Research on Stiffness of Multibackbone Continuum Robot Based on Screw Theory and Euler-Bernoulli Beam

Continuum robots have become a focus for extensive research, since they canworkwell in complex and confined environments.The main contribution of this paper is to establish a stiffness model of a single section multibackbone continuum robot and analyze the effect of the structural parameters of continuum robot on the overall rotation and translation stiffness. First, a stiffnessmodel which indicates the end configuration of continuum robot under external load is deduced by the screw theory and Euler-Bernoulli beam. Then, the stiffness elements are fully analyzed, therefore, obtaining the influence of the structural parameters of continuum robot on the stiffness elements. Meanwhile, a numerical analysis of stiffness elements is given. Furthermore, the minimum and maximum rotation/translation stiffness are introduced to analyze the effect of the structural parameters of continuum robot on the overall rotation and translation stiffness. Finally, the experiments are used to validate the proposed stiffness model. The experimental results show that the proposed stiffness model of continuum robot is correct and the errors are less than 7%.


Introduction
Continuum robot is a new kind of bionic robot, inspired by elephant trunks and octopus tentacles.Continuum robot does not have its own joints, which can produce flexible deformation in any part, so it has a strong ability to avoid obstacles and better adapt to the complex unstructured environments.Continuum robots offer a number of potential advantages over the traditional rigid-link robots in applications involving disaster relief [1], industrial applications [2], and medical aid [3].
The main types of continuum robots include the roddriven continuum robot [4], cable-driven continuum robot [5], pneumatic continuum robot [6], and concentric tube continuum robot [7].Du et al. [8] developed an optimization notched continuum manipulator based on performance evaluation indices.Tian et al. [9] proposed the kinematic analysis of a continuum bionic robot with three flexible actuation rods.Moreover, Bergeles et al. [10] presented optimization framework based on anatomical and surgical task constraints.Li et al. [11] developed a new constrained wire-driven flexible mechanism which has a larger workspace and is more dexterous compared to the existing surgical arms.He et al. [12] proposed a multibackbone continuum robot which is driven by NiTi alloy wire.Trivedi et al. [13] discussed the novel capabilities of soft robots, described examples from nature that provide biological inspiration, surveyed the state of the art, and outlined existing challenges in soft robot design, modeling, fabrication, and control.Webster III and Jones [14] reviewed several modeling approaches in a common frame and notational convention, illustrating that, for piecewise constant curvature, they produce identical results.
The analysis of stiffness is an important step in the design and control of continuum robots, since it determines the relationship between the deformation and the force of continuum robots.Selig and Ding [15] applied the screw theory [16] to analyze compliance and stiffness matrices of a beam.Pei et al. [17] studied the compliance of cartwheel flexural hinges.Ding and Dai [18] investigated spatial continuous compliance on both serial and parallel mechanisms based on screw theory and Lie groups, applying eigenvectors and eigenvalues to identify principal screws in the mechanisms.Awtar and Sen [19] proposed a generalized constraint model for compliance and stiffness analysis of 2D beam flexures.Krishnan et al. [20] studied serial and parallel concatenation of building blocks.Tunay [21] proposed a concept of equivalent bending stiffness to establish the kinematic model of continuum robot.Gao et al. [22] established a mathematical model for predicting the loaded posture of a single section continuum manipulator.Qi et al. [23] analyzed the compliance characteristics of a new planar spring continuum robot.Gravagne et al. [24] discussed the dynamics of a planar continuum backbone section, incorporating a large-deflection dynamic model.Trivedi et al. [25] presented a new approach for modeling soft robotic manipulators that incorporates the effect of material nonlinearities and distributed weight and payload.Camarillo et al. [26] proposed a new linear model for transforming desired beam configuration to tendon displacements and vice versa.Fras et al. [27] described the design and implementation of a static model used for position estimation of a flexible modular medical manipulator equipped with optic-fiber based sensors.Till and Rucker [28] exhibited low numerical damping, handled arbitrarily large time steps, and provided an accurate, high-order representation of the rod shape in steady state.Hadi Sadati et al. [29] introduced a novel series solution for variable-curvature Cosserat rod static and Lagrangian dynamic method.
Several different performance indices have been proposed for stiffness evaluation, including determinant of stiffness matrix, average stiffness, and minimum and maximum stiffness [30,31].The determinant of stiffness matrix does not distinguish specific stiffness values, and the average stiffness cannot give enough information of stiffness values [32].The minimum and maximum stiffness are minimum and maximum eigenvalue of stiffness matrix, respectively, which can indicate variation range of stiffness values, and the corresponding eigenvector directions represent the minimum and maximum stiffness directions, respectively.However, it is well known that the conventional minimum and maximum stiffness cannot be applied to a 6 × 6 symmetric stiffness/compliance matrix.This is due to the fact that the eigenvalues of a stiffness/compliance matrix are not consistent unit and no way to compare the sizes of eigenvalues.As a consequence, we define the minimum and maximum rotation/translation stiffness, to evaluate the influence of the structural parameters of continuum robot on the rotation/translation stiffness of continuum robot.
In the paper, a general loading is represented by a wrench, while a general deformation is represented by a twist.The main contribution is to present a method to establish the stiffness model of a single section multibackbone continuum robot based on screw theory and Euler-Bernoulli beam, as well as analyzing the effect of the structural parameters of continuum robot on the overall rotation and translation stiffness by the minimum and maximum rotation/translation stiffness.The remainder of the paper is organized as follows: In Section 2, we give structure overview of a single section multibackbone continuum robot.In Section 3, based on the screw theory and Euler-Bernoulli beam, a brief stiffness model which illustrates the end configuration of the continuum robot under external load is deduced.In Section 4, the stiffness elements are analyzed to fully indicate relationship between wrench and twist of continuum robot and obtain the influence of the structural parameters of continuum robot on the stiffness elements.In addition, the minimum

Overview of a Single Section Multibackbone Continuum Robot
The simplified structure of a single section multibackbone continuum robot is shown in Figure 1.The continuum robot is composed of a base disk, several spacer disks, an end disk, and four super elastic NiTi wires as its backbones.
The central NiTi wire is the primary backbone and the remaining NiTi wires are the secondary backbones.The primary backbone is attached to all disks, and all disks are distributed in equal distance.The secondary backbones are equidistant from primary backbone and from each other.The secondary backbones are only attached to the end disk and slide through holes in base disk and spacer disks, which have double effect: driving robot to achieve two degrees of freedom bending motion and providing auxiliary support to increase the stiffness of continuum robot.

Stiffness Model of Continuum Robot
In order to facilitate the derivation of the presented theory, the model assumptions of continuum robot are summarized in this paper as follows: (1) The disks of robot are thin, so the thickness of the disks is ignored.(2) The friction between the disks and backbones is neglected.
(3) The continuum robot is under static balance.(4) The gravity of all the disks and the four backbones, which is ignored, is very smaller.(5) We assume that the deformation of backbones is sufficiently small so that the principle of Euler-Bernoulli beam theory applies.When the continuum robot is not exerted by an external load and the secondary backbone does not pull and push the continuum robot, then the simplified geometric configuration of continuum robot is shown in Figure 2(a) and the length of all the backbones are equal.The global coordinate system - is placed at the end disk, as we are interested in the motion of the end of continuum robot.Point  is geometric center of end disk, the axis  points from point  to the first secondary backbone, the axis  is perpendicular to the end disk, and the axis  is then defined according to the right-hand rule.Moreover, the coordinate system   -      is established at point   , the axis   is parallel to the axis , the axis   is parallel to the axis , and the axis   is parallel to the axis .The secondary backbones are the average interval 120 degrees around the primary backbone.Assuming that the continuum robot is subjected to the external load at the end of continuum robot, the configuration of continuum robot is shown in Figure 2(b).
For easy comprehension of the deduction of the presented theory, in this paper, symbols are defined as follows: : the index of secondary backbones  = 1, 2, 3. : the distance from the primary backbone to each secondary backbone on the disk.
: the length of primary backbone.  : the length of secondary backbones  = 1, 2, 3.   : the radius of backbones.
In the framework of a screw theory [14] and the global coordinate system -, a small deformation of continuum robot is defined as a twist  = (, )  in axis coordinates. is an element of Lie algebra se(3) of Lie group SE (3), where  = (  ,   ,   )  represents the corresponding rotational deflections around their corresponding axes and  = (  ,   ,   )  reveals the three translational deflections along their corresponding axes. = (, )  is defined as the wrench at the end of the continuum robot in ray coordinates, which is an element of the dual space of Lie algebra se * (3), where  = (  ,   ,   )  represents three moments around the axes , , and , respectively;  = (  ,   ,   )  reveals three forces along axes , , and , respectively.According to the elastic theory, wrench  and twist  are related by where   and   are the stiffness and compliance matrices of continuum robot in the global coordinate system -, respectively.
In the following, we derive stiffness model of continuum robot.First, the compliance matrix of primary backbone in the global coordinate system - can be derived by screw theory [17] and Euler-Bernoulli beam [13], which is expressed as where  =  2  is the area of the cross section,   = (1/4) 4

𝑏
and   = (1/4) 4  are the area moments,  = (1/2) 4  is the torsional moment of inertia,  denotes the elastic module of the material, and  denotes the shear module of the material.
For convenience, by extracting common factor, ( 2) is simplified as where and ] is the Poisson's ratio.
It is assumed that the  0 is stiffness matrix of primary backbone in the global coordinate system -, and then the stiffness matrix  0 is calculated as The local coordinate system   -      is established for the  ( = 1, 2, 3)th secondary backbone at the end disk, as shown in Figure 2(a).Point   is the intersection point of the th secondary backbone and the end disk, the axis   is parallel to axis , the axis   is parallel to axis , and the axis   is parallel to axis .Suppose that   is the compliance matrix of the th secondary backbone in the local coordinate system   -      , since   =  ( = 1, 2, 3), which can be obtained as where  = 1, 2, 3.
Since compliance matrix   ( = 1, 2, 3) is represented in the different local coordinate system, in order to establish the stiffness model of continuum robot, we should transform the compliance matrix   ( = 1, 2, 3) to the global coordinate system by coordinate transformation.Here, to shift the local coordinate system   -      of each secondary backbone into the global coordinate system -, an adjoint action of Lie group SE(3) on its Lie algebra is introduced by a 6 × 6 matrix representation [21]: where  is a 3 × 3 rotation matrix representing the orientation of the local coordinate system   -      , relative to the global coordinate system -; for instance, the 1st secondary backbone  is a unit matrix. is a skew-symmetric matrix spanned by the position vector  from the origin of frame - to the origin of frame   -      ; for example, the 1st secondary backbone  = [, 0, 0].
Here, the adjoint transformation matrix of the th ( = 1, 2, 3) secondary backbone is given by where In addition, the transpose of the adjoint transformation matrix is given by It is supposed that   is the compliance matrix of the th secondary backbone in the global coordinate system - and    and    are twist and wrench at the end of the th secondary backbone in the global coordinate system -, respectively.  and   are twist and wrench at the end of the th secondary backbone in the local coordinate system   -      , yielding Moreover, the coordinates of a twist and a wrench in the global coordinate system - are calculated as [16] To obtain the compliance matrix   in the global coordinate system -, we deduct it as follows based on (11) and (12): Accordingly, the compliance matrix   can be obtained according to the relation where  = 1, 2, 3.
Similarly, we can derive the stiffness matrix   of the  ( = 1, 2, 3)th secondary backbone in the global coordinate system - as Since the continuum robot is composed of four NiTi wires in parallel, if all deformations are presented into the same global coordinate system -, then the overall stiffness matrix of the parallel flexible mechanism is the sum of the stiffness matrix of each flexible mechanism [16]; the overall stiffness matrix   of continuum robot can be obtained as where Substituting ( 17) into ( 16) yields where  = 1, 2, 3, 4, 5, 6;  = 1, 2, 3, 4, 5, 6; and It can be seen that   in ( 18) is determined by six independent design parameters, ,   , , ,   , and .Simultaneously, the stiffness matrix   gives the relationship between the wrench and twist at the end of the continuum robot.Furthermore, the compliance matrix is the inverse of the stiffness matrix which is calculated as where

Stiffness Analysis of Continuum Robot
Based on the stiffness model of continuum robot in Section 3, we will first analyze the effect of the structural parameters of continuum robot on the stiffness elements in this section.
Then, the influence of the structural parameters on the overall rotation and translation stiffness is analyzed.Finally, to more intuitively understand the stiffness elements of continuum robot, the stiffness elements are analyzed by the numerical example.

Analysis of Stiffness Elements.
Each element in the stiffness matrix reflects the mapping relationship between each element in the wrench and each element in the twist.Hence, through the analysis of the stiffness elements of stiffness matrix   in (18), we can obtain the following conclusions.
(1) The rotational stiffness around axes x, , and  is calculated as By ( 22), (23), and (24), assuming that  and   are constant values, it can be seen that when length  increases, with a decreasing of the rotational stiffness  11 ,  22 , and  33 , obviously,  11 and  22 are inversely proportional to .By ( 22) and (24), supposing that  and  are constant values, when length   increases,  11 and  33 also increase.Assuming that  and   are constant values, when length  increases,  11 and  33 also increase.By (23), assuming that  and   are constant values, when length  increases,  22 also increases, as shown Figure 3.
(2) The translational stiffness along axes x, , and  is expressed as By ( 25) and ( 26), changing  has no effect on the  44 and  55 .Assuming that   is constant value, when length  increases, with a decreasing of the translational stiffness  44 and  55 , which are inversely proportional to  3 .Assume that   is constant value, when length  increases, with decreasing of the translational stiffness  44 and  55 , which are inversely proportional to  3 .Supposing that the length  is constant value, when length   increases,  44 and  55 also increase, which are directly proportional to  4   .By (27), assuming that  and   are constant values, it can be seen that when length  increases, but  66 decreases, obviously,  66 is inversely proportional to .Supposing that  and   are constant values, when length  increases,  66 decreases.By (18), the translational stiffness of  44 ,  55 , and  66 is decoupled; that is, the forces   ,   , and   only produce translation along axes x, , and , respectively.
(3) The rotational stiffness which is produced by force   is calculated as By (28),   only produce rotation around axis .Dividing (25) by (28) By ( 30) and ( 31),   produce two rotational stiffness elements.One is the rotation stiffness  51 around the axis x, which is a function of the variables  and   , The effects of variables  and   on  51 can be easily obtained by (30).The other is the rotational stiffness  53 around the axis , which is a function of the variables , , and   , The effects of variables , , and   on  53 can be easily given by (31).Equation ( 26) is divided by (30), which is defined as  2 .Equation ( 26) is divided by (31), taking the absolute value, which is defined as  3 .Equation ( 31) is divided by (30), taking the absolute value, which is defined as  4 .This yields By (32), when length  increases,  2 decreases.The result explains that when the force   is a constant value and length  increases, the value   /  decreases.By (33), when length  increases,  3 decreases, indicating that when the force   is a constant value and length  increases, the value |  /  | decreases.By (34), assuming that  is constant value, it can be seen that when  increases,  4 also increases.The result indicates that when   ,  are constant values and  increases, the value |  /  | increases.Supposing that  is constant value, when  increases,  5 also decreases.Indicating that when   ,  are a constant values and  increases, the value |  /  | decreases.
(5) The rotational stiffness which is produced by force   is calculated as Dividing ( 27) by ( 35), which is defined as  5 , we obtain When   ≪ , (36) is simplified as By (37), when   ≪  and  increases,  5 decreases.The result explains that when the force   is a constant value,   ≪ , and  increases, the value   /  decreases.When  ≪   , (36) is simplified as By (38), when  ≪   ,  5 tends to positive infinity.The result indicates that when the force   is a constant value and  ≪   , the value   /  tends to positive infinity.

Eigen-Stiffness Analysis of Continuum Robot.
Since the eigenvalues of the stiffness matrix in (18) are not consistent unit, there is no way to compare the sizes of eigenvalues; the conventional minimum/maximum stiffness cannot be applied to a 6 × 6 symmetric stiffness/compliance matrix.Here, we define the minimum and maximum rotation/translation stiffness to evaluate the effect of the structural parameters of continuum robot on the rotation/translation stiffness.In order to facilitate the presentation of this definition, we first block the stiffness matrix   in (18), which can be generally represented in the form of where the symmetric 3 × 3 block matrices  and  denote the pure rotation and translation matrices,  is the coupling one,   is transpose of the matrix , and Then, by (1), we obtain Since the pure translation matrix  is already a diagonal matrix, the eigenvalues of matrix  are (  /) 44 , (  /) 55 , and (  /) 66 , respectively.Define  min = min{(  /) 44 , (  /) 55 , (  /) 66 } and  max = max{(  /) 44 , (  /) 55 , (  /) 66 } as the minimum and maximum translation stiffness for the matrix , respectively.The pure rotation matrix  is symmetric, but not a diagonal matrix; thus, there is an orthogonal matrix , making    a diagonal matrix [33]; that is, where  = [ 1 ,  2 ,  3 ] ∈ SO(3) represents an orthogonal matrix whose columns are just the eigenvectors of matrix  and    = diag[ 1 ,  2 ,  3 ] is a diagonal matrix consisting of the corresponding eigenvalues of matrix .We define  min = min{ 1 ,  2 ,  3 } and  max = max{ 1 ,  2 ,  3 } as the minimum and maximum rotation stiffness for the matrix , respectively.The matrix  is diagonal matrix, it is easy to obtain the minimum and maximum translation stiffness; therefore, we do not do a lot of analysis here.However, the matrix  is not a diagonal matrix and is coupled, In the following, we analyze the effect of mainly structural parameters  and  on minimum rotation stiffness  min and maximum rotation stiffness  max .The parameters of the NiTi wire are shown in Table 1.
First, the influence of parameter  on the minimum and maximum rotation stiffness is analyzed.For ease of research, assuming that  is a constant and  = 3 mm, the images of minimum and maximum rotation stiffness with the change  are shown in Figure 4.
In Figure 4(a), it can be seen that when  increases,  min decreases.When  ∈ [50, 125] mm, the speed of change of  min is bigger.When  ∈ [125, 400] mm, the speed of change of  min is smaller.In Figure 4   In Figure 5(a), if  increases,  min decreases.When  ∈ [1,17] mm, the speed of change of  min is bigger.When  ∈ [17,30] mm, the speed of change of  min is smaller.In Figure 5(b), when  increases,  max decreases.When  ∈ [1,10] mm, the speed of change of  max is bigger.When  ∈ [10,30] mm, the speed of change of  min is smaller.In summary, if  ∈ [1,15] mm, the influence of  on the minimum and maximum rotation stiffness is bigger.If  ∈ [15,30] mm, the influence of  on the minimum and maximum rotation stiffness is smaller.The block matrices  and  have been analyzed above.Next, the matrix  is researched.In order to facilitate the study, the lemma is first given as follows.

Mathematical Problems in Engineering
Lemma 1.If a matrix  is an -order asymmetric matrix, then the matrix  must not be orthogonal to diagonalization.
Proof.Set a matrix  as an -order asymmetric matrix, and suppose there is an -order orthogonal matrix , making where Λ is a diagonal matrix and the diagonal elements are the eigenvalues of the matrix .Transposing (43) yields Since by ( 43), (49), and (50), we obtain Multiplying  and its transpose   of (51) yields Since the matrix  is an orthogonal matrix, that is,   is a unit matrix, then, this yields   = .
(48) Thus, it is obtained that the matrix  is symmetric matrix.It is inconsistent with the hypothetical matrix  which is an -order asymmetric matrix at the beginning of the proof.Therefore, the matrix  cannot be orthogonal to diagonalization.
The matrix  reflects the relationship between  and , and the units of each element in the matrix  are consistent.However, the matrix  is not a symmetric matrix in (40), by Lemma 1, obtaining that matrix  cannot be orthogonal to diagonalization.Therefore, the matrix  has no minimum eigenvalue and maximum eigenvalue.Here, we cannot define the minimum stiffness and maximum stiffness of matrix .The transpose matrix   also cannot define the minimum stiffness and maximum stiffness.

Numerical Analysis of Stiffness Elements.
In this section, in order to understand the stiffness elements of continuum robot more intuitively, the stiffness elements of continuum robot are analyzed by the numerical example.Here, we select the structural parameters of continuum robot such as  = 210 mm,  = 5 mm.The parameters of NiTi wire are shown in Table 1.
The results are tabulated in Table 2, where the unit for force component is Newton, the unit of moment is N•m, and the units for translational and rotational displacements are meter and radian, respectively.By analyzing the numerical results, the following conclusions can be obtained.
(1) The rotational stiffness elements   /  and   /  are approximately equal.It is indicated that the same moment is, respectively, applied around the axes  and  at the end of continuum robot; then the rotational angles of the end are the approximately equal.Moreover,   /  is about 94.88 times larger than   /  , and   /  is about 94.47 times larger than   /  .It is shown that the rotation around the axes  and  is harder than that around the axis .

Continuum robot Tension sensor
Stepping motor (2) The translational stiffness elements   /  and   /  are equal.It is shown that the same force is, respectively, applied along the axes  and  at the end of continuum robot; then the translational distances of the end are equal.Meanwhile, the translational stiffness element   /  = 1.05 × 10 6 (N/m) is significantly larger than translational stiffness element   /  =   /  = 3.43 × 10 (N/m).It is indicated that the translation along the axis  is harder than translation along the axes  and , respectively.It all fits our intuition.

Experiments
The proposed stiffness model is validated by a single section NiTi wire driving continuum robot, as shown in Figure 6.The structural parameters of continuum robot are  = 210 mm,  = 5 mm.The parameters of NiTi wire are shown in Table 1.The disks are made of 3D printed machine (EDEN-260V) with hard material (VeroClear RGD810), and the disks are very hard.The secondary backbones are connected with three stepping motors, which produce the driving force to bend the continuum robot.The tension sensor is used to record the driving force which is produced by motors.
The experimental setup is shown as Figure 7, which consists of a continuum robot, a control cabinet, a teaching apparatus, a robotic arm, a computer, and a force/torque sensor.The robotic arm (STEP-SD500) is used to provide an external load at the end of continuum robot, and the 6 DOF force/torque commercial sensor (OPTOFORCE-HEX-70-XE-450) is mounted at the end of robotic arm to record the external load force.
Here, we facilitate the experimental operations by verifying the compliance matrix to indicate that the stiffness matrix is correct.Meanwhile, we pay more attention to the relations between forces and deformations at the end of continuum robot along the three translational directions.Therefore, we operate the experiment to verify the relations between forces and deformations in three translational directions.It is assumed that the external forces applied at the end of continuum robot are twists  1 ,  2 , and  3 , where  1 = (0, 0, 0,   , 0, 0)  ,  2 = (0, 0, 0, 0,   , 0)  , and  3 = (0, 0, 0, 0, 0,   )  .Substituting twists  1 ,  2 , and  3 into (20), respectively, we obtain external load force and translational distance that meet the following relationship: where   (N),   (N),   (N),   (cm),   (cm), and   (mm).Next, through the experiments we verify that (49), (50), and (51) are correct.The experiments were done at low moving speed to reduce the effect of vibration.First, the experiments along the axis  are shown in Figure 8.The force and translational distance at the end of continuum robot is recorded by 6 DOF force/torque commercial sensor and the operation of robotic arm, respectively.The experimental values and the functional image of (49) are shown in Figure 9.The experimental values are distributed around the theoretical image, and the errors between the theoretical values and the experimental values are no more than 4%, thus, indicating that (49) is correct.
Then, the experiments along the axis  are shown in Figure 10.
The experimental values and the functional image of (50) are shown in Figure 11.The errors between the theoretical values and the experimental values are no more than 5%, therefore, demonstrating that (50) is correct.
Finally, the experiments along the axis  are shown in Figure 12.
The experimental values and the functional image of (51) are shown in Figure 13.The errors between the theoretical values and the experimental values are no more than 7%, thence, showing that (51) is correct.
Through the above experiments, we verify that the translational stiffness of continuum robot along the axes , , and , respectively, is correct.By comparing the translational experimental values along the axes  and , respectively, if the external force   =   , the translational distances   and   are approximately equal.By comparing the translational experimental values along the axes  and , respectively, if the external force   =   , the translational distance   is much larger than translational distance   ; it also fits our intuitive perception.

Discussions
In this paper, we mainly establish the stiffness model of a single section multibackbone continuum robot, with the initial configuration of continuum robot being a straight line and   =  ( = 1,2,3).If the continuum robot is multisection and the initial configuration of continuum robot is a straight line, then the stiffness model of a multisection continuum robot can be studied by each lower section as a payload of its upper sections.In addition, the stiffness model of continuum robot is based on the initial configuration which is a straight line.If the initial configuration of continuum robot is a bending shape, then the length of the four NiTi wires of continuum robot is not equal.Thus, the stiffness model in (18) is not suitable for analyzing the load model of continuum robot.Simultaneously, it should be pointed out that the deformations of all backbones should be small; otherwise, the elastic properties of the backbone are damaged; then the stiffness model in (18) is also not suitable for analyzing the load model of continuum robot.
In our model, the 6 × 6 stiffness matrix in (18) indicates relationship between force/moment and motion/rotation at

Figure 2 :
Figure 2: The continuum robot is subjected to the external load at the end of continuum robot.

Figure 4 :
Figure 4: (a) The influence of parameter  on the minimum rotation stiffness.(b) The influence of parameter  on the maximum rotation stiffness.

J 2 .
034 × 10 −13 m 4 G 2.5 × 10 10 Pa E 6.5 × 10 10 Pa the speed of change of  max is smaller.Considering the minimum and maximum rotation stiffness comprehensively, if  ∈ [50, 125] mm, the influence of  on the minimum and maximum rotation stiffness is bigger.If  ∈ [125, 400] mm, the influence of  on the minimum and maximum rotation stiffness is smaller.Next, we analyze the influence of parameter  on the minimum and maximum rotation stiffness.Here, it is supposed that  is a constant and  = 200 mm; the minimum and maximum rotation stiffness of the image with the change  are shown in Figure 5.

Figure 5 :
Figure 5: (a) The influence of parameter  on the minimum rotation stiffness.(b) The influence of parameter  on the maximum rotation stiffness.

Figure 6 :
Figure 6: A single section NiTi wire continuum robot.

Figure 8 :
Figure 8: The experiments along the axis .

Figure 9 :
Figure 9: Theoretical and experimental values along the axis .

Figure 10 :
Figure 10: The experiments along the axis .

Figure 11 :
Figure 11: Theoretical and experimental values along the axis .

Figure 12 :
Figure 12: The experiments along the axis .

Table 1 :
The parameters of the NiTi wire.

Table 2 :
Numerical results of stiffness elements.