} \omega \\ \dot{\boldsymbol{c}} \end{array}\right] \end{equation}

where J is the screw-based velocity Jacobian matrix for UPS configuration

(15) \begin{equation} J=\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c} P_{0}\times { }^{0}{P}{}_{6} & P_{1}\times { }^{1}{P}{}_{6} & P_{2} & P_{3}\times { }^{3}{P}{}_{6} & P_{4}\times { }^{4}{P}{}_{6} & P_{5}\times { }^{5}{P}{}_{6}\\ P_{0} & P_{1} & 0 & P_{3} & P_{4} & P_{5} \end{array}\right] \end{equation}

$P_{i}$ is the joint i’s direction vector. ${ }^{n}{P_{i}}{}$ is the position vector pointing from the i ^th joint to the end-effector. The values of position vectors and the direction of joint axes are imported from Section 2 for Jacobian calculations.

For the 1^st joint (revolute),

(16) \begin{equation} J_{1}=\left[\begin{array}{c} P_{0}\times { }^{0}{P}{}_{6}\\ P_{0} \end{array}\right] \end{equation}

(17) \begin{equation} J_{1}=\left[\begin{array}{c} \left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]\times \left[\begin{array}{c} -{S\theta _{1}}C\theta _{2}d_{3}\\ C\theta _{1}C\theta _{2}d_{3}\\ S\theta _{2}d_{3} \end{array}\right]\\[20pt] \left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right] \end{array}\right]=\left[\begin{array}{c} {-C\theta _{1}}C\theta _{2}d_{3}\\ -S\theta _{1}C\theta _{2}d_{3}\\ 0\\ 0\\ 0\\ 1 \end{array}\right] \end{equation}

The 3^rd joint (prismatic),

(18) \begin{equation} J_{i}=\left[\begin{array}{c} P_{i-1}\\ 0 \end{array}\right];\, J_{3}=\left[\begin{array}{c} \begin{array}{c} -{S\theta _{1}}C\theta _{2}\\ C\theta _{1}C\theta _{2}\\ S\theta _{2} \end{array}\\ 0\\ 0\\ 0 \end{array}\right] \end{equation}

Similarly, the Jacobian for the other joints $J_{4},\,J_{5},\,\text{and}\,J_{6}$ is computed.

(19) \begin{equation}J_{2}=\left[\begin{array}{c} {S\theta _{1}}S\theta _{2}d_{3}\\ -C\theta _{1}S\theta _{2}d_{3}\\ C\theta _{2}d_{3}\\ C\theta _{1}\\ S\theta _{1}\\ 0 \end{array}\right] J_{4}=\left[\begin{array}{c} 0\\ 0\\ 0\\ S\theta _{1}S\theta _{2}\\ -C\theta _{1}S\theta _{2}\\ C\theta _{2} \end{array}\right] ;\, J_{5}=\left[\begin{array}{c} 0\\ 0\\ 0\\ C\theta _{1}\\ S\theta _{1}\\ 0 \end{array}\right] ;\, J_{6}=\left[\begin{array}{c} 0\\ 0\\ 0\\ -{S\theta _{1}}C\theta _{2}\\ C\theta _{1}C\theta _{2}\\ S\theta _{2} \end{array}\right]\end{equation}

The Jacobian matrix for the UPS configuration is given below, $J=\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c} J_{1} & J_{2} & J_{3} & J_{4} & J_{5} & J_{6} \end{array}\right]$

(20) \begin{equation}J_{UPS}=\left[\begin{array}{c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c@{\kern+12pt}c} \begin{array}{c} {-C\theta _{1}}C\theta _{2}d_{3}\\ -S\theta _{1}C\theta _{2}d_{3}\\ 0\\ 0\\ 0\\ 1 \end{array} & \begin{array}{c} {S\theta _{1}}S\theta _{2}d_{3}\\ -C\theta _{1}S\theta _{2}d_{3}\\ C\theta _{2}d_{3}\\ C\theta _{1}\\ S\theta _{1}\\ 0 \end{array} & \begin{array}{c} \begin{array}{c} -{S\theta _{1}}C\theta _{2}\\ C\theta _{1}C\theta _{2}\\ S\theta _{2} \end{array}\\ 0\\ 0\\ 0 \end{array} & \begin{array}{c} 0\\ 0\\ 0\\ S\theta _{1}S\theta _{2}\\ -C\theta _{1}S\theta _{2}\\ C\theta _{2} \end{array} & \begin{array}{c} 0\\ 0\\ 0\\ C\theta _{1}\\ S\theta _{1}\\ 0 \end{array} & \begin{array}{c} 0\\ 0\\ 0\\ -{S\theta _{1}}C\theta _{2}\\ C\theta _{1}C\theta _{2}\\ S\theta _{2} \end{array} \end{array}\right]\end{equation}

3.1. Stiffness map results

The parallel manipulator configuration proposed in this article comprises two main components: the constraining leg, which can be considered a serial (passive), and the 3-PPRS manipulator as a parallel (active) mechanism. As a result, the overall stiffness matrix of the parallel mechanism with a constraining leg can be obtained by adding the stiffness matrices of the parallel and serial mechanisms [Reference Zhang and Gosselin20]. The stiffness matrix for the manipulator, including the UPS constraining leg, is given by,

(21) \begin{equation}K=K_{UPS}+K_{\textit{parallel}};\end{equation}

where

(22) \begin{equation} K_{UPS}=\left(J_{UPS}^{\prime}\right)^{-T}K_{dia}\left(J_{UPS}^{\prime}\right)^{-1} \end{equation}

By the principle of the virtual joint method [Reference Pashkevich, Klimchik and Chablat21], the diagonal stiffness matrix $K_{dia}$ is given as follows:

(23) \begin{equation} K_{dia}=diag\left[0,K_{6},0,0,0,0\right] \end{equation}

Within the workspace, the stiffness of the UPS leg $K_{UPS}$ in the X-Y plane is determined. The D-H methodology discussed in Section 2 is used to solve the inverse kinematics of the UPS configuration. With the solved UPS parameters (joint angles and displacement), the Jacobian $J_{UPS}$ is computed, and hence the stiffness of the UPS leg.

As the spatial motion actuators of the 3-PPRS manipulator are aligned along Z axis, the configuration has high torsional stiffness. Hence, the translational stiffness alone is studied. The stiffness values Kx and Kz in the X-Y plane are plotted in Fig. 5. The stiffness values Kx are high around the region of the home position, and the values are five times larger compared to Kz values. The stiffness in X-axis is a valuable addition to this parallel manipulator (3-PPRS) because the existing parallel manipulator has one of the planar motion actuators aligned along the X-axis. Including UPS leg in the 3-PPRS parallel manipulator will enhance its translational stiffness.

Figure 5.

Stiffness plot for Kx and Kz in the X-Y plane for the UPS configuration.

4. Static structural analysis of 3-PPRS parallel manipulator

Using ANSYS Workbench software, the structural analysis of the three different 3-PPRS parallel manipulator architectures, namely the 3-PPRS manipulator with planar transmission links, the 3-PPRS manipulator with spatial transmission links, and the 3-PPRS manipulator with spatial transmission links in addition to the passive load-balancing UPS leg, is compared. In this article, the load balancing of the entire manipulator is of prime interest. The vertical hydraulic actuators predominantly carrying the load against gravity are only considered for static structural analysis. On the other hand, the horizontal electrical actuators not carrying any load against gravity are excluded from the analysis, simplifying the computer-aided design (CAD) model.

Three architectures are structurally compared in the static structural analysis [Reference Zhang, Tang, Yan, Cui, Zhang and Yao22] using total deformation and equivalent stress (Von Mises). In the ANSYS Workbench Static Structural tool, the CAD models of the 3-PPRS manipulator are imported from SolidWorks as step files, and the manipulator’s material was set to mild steel (MS). The fixed platform is fastened to the ground, and a load of 5000 N is applied to the center of the mobile platform. The results of the static structural simulation are shown in Figs. 6–7.

Figure 6.

The total deformation for the three different architectures of the 3-PPRS manipulator.

Figure 7.

The equivalent stress for three different architectures of the 3-PPRS manipulator.

4.1. Results of structural analysis

From Tables II-III, it is clear that compared to the other two 3-PPRS manipulator architectures, the one with both spatial transmission links and a passive load balancing leg has the least total deformation value (0.02581 mm) and the lowest Von Mises equivalent stress (72.30 MPa). The proposed architecture is designed safe because the selected material (MS) has a tensile yield strength of 250 MPa [Reference Jha, Ranjan, Kumar, Sharma and Scholar23]. Thus, the architecture of a 3-PPRS manipulator with spatial transmission links and a passive load-balancing UPS leg is determined to be effective based on the analytical results and selected for manufacturing.

Table II.

Total deformation of the 3-PPRS manipulator.

Table III.

Equivalent stress of 3-PPRS manipulator.

5. Dynamic system modeling

Due to the manipulator assembly’s symmetrical nature, a single-leg assembly model is considered [Reference Plummer24]. The specimen model includes the model of the transmission link, push rod, and specimen. Hence, the stiffness of the link, push rod, and specimen are added, as shown in Fig. 8. Specimen stiffness indicates the stiffness of the specimen, connecting link, and the pushrod, and it is represented as $K_{sp}$ . Similar is the case with specimen damping $C_{sp}$ . The oil column in the cylinder is compressible, and it is represented by a stiffness of $K_{a}$ and damping of $C_{a}.$

Figure 8.

The mechanical model of a single actuator leg and a passive hydraulic cylinder.

The UPS leg is modeled as a passive hydraulic cylinder in Case 1 and a passive damper in Case 2. In the case of a passive damper, it is modeled as a non-linear spring. It means the spring force will be larger at the compressed state and minimum at the extended state [Reference Kluever25]. The spring force is proportional to the displacement by the passive damper, which will be the difference between the total spring length and actual displacement, that is, ( $z_{max}-z_{t}$ ). The mass is considered negligible because of its passive nature.

5.1. Case 1 modeling of UPS leg as a passive hydraulic cylinder

The mechanical model of the single actuator leg and the UPS leg as a passive hydraulic cylinder is shown in Fig. 8. The passive hydraulic cylinder contains a linear spring ( $K_{b}$ ) and damper ( $C_{b}$ ).

The free-body diagram of the above mechanical model is shown in Fig. 9.

Figure 9.

The free-body diagram of the mechanical model with UPS leg as a passive hydraulic cylinder.

Upward direction is the positive sign convention for displacements $z_{p}$ and $z_{t}$ . Summing all external forces with an upward direction as the positive sign convention and applying Newton’s second law, for

(24) \begin{equation} \text{Mass }(M_{p}): +\uparrow \sum F=-K_{sp}(z_{p}-z_{t})-C_{sp}(\dot{z_{p}}-\dot{z_{t}})-K_{a}z_{p}-C_{a}\dot{z_{p}}+F_{h}=M_{p}\ddot{z_{p}}\end{equation}

(25) \begin{equation} \text{Mass }(M_{t}): +\uparrow \sum F=K_{sp}(z_{p}-z_{t})+C_{sp}(\dot{z_{p}}-\dot{z_{t}})-K_{sa}z_{t}-C_{sa}\dot{z_{t}}=M_{t}\ddot{z_{t}}\end{equation}

Rearranging these equations with dynamic variables ( $z_{p}$ and $z_{t}$ ) on the left-hand side and the input variable $F_{h}$ on the right-hand side,

(26) \begin{equation} M_{p}\ddot{z_{p}}+C_{sp}\left(\dot{z_{p}}-\dot{z_{t}}\right)+K_{sp}\left(z_{p}-z_{t}\right)+K_{a}z_{p}+C_{a}\dot{z_{p}}=F_{h} \end{equation}

(27) \begin{equation} M_{t}\ddot{z_{t}}+C_{sp}\left(\dot{z_{t}}-\dot{z_{p}}\right)+K_{sp}\left(z_{t}-z_{p}\right)+K_{sa}z_{t}+C_{sa}\dot{z_{t}}=0 \end{equation}

Equations (26) and (27) represent the mathematical model of the single actuator leg and the load-balancing UPS leg. As there are two inertia elements, $M_{t}\& M_{p}$ , the model consists of two second-order ordinary differential equations (ODEs). Since both ODEs are coupled, solving one ODE separately from the other is impossible. In Case 1, the model is linear because we have assumed linear stiffness and damper elements. In general, this condition is suitable for an inherently stable system. Intuition tells us that the shake table model is stable and will always return to static equilibrium when input hydraulic force $F_{h}$ is stopped.

Writing equation (27) in the Laplace form,

(28) \begin{equation} M_{t}s^{2}z_{t}+C_{sp}s\left(z_{t}-z_{p}\right)+K_{sp}\left(z_{t}-z_{p}\right)+K_{sa}z_{t}+C_{sa}sz_{t}=0 \end{equation}

(29) \begin{equation} z_{t}\left(M_{t}s^{2}+C_{sp}s+K_{sp}+K_{sa}+C_{sa}s\right)=-z_{p}\left(-C_{sp}s-K_{sp}\right) \end{equation}

By rearranging the above equation, the relation between pushrod displacement and table displacement is obtained in equation (30),

(30) \begin{equation} \frac{z_{t}}{z_{p}}=\frac{C_{sp}s+K_{sp}}{\left(M_{t}s^{2}+\left(C_{sp}+C_{sa}\right)s+\left(K_{sp}+K_{sa}\right)\right)} \end{equation}

Writing equation (26) in the Laplace form,

(31) \begin{equation} M_{p}s^{2}z_{p}+C_{sp}s\left(z_{p}-z_{t}\right)+K_{sp}\left(z_{p}-z_{t}\right)+K_{a}z_{p}+C_{a}sz_{p}=F_{h} \end{equation}

(32) \begin{equation} z_{p}\left( M_{p}s^{2}+C_{sp}s+K_{sp}+C_{a}s+K_{a}\right)+z_{t}\left(-C_{sp}s-K_{sp}\right)=F_{h} \end{equation}

(33) \begin{equation} z_{p}=\frac{z_{t}\left(C_{sp}s+K_{sp}\right)+F_{h}}{M_{p}s^{2}+\left(C_{sp}+C_{a}\right)s+\left(K_{sp}+K_{a}\right)} \end{equation}

Substituting $z_{p}$ value in equation (30), and upon simplification, the following relations between displacements $z_{t}$ & $z_{p}$ and the hydraulic force ( $F_{h}$ ) applied to the piston are obtained

(34) \begin{equation} \frac{z_{t}}{F_{h}}=\frac{A^{\prime}}{\left(M_{p}s^{2}+C^{\prime}\right)B^{\prime}-{A^{\prime}}^{2}} \end{equation}

Similarly,

(35) \begin{equation} \frac{z_{p}}{F_{h}}=\frac{B^{\prime}}{\left(M_{p}s^{2}+C^{\prime}\right)B^{\prime}-{A^{\prime}}^{2}} \end{equation}

where

(36) \begin{equation} A^{\prime}=C_{sp}s+K_{sp} \end{equation}

(37) \begin{equation} B^{\prime}=M_{t}s^{2}+\left(C_{sp}+C_{sa}\right)s+\left(K_{sp}+K_{sa}\right) \end{equation}

(38) \begin{equation} C^{\prime}=\left(C_{sp}+C_{a}\right)s+\left(K_{sp}+K_{a}\right) \end{equation}

The graphical representation of equation (30) is shown in Fig. 10.

Figure 10.

The conceptual realization of the passive hydraulic cylinder transfer function is shown in equation (30).

5.2. Case 2 modeling of UPS leg as a passive damper

The mechanical model of the single actuator leg and the UPS leg as a passive damper is shown in Fig. 11. The passive damper contains the non-linear spring ( $K_{n}$ ) and damper ( $C_{n}$ ). The non-linear spring and damper are differentiated structurally from linear spring and damper, as shown in Fig. 11.

Figure 11.

The mechanical model of a single actuator leg and a passive hydraulic damper.

The free-body diagram of the above mechanical model is given in Fig. 12. The initially compressed spring and damper of the passive damper element are indicated in the free-body diagram by blue dotted lines. The displacement offered by the passive damper will be ( $z_{max}-z_{t}$ ). Since stroke length is constant $(\dot{z_{max}}=0).$

Figure 12.

The free-body diagram of the mechanical model with UPS leg as a passive damper.

Equation (27) becomes

(39) \begin{equation} M_{t}\ddot{z_{t}}+C_{sp}\left(\dot{z_{t}}-\dot{z_{p}}\right)+K_{sp}\left(z_{t}-z_{p}\right)-K_{sa}\left(z_{max}-z_{t}\right)-C_{sa}\left(\dot{z_{max}}-\dot{z_{t}}\right)=0 \end{equation}

Writing the above equation in the Laplace form,

(40) \begin{equation} M_{t}s^{2}z_{t}+C_{sp}s\left(z_{t}-z_{p}\right)+K_{sp}\left(z_{t}-z_{p}\right)-K_{sa}z_{max}+K_{sa}z_{t}+C_{sa}sz_{t}=0 \end{equation}

(41) \begin{equation} z_{t}\left(M_{t}s^{2}+\left(C_{sp}+C_{sa}\right)s+K_{sp}+K_{sa}\right)-K_{sa}z_{max}=z_{P}\left(C_{sp}s+K_{sp}\right) \end{equation}

On further simplification, the relation between the table displacement ( $z_{t}$ ) and the pushrod displacement ( $z_{P}$ ) is obtained in equation (43).

(42) \begin{equation} z_{t} B^{\prime}=z_{P}A^{\prime}+K_{sa}z_{max} \end{equation}

(43) \begin{equation} z_{t}=z_{P}\frac{A^{\prime}}{B^{\prime}}+\frac{K_{sa}z_{max}}{B^{\prime}} \end{equation}

The inclusion of a passive damper causes an additional term in the transfer function model, $K_{sa}z_{max}$ , which appears to be a disturbance in equation 42. The passive damper element comprises a non-linear spring [Reference Abebe, Santhosh, Ahmed and Ponnusamy26] which is initially compressed to provide high spring force. This nonlinearity in the model provides force compensation during the upward displacement of the mobile platform $z_{t}$ (load against gravity) which was captured as DC gain in Fig. 17.

The graphical representation of equation (43) is shown in Fig. 13.

Figure 13.

The conceptual realization of the passive damper model transfer function is shown in equation (43).

5.3. Stability results and damping analysis

The following sections present the stability results and damping analysis for Cases 1 and 2.

5.3.1. For Case 1: passive hydraulic cylinder

From equation (30), it can be inferred that $K_{sa}$ appears on the denominator of the transfer function. Hence, the value of $K_{sa}$ should be chosen 0.1 times the value of $K_{sp}$ . This will ensure that the DC gain remains unaffected. A step signal is given as input to the transfer function of the passive hydraulic cylinder given by equation (30). The Simulink model for the transfer function is displayed in Fig. 14.

Figure 14.

The Simulink transfer function model for the passive cylinder.

The parameter values for the transfer functions are chosen from Table IV.

Table IV.

The parameter values for the transfer functions.

In the case of a passive hydraulic cylinder as the load-balancing UPS leg, the output displacement of the specimen/table shows overshoot and vibrations, as illustrated in Fig. 15.

Figure 15.

The response plot of the passive hydraulic cylinder.

As shown in Table V, the system is stable (since all the complex poles of the transfer function are in the left half of the s-plane), but it is underdamped (damping ratio $\zeta \lt 1$ ).

Table V.

Transfer function parameter table for the passive hydraulic cylinder.

5.3.2. For Case 2: passive damper element

Equation (43) represents the transfer of the passive damper element. The $K_{sa}$ value is ( $K_{n}+K_{b}$ ), where $K_{n}$ represents the non-linear part of the damper (Fig. 11). After estimating the non-linear value of $K_{n}$ , the $K_{sa}$ value was determined to be 1 $\times 10^{8}\,\textrm{N}/\textrm{m}$ . Similarly, the $C_{sa}$ value is estimated to be 3.18 $\times 10^{6}\,\textrm{Ns}/\textrm{m}$ (Table IV), which is ten times larger than $C_{sp}$ . Due to this, the damping of the model improved, as shown in Fig. 17.

A step signal is given as input to the transfer function of the passive damper model given by equation (43). The Simulink model for the transfer function is displayed in Fig. 16.

Figure 16.

The Simulink transfer function model for the passive damper.

With the inclusion of a passive hydraulic damper in the model, the vibrations are suppressed, and the DC gain factor increases, as shown in Fig. 17.

Figure 17.

The response plot of the passive damper model.

As shown in Table VI, the system is stable (the two complex poles and two real poles of the transfer function are in the left half of the s-plane) and critically damped (damping ratio $\zeta =1$ ).

Table VI.

Transfer function parameter table for passive damper model.