# Predicting Nonlinear Stiffness, Motion Range, and Load-Bearing Capability of Leaf-Type Isosceles-Trapezoidal Flexural Pivot Using Comprehensive Elliptic Integral Solution

Authors: Aimei Zhang; Yanjie Gou; Xihui Yang
Journal: Mathematical Problems in Engineering
 (2020)

---

## Abstract

A leaf-type isosceles-trapezoidal flexural (LITF) pivot consists of two leaf springs that are situated in the same plane and intersect at a virtual center of motion outside the pivot. The LITF pivot offers many advantages, including large rotation range and monolithic structure. Each leaf spring of a LITF pivot subject to end loads is deflected into an S-shaped configuration carrying one or two inflection points, which is quite difficult to model. The kinetostatic characteristics of the LITF pivot are precisely modeled using the comprehensive elliptic integral solution for the large-deflection problem derived in our previous work, and the strength-checking method is further presented. Two cases are employed to verify the accuracy of the model. The deflected shapes and nonlinear stiffness characteristics within the range of the yield strength are discussed. The load-bearing capability and motion range of the pivot are proposed. The nonlinear finite element results validate the effectiveness and accuracy of the proposed model for LITF pivots.


---

## Body

## 1. Introduction

A leaf-type isosceles-trapezoidal flexural (LITF) pivot consists of two leaf springs located in the same plane [1]. The two leaf springs are arranged symmetrically and intersect at a virtual center of motion outside the pivot, as shown in Figure 1. The parallelogram flexure is a special type in which the two leaf springs intersect at infinity. LITF pivots have been utilized in many accurate mechanisms [2–4] due to their obvious advantages such as low cost, monolithic manufacturing, reduced weight, and smooth motion [5–7].Figure 1
Classification of the leaf-type isosceles-trapezoidal flexural (LITF) pivot (a) and parallelogram flexure (b).
(a)
(b)When delivering movement, the leaf springs of a LITF pivot undergo nonlinear large deflection that may carry one or two inflection points (where the resultant moment is equal to zero [8]), which complicates the accurate modeling of LITF pivots. The remote center location and stiffness of a LITF pivot are presented by the model of screw theory based on the small-deflection assumption [9], which limits the application range of the model. The two pseudorigid-body models with small-deflection assumption [10], i.e., a four-bar model and a pin-joint model, were proposed for the analysis of the moment-angle characteristics of LITF pivots subject to horizontal force and moment. However, the influence of vertical force to LITF pivots was neglected. The analytic models for stiffness and center shift were presented by using the beam constraint model (BCM) method [11], which can be used to solve the nonlinear characteristics of LITF pivots, i.e., the rotation angle is in the range of ±15∘. The efficiency of uniform-strength composite leaf springs under various loading conditions [12] was analyzed. Therefore, the accurate nonlinear analysis and load-bearing capability solution within the entire stress range are indispensable for the application of LITF pivots.Because the leaf spring is so thin and flexible that the effects on axial elongation and shear are negligible, the elliptic integral solution is often considered to be the most accurate model for modeling this kind of large deflection beams. Howell [13] presented the elliptic integral solutions for the large deflection beam with no inflection point. An elliptic integral solution for the beam with an inflection point was derived by Kimball and Tsai [14]. In our previous research [15], we developed the comprehensive elliptic integral solution to solve the large deflections of beams with multiple inflection points and subject to any kinds of load cases. Because each of the deflected leaf spring carry one or two inflection points, LITF pivots can be modeled by the comprehensive elliptic integral solution. The model can be used to solve the exact deflected shapes and nonlinear stiffness of LITF pivots subject to different loads. Through the stress analysis for deflected leaf springs, the maximum motion range and allowable loads of LITF pivots are solved.The rest of this paper is organized as follows. In Section2, the accurate kinetostatic model and stress check for LITF pivots are proposed. In Section 3, two examples are calculated to demonstrate the accuracy of the model for LITF pivots. The nonlinear stiffness and workspace evaluation of the two examples are then discussed. In Section 4, concluding remarks are presented.


## 2. Modeling

### 2.1. LITF Pivot

As shown in Figure2, two springs (O1A and O2A, length L) of a LITF pivot intersect at point O′ and the angle between two leaf springs is 2β. The lengths of O1O2 and AB are w1 and w2, respectively. Letting N=w2/w1, when N≠1, we then have(1)w1=2LsinβN−1w2=2NLsinβN−1.Figure 2
Deformation diagram of LITF pivot. (a)N<1. (b) N > 1. (c) N = 1.
(a)
(b)
(c)WhenN=1, the LITF pivot becomes a parallelogram flexure, as shown in Figure 2(c), β=0∘ and w1=w2.The global coordinate systemOXY is established for the LITF pivot with the X axis oriented along O2O1 and the origin located at the midpoint of O2O1, as shown in Figure 2. The initial angle between leaf spring O1A and the X axis is θ1, and the angle between leaf spring O2B and the X axis is θ2. For N<1, θ1=90∘+β and θ2=90∘−β. For N>1, θ1=90∘−β and θ2=90∘+β.The local coordinate systemsO1X1Y1 and O2X2Y2 for leaf springs O1A and O2B are established with the origins placed at the fixed end and the X1 and X2 axes oriented along the leaf springs, respectively. The deflected end coordinates and the end angle of spring O1A with respect to the local coordinate O1X1Y1 are a1, b1, and θ1o, respectively. Similarly, the corresponding end coordinates and angle of O2B with respect to the local coordinate O2X2Y2 are denoted a2, b2, and θ2o, respectively. The horizontal displacement ΔX, vertical displacement ΔY, and rotation angle Δθ of the freedom for the LITF pivot in the global coordinate system are expressed as(2)Δθ=θ1o=θ2oΔX=12a1cosθ1+a2cosθ2−b1sinθ1+b2sinθ2ΔY=12a1sinθ1+a2sinθ2+b1cosθ1+b2cosθ2−Lcosβ.The loop closure equations are given as(3)cosθ1−sinθ1sinθ1cosθ1a1b1=cosθ2−sinθ2sinθ2cosθ2a2b2−w10+w2cosΔθsinΔθ.Figure3 shows the free-body diagrams for link AB and leaf springs O1A and O2B. When the pivot is subject to horizontal force Fx, vertical force Fy, and moment M at the midpoint C of link AB, the horizontal and vertical components of the end force and moment loaded at the O1A and O2B are F1x, F1y, M1o and F2x, F2y, M2o, respectively. Applying the static equilibrium for link AB yields(4)F1x+F2x+Fx=0−F1y−F2y+Fy=0−M1o−M2o+M−F1xw2sinΔθ−F1yw2cosΔθ−Fxw2sinΔθ2+Fyw2cosΔθ2=0.Figure 3
Force analysis of the LITF pivot. (a)N ≤ 1 and (b) N > 1.
(a)
(b)P
1 and n1P1 are the components of F1x and F1y along O1Y1 and X1O1 and have(5)F1xF1y=cosθ1sinθ1−sinθ1cosθ1n1P1P1.Similarly, forO2B,(6)F2xF2y=cosθ2sinθ2−sinθ2cosθ2n2P2P2.The deflection of each leaf spring can be modeled using the comprehensive elliptic integral solution summarized in the following section. The comprehensive elliptic integral solution for each leaf spring, together with the loop closure equation equation (3) and the static equilibrium equation (4), constitute the kinetostatic model for LITF pivots.


### 2.2. Comprehensive Elliptic Integral Solution

Each of the two springs of a LITF pivot can be viewed as a cantilever beam subject to an end vertical forceP, an end horizontal force nP, and an end moment Mo, as shown in Figure 4. The tip coordinates and tip angle of the deflected beam are denoted as a, b, and θo, respectively. Each deflected spring may carry m inflection points (m=1 or 2). The comprehensive solution [15] for the beam with inflection points, as summarized in the following, formulates the load parameters (κ, n, and α) and deflection parameters (a, b, and θo) by introducing m as the shape parameter:(7)α=SrηfaL=Srαη5/2−nηf+2nηe+2ηcbL=Srαη5/2ηf−2ηe+n2ηc,where α is defined as the force index (EI is the flexural rigidity of the beam),(8)α=PL2EI,Figure 4
Cantilever beam subject to combined force and moment loads.κ is the load ratio,(9)κ=Mo22PEI,and Sr is the sign of the resulting moment at the fixed end of the beam,(10)Sr=−1m,Mo≥0,−1m+1,M0<0.Moreover,(11)λ=sinθo−ncosθo+κ,η=1+n2,f=−1mFγ2,t−Fγ1,t+2mSrFt,e=−1mEγ2,t−Eγ1,t+2mSrEt,γ1=sin−1η−nλ+η,γ2=sin−1η+sinθo−ncosθoλ+η−π+ϕ<θo≤ϕ,−sin−1η+sinθo−ncosθoλ+η−2π+ϕ<θo≤−π+ϕ,t=λ+η2η,c=λ+n−−1mλ−sinθo+ncosθo.Fγ,t and Eγ,t=∫0γ1−tsinδdδare the incomplete elliptic integrals of the first and second kinds [16], respectively, in which γ is called the amplitude and t−1≤t≤1 is the modulus. When γ=π/2, they become the complete elliptic integrals of the first and second kinds and are denoted as Ft and Et, respectively. ϕ represents the angle of the force applied at the free end (as marked in Figure 4):(12)ϕ=π2+arctann.The coordinates (x, y) of an arbitrary point A on the beam (shown in Figure 4) can be written as(13)xL=Srαη5/2−nηfθ+2nηeθ+2ηcθyL=Srαη5/2ηfθ−2ηeθ+n2ηcθ,where θ is the deflected angle at point A and(14)fθ=−1mθFγ,t−Fγ1,t+2mθSrFt,eθ=−1mθEγ,t−Eγ1,t+2mθSrEt,γ=sin−1η+sinθ−ncosθλ+η−π+ϕ<θ≤ϕ,−sin−1η+sinθ−ncosθλ+η−2π+ϕ<θ≤−π+ϕ,cθ=λ+n−−1mθλ−sinθ+ncosθ.The valuemθ is equal to the number of the inflection points between the fixed end and point A.Most deflected springs mainly loaded by the horizontal forceFx and the moment M have one inflection point m=1. When the vertical force dominates in the applied loads, the deflected spring may contain two inflection points m=2.The kinetostatic model applied for LITF pivots can be calculated by the built-in function “fsolve” in MatLab (MathWorks, USA). The deflected shapes and the nonlinear stiffness characteristics of LITF pivots are further shown in Section3. However, the stress of deflected leaf springs should be less than the maximum yield stress, which determines the motion range and load-bearing capacity of LITF pivots.


### 2.3. Stress Analysis

The maximum bending stress, the important index of checking the LITF pivot, decides the maximum rotation angle and the maximum allowable loads. For deflected LITF pivots, the maximum bending stressσmax of each spring should be less than the yield strength Sy of the material used; thus, for the rectangular cross section I=wh3/12 (w is the width and h is the height), we have [17](15)σmax=hMmax2I≤Sy,where Mmax is the maximum resultant moment distributing in leaf springs. Substituting curvature K=M/EI into equation (15) yields(16)σmax=EhKmax2≤Sy,where Kmax is the maximum curvature occurring in the two deflected leaf springs.The motion range and load-bearing capability of LITF pivots subject to different loads, as two of the most important criteria with which to compare the compliant joints, can be obtained by the kinetostatic model and stress-checking equation (16).The flowchart of the solution to the deflected shape and workspace evaluation of LITF pivots is shown in Figure5.Figure 5
The flowchart of the solution to LITF pivots.


## 2.1. LITF Pivot

As shown in Figure2, two springs (O1A and O2A, length L) of a LITF pivot intersect at point O′ and the angle between two leaf springs is 2β. The lengths of O1O2 and AB are w1 and w2, respectively. Letting N=w2/w1, when N≠1, we then have(1)w1=2LsinβN−1w2=2NLsinβN−1.Figure 2
Deformation diagram of LITF pivot. (a)N<1. (b) N > 1. (c) N = 1.
(a)
(b)
(c)WhenN=1, the LITF pivot becomes a parallelogram flexure, as shown in Figure 2(c), β=0∘ and w1=w2.The global coordinate systemOXY is established for the LITF pivot with the X axis oriented along O2O1 and the origin located at the midpoint of O2O1, as shown in Figure 2. The initial angle between leaf spring O1A and the X axis is θ1, and the angle between leaf spring O2B and the X axis is θ2. For N<1, θ1=90∘+β and θ2=90∘−β. For N>1, θ1=90∘−β and θ2=90∘+β.The local coordinate systemsO1X1Y1 and O2X2Y2 for leaf springs O1A and O2B are established with the origins placed at the fixed end and the X1 and X2 axes oriented along the leaf springs, respectively. The deflected end coordinates and the end angle of spring O1A with respect to the local coordinate O1X1Y1 are a1, b1, and θ1o, respectively. Similarly, the corresponding end coordinates and angle of O2B with respect to the local coordinate O2X2Y2 are denoted a2, b2, and θ2o, respectively. The horizontal displacement ΔX, vertical displacement ΔY, and rotation angle Δθ of the freedom for the LITF pivot in the global coordinate system are expressed as(2)Δθ=θ1o=θ2oΔX=12a1cosθ1+a2cosθ2−b1sinθ1+b2sinθ2ΔY=12a1sinθ1+a2sinθ2+b1cosθ1+b2cosθ2−Lcosβ.The loop closure equations are given as(3)cosθ1−sinθ1sinθ1cosθ1a1b1=cosθ2−sinθ2sinθ2cosθ2a2b2−w10+w2cosΔθsinΔθ.Figure3 shows the free-body diagrams for link AB and leaf springs O1A and O2B. When the pivot is subject to horizontal force Fx, vertical force Fy, and moment M at the midpoint C of link AB, the horizontal and vertical components of the end force and moment loaded at the O1A and O2B are F1x, F1y, M1o and F2x, F2y, M2o, respectively. Applying the static equilibrium for link AB yields(4)F1x+F2x+Fx=0−F1y−F2y+Fy=0−M1o−M2o+M−F1xw2sinΔθ−F1yw2cosΔθ−Fxw2sinΔθ2+Fyw2cosΔθ2=0.Figure 3
Force analysis of the LITF pivot. (a)N ≤ 1 and (b) N > 1.
(a)
(b)P
1 and n1P1 are the components of F1x and F1y along O1Y1 and X1O1 and have(5)F1xF1y=cosθ1sinθ1−sinθ1cosθ1n1P1P1.Similarly, forO2B,(6)F2xF2y=cosθ2sinθ2−sinθ2cosθ2n2P2P2.The deflection of each leaf spring can be modeled using the comprehensive elliptic integral solution summarized in the following section. The comprehensive elliptic integral solution for each leaf spring, together with the loop closure equation equation (3) and the static equilibrium equation (4), constitute the kinetostatic model for LITF pivots.


## 2.2. Comprehensive Elliptic Integral Solution

Each of the two springs of a LITF pivot can be viewed as a cantilever beam subject to an end vertical forceP, an end horizontal force nP, and an end moment Mo, as shown in Figure 4. The tip coordinates and tip angle of the deflected beam are denoted as a, b, and θo, respectively. Each deflected spring may carry m inflection points (m=1 or 2). The comprehensive solution [15] for the beam with inflection points, as summarized in the following, formulates the load parameters (κ, n, and α) and deflection parameters (a, b, and θo) by introducing m as the shape parameter:(7)α=SrηfaL=Srαη5/2−nηf+2nηe+2ηcbL=Srαη5/2ηf−2ηe+n2ηc,where α is defined as the force index (EI is the flexural rigidity of the beam),(8)α=PL2EI,Figure 4
Cantilever beam subject to combined force and moment loads.κ is the load ratio,(9)κ=Mo22PEI,and Sr is the sign of the resulting moment at the fixed end of the beam,(10)Sr=−1m,Mo≥0,−1m+1,M0<0.Moreover,(11)λ=sinθo−ncosθo+κ,η=1+n2,f=−1mFγ2,t−Fγ1,t+2mSrFt,e=−1mEγ2,t−Eγ1,t+2mSrEt,γ1=sin−1η−nλ+η,γ2=sin−1η+sinθo−ncosθoλ+η−π+ϕ<θo≤ϕ,−sin−1η+sinθo−ncosθoλ+η−2π+ϕ<θo≤−π+ϕ,t=λ+η2η,c=λ+n−−1mλ−sinθo+ncosθo.Fγ,t and Eγ,t=∫0γ1−tsinδdδare the incomplete elliptic integrals of the first and second kinds [16], respectively, in which γ is called the amplitude and t−1≤t≤1 is the modulus. When γ=π/2, they become the complete elliptic integrals of the first and second kinds and are denoted as Ft and Et, respectively. ϕ represents the angle of the force applied at the free end (as marked in Figure 4):(12)ϕ=π2+arctann.The coordinates (x, y) of an arbitrary point A on the beam (shown in Figure 4) can be written as(13)xL=Srαη5/2−nηfθ+2nηeθ+2ηcθyL=Srαη5/2ηfθ−2ηeθ+n2ηcθ,where θ is the deflected angle at point A and(14)fθ=−1mθFγ,t−Fγ1,t+2mθSrFt,eθ=−1mθEγ,t−Eγ1,t+2mθSrEt,γ=sin−1η+sinθ−ncosθλ+η−π+ϕ<θ≤ϕ,−sin−1η+sinθ−ncosθλ+η−2π+ϕ<θ≤−π+ϕ,cθ=λ+n−−1mθλ−sinθ+ncosθ.The valuemθ is equal to the number of the inflection points between the fixed end and point A.Most deflected springs mainly loaded by the horizontal forceFx and the moment M have one inflection point m=1. When the vertical force dominates in the applied loads, the deflected spring may contain two inflection points m=2.The kinetostatic model applied for LITF pivots can be calculated by the built-in function “fsolve” in MatLab (MathWorks, USA). The deflected shapes and the nonlinear stiffness characteristics of LITF pivots are further shown in Section3. However, the stress of deflected leaf springs should be less than the maximum yield stress, which determines the motion range and load-bearing capacity of LITF pivots.


## 2.3. Stress Analysis

The maximum bending stress, the important index of checking the LITF pivot, decides the maximum rotation angle and the maximum allowable loads. For deflected LITF pivots, the maximum bending stressσmax of each spring should be less than the yield strength Sy of the material used; thus, for the rectangular cross section I=wh3/12 (w is the width and h is the height), we have [17](15)σmax=hMmax2I≤Sy,where Mmax is the maximum resultant moment distributing in leaf springs. Substituting curvature K=M/EI into equation (15) yields(16)σmax=EhKmax2≤Sy,where Kmax is the maximum curvature occurring in the two deflected leaf springs.The motion range and load-bearing capability of LITF pivots subject to different loads, as two of the most important criteria with which to compare the compliant joints, can be obtained by the kinetostatic model and stress-checking equation (16).The flowchart of the solution to the deflected shape and workspace evaluation of LITF pivots is shown in Figure5.Figure 5
The flowchart of the solution to LITF pivots.


## 3. Case Studies

In this section, a LITF pivot and a parallelogram flexure are employed as two cases to demonstrate the effectiveness of the comprehensive elliptic integral model. The parameters of the two pivots are given in Table1, and the materials are polypropylene in which E=1.4GPa and Sy=34MPa [13].Table 1
Parameters of pivot.
w(m)
h(m)
L(m)
N
β(°)
w
1(m)
w
2(m)
LITF pivot
0.0252
0.001
0.07
0.4
30
∘
—
—
Parallelogram flexure
0.0252
0.001
0.07
1
0
∘
0.1
0.1


### 3.1. Solution for LITF Pivot

The parameters of the LITF pivot are shown in Table1. The lengths of link O1O2 and AB solved by equation (1) are w1=0.1167m and w2=0.0467m, respectively. The deflected shapes of the pivot subject to different loads, the load-bearing capacity, and the corresponding motion range of the pivot will be discussed here.


#### 3.1.1. Deflected Shapes under Different Loads

The deflected results of the pivot subject to different loads are obtained separately by using the comprehensive elliptic integral solution and a nonlinear finite element analysis (NFEA) model, as shown in Figures6–10. For the NFEA model built with the ANSYS software, springs O1A and O2B are meshed into 100 elements with BEAM188, respectively, and the large displacement analysis option is turned on. BEAM188 is suitable for analyzing slender to moderately stubby beam structures. This element is based on the Timoshenko beam theory. Shear deformation is included. The results of the comprehensive elliptic integral solution agree well with NFEA.Figure 6
Plots of momentM vs angle Δθ.Figure 7
Deflected shape of positive pivot subject toFx=0N, Fy=0N, and M=0.49N⋅m.Figure 8
Plots of momentM vs horizontal displacement ΔX.Figure 9
Plots of momentM vs vertical displacement ΔY.Figure 10
Plots of momentM vs angle Δθ.For the LITF pivot subject to pure moment loaded at point C, the relationship between the rotation angleΔθ and the moment M is shown in Figure 6. The LITF pivot reveals fine linearity for Δθ less than 10∘ (the dashed line in Figure 6 expresses the linear approximation of the LITF pivot with small deformation). However, when Δθ is larger than 10∘, the nonlinearity of the stiffness for this kind of pivot becomes remarkable. For M=0.49N⋅m, the maximum curvature of the pivot occurring at point A is equal to 48.5519·m−1, which is substituted into equation (16) to obtain the maximum stress, σmax=33.986MPa, close to the yield strength. Meanwhile, Δθ attains 27.04∘, for which the deflected shape of the pivot is shown in Figure 7.IfFx=5N and M are loaded at point C simultaneously, the relationships of ΔX, ΔY, and Δθ with M before yield failure have slight nonlinearities, as shown in Figures 8–10. Otherwise, for Fx=−5N, Fy=0N, and different M, the nonlinearities of the pivot become obvious. For M from 0 to 0.5N·m, ΔX varies from −0.005m m to 0.011 m and ΔY increases from −0.0007m to 0 m and then decreases to −0.004m again and Δθ changes from −7.6∘ to 18.9∘, as shown in Figures 8–10. The curve of Fx=−5N and Fy=0N is intersected at M=0.2 with that of Fx=−5 and Fy=−5, where the resultant moments for the rotation center O′ are equal to zero, so that the pivot returns to the original position. The corresponding deflected shapes of the pivot for Fx=−5N, Fy=0N, and M=0∼0.5N⋅m, as shown in Figure 11, incline to the left and then to the right.Figure 11
Deflected shapes of the positive pivot subject toFx=−5N, Fy=0N, and M=0∼0.5N⋅m.


#### 3.1.2. Workspace Evaluation

The stress of the deflected pivot solved by the kinetostatic model is checked by equation (16), and then the load-bearing capacity in different load cases and the motion range of the pivot are obtained.(1) Horizontal Force and Moment. Figure 12 shows that the pivot subject to different M and Fx can bear a range of horizontal force. The arrows drawn in Figure 12 roughly mark the descending direction of the stress, and the covered area is the safe working region.Figure 12
Allowable maximum horizontal force of the pivot subject to different moments.The pivot only subjected to horizontal force, i.e.,M=0, can bear the maximum horizontal force reaching Fx=±13.6N, for which the corresponding rotation angles are Δθ=±16.9∘, as shown in Figure 13. With the incremental moment, the maximum positive horizontal force of the pivot gradually decreases and the anticlockwise rotation angle of link AB shows the increasing tendency.Figure 13
Corresponding rotation angle of the pivot subject to different moments and maximum horizontal force depicted in Figure12.ForM=0.49N⋅m, the maximum stress of the pivot without horizontal force reaches the yield strength. When negative horizontal force and moment act on the pivot, the negative allowable horizontal force increases gradually and the corresponding angle decreases slightly with the increasing moment, as shown in Figures 12 and 13. For M=0.5N⋅m, the allowable negative force is Fx=−24.12N and Δθ=−13.06∘, as shown in Figure 14, and the maximum curvature also happens at point A.Figure 14
Deformed shape of the positive pivot subject toFx=−24.21N, Fy=0N, and M=0.5N⋅m.The relative errors of the rotation angles between the comprehensive solutionΔθCS and the nonlinear finite element ΔθFEA results are expressed as(17)Error=ΔθCSΔθFEA−1.The errors of the positive rotation angles depicted in Figure13 between the comprehensive solution and the nonlinear finite element results are less than 1.5%, which is shown in Figure 15.Figure 15
The errors of the rotation angles between the comprehensive solution and the nonlinear finite element results.(2) Vertical Force and Moment. For the pivot subject to M and Fy, Figure 16 draws the maximum vertical force that the pivot subject to different moments can bear. Similarly, the declining direction of the stress is masked roughly by the arrows in Figure 16. Positive vertical force can counteract the rotation angle of the pivot caused by the moment. It should be noted that the tensile stress might lead to the failure of the pivot when positive vertical force reaches a certain value because the Bernoulli–Euler beam theory neglects the effect of axial elongation and the maximum positive vertical force cannot be predicted, the discussion of which is outside the scope of this paper.The corresponding rotation angles subject to different moments and the maximum vertical forces depicted in Figure16 are shown in Figure 17. The rotation angle slightly increases for M<0.2N⋅m and then decreases for M>0.2N⋅m. When M=0.2N⋅m, the allowable negative vertical force attains Fy=−11.42N and the rotation angle is Δθ=31.86∘, for which the deflected shape agrees well with the result calculated by NFEA, as shown in Figure 18. The maximum stress of the pivot subject to Fy=−11.42N and M=0.2N⋅m is σmax=33.995MPa and occurs in the deflected spring O2B shown by the diamond shape in Figure 18. When M is greater than 0.49N⋅m, the pivot subject to the negative force directly leads to the failure of the spring, so in this case, the pivot can only withstand the positive vertical force.Figure 16
Allowable maximum vertical force of the pivot subject to different moments.Figure 17
Corresponding rotation angle of the pivot subject to different moments and maximum vertical force depicted in Figure16.Figure 18
Deformed shape of the positive pivot subject toFx=0N, Fy=−11.42N, and M=0.2N⋅m.ForM=0N⋅m, the pivot only subject to vertical force and the buckling of the spring may take place. For the buckled LITF pivot, the maximum bending stress σmax may be less than the yield strength of the material used, but the LITF pivot has been invalidated, so the maximum negative vertical force for M=0N⋅m is equal to the critical buckling force.The buckled springs can be seemed as the fixed-guided beams with two inflection pointsm=2 that perhaps have two deformed shapes (I) and (II), as shown in Figure 19. The vertical displacement of the freedom is δ, and the end slope of the buckled spring remains constant, i.e., θo≡0.Figure 19
Buckling deformation of the LITF pivot.For the buckled springs, the coordinates of the free end are given as(18)a=L−δcosβb=−δsinβ.The vertical forceFy can be solved as(19)Fy=2Psinβ−ncosβ.Substitutingm=2, θo≡0, and equation (8) into equation (9) yields(20)P=16EI1+n2L2F2t,(21)aL=−n1+n21−2EtFt,(22)bL=11+n21−2EtFt.From equations (18), (21), and (22), n is(23)n=−ab=L−δcosβδsinβ.Substituting equations (20) and (23) into equation (19) yields(24)Fy=32EIL2δ−Lcosβδ2+L2−2LδcosβF2t.WhenFy reaches the critical buckling force, we have n⟶∞ and a/L⟶0, and then equation (21) reduces to Ft/Et⟶1 and has(25)t=0.We haveFt=π/2; then, the critical buckling force Fcr from equation (24) is(26)Fcr=−32EIcosβL2F20=−8π2EIcosβL2.Thus, forM=0, the maximum negative vertical force of the pivot Fy is determined by the critical buckling force solved by equation (26) and equal to −41.0272N, as shown in Figure 16. When the pivot is loaded only by the vertical force, the leaf with two inflection points includes two deflection paths, which are shown in the left-hand leaf and right-hand leaf of Figure 19. The choice of the two solutions is decided by the processing factor of the leaf.


### 3.2. Parallelogram Flexure

A parallelogram flexure is a one-degree-of-freedom device that obtains accurate motion by the bending of the springs. Many authors have contributed to this problem; for example, Awtar et al. [18] proposed a beam constraint model and Dibiasio et al. [19] presented a pseudorigid-body model to simplify the derivation and calculation. In the paper, the kinetostatic model is also suitable to analyze the parallelogram flexure.For the parameters of the mechanism given in Table1, the leaf springs O1A and O2B with one inflection point guide the motion of link AB with minimal rotation. When Fx is applied at point C, the horizontal displacement ΔX is obtained to arrive at a static equilibrium state, as shown in Figure 20. With increasing horizontal force, the nonlinear characteristics of the curve are gradually obvious and the rotation angle Δθ is slowly increasing, as shown in Figure 21. When Fx=8.5N is loaded at point C, ΔX=0.032m and Δθ=−0.145∘, for which the maximum curvature occurring at point O2 is Kmax=46.9396m−1 and the maximum bending stress σmax solved by equation (16) is slightly less than Sy.Figure 20
Plots of horizontal forceFx vs horizontal displacement ΔX.Figure 21
Plots of rotation angleΔθ of parallelogram flexure subject to horizontal force Fx.The rotation angleΔθ of link AB is a parasitic error motion that is undesirable in response to the horizontal force Fx, which may be eliminated by an appropriate combination of moment M or vertical force Fy [18]. When Fx and M are loaded simultaneously at point C to ensure Δθ≡0, we have, from equations (4)–(6),(27)MFx=−M1oP1.For a parallelogram flexure because each deflected leaf spring carries one inflection point, where the resultant moment is equal to zero and the rotation angles at the fixed and free ends of each deflected leaf spring are both equal to zero, the inflection point occurs at the middle of the deflected spring, i.e.,x=a1/2. The moment at the inflection point is(28)a12P1+M1o=0.Substituting equation (28) into equation (27) yields(29)MFx=a12=a22=L+ΔY2.ForFy=0, the ratios in equation (29) during the intermediate stage are approximatively constant, which agree well with the results of Ref. [18] equal to 0.5L, as shown in Figures 22 and 23. Then, the ratios between M and Fx are less than 0.5L with increasing horizontal force Fx and the corresponding transverse stiffness gradually increases for protecting Δθ≡0. As listed in Table 2, the corresponding moments and displacements for Fx=2∼10N solved by the elliptic integral solution agrees well with the load-deflection relationship expressed in equation (29) and the deflected shapes of the pivot subject to Fx and M are shown in Figure 24.Figure 22
MomentM vs Fx for end angle Δθ≡0 of parallelogram flexure.Figure 23
Horizontal displacementΔx vs Fx for end angle Δθ≡0 of parallelogram flexure.Table 2
Loads and deflections of parallelogram flexure.
F
x(N)
F
y(N)
M (N⋅m)
Δ
X (m)
Δ
Y (m)
2
0
0.0692
0.0095
−
7.84
×
10
−
4
4
0
0.1343
0.0181
−
2.87
×
10
−
3
6
0
0.1929
0.0252
−
5.71
×
10
−
3
8
0
0.2448
0.0308
−
0.0088
10
0
0.2908
0.0354
−
0.0118Figure 24
Deflected shapes of parallelogram flexure loaded byFx and M for Δθ≡0.WhenFx and Fy are loaded at point C simultaneously, M is needed to ensure that Δθ≡0. In this case, the inflection points also appear in the middle of the deflected leaf springs. However, if Fy is a tensile force, the ratios between M and Fx are less than a1/2 because of n being less than zero. On the contrary, for Fy as a pressure, the ratios between M and Fx are greater than a1/2. The more obvious nonlinearity of the pivot appears with increasing pressure Fy, as shown in Figure 23. Until the pressure reaches the critical buckling force calculated by equation (26) Fy=−47.3741N, the buckling of the pivot leads directly to failure.


## 3.1. Solution for LITF Pivot

The parameters of the LITF pivot are shown in Table1. The lengths of link O1O2 and AB solved by equation (1) are w1=0.1167m and w2=0.0467m, respectively. The deflected shapes of the pivot subject to different loads, the load-bearing capacity, and the corresponding motion range of the pivot will be discussed here.


### 3.1.1. Deflected Shapes under Different Loads

The deflected results of the pivot subject to different loads are obtained separately by using the comprehensive elliptic integral solution and a nonlinear finite element analysis (NFEA) model, as shown in Figures6–10. For the NFEA model built with the ANSYS software, springs O1A and O2B are meshed into 100 elements with BEAM188, respectively, and the large displacement analysis option is turned on. BEAM188 is suitable for analyzing slender to moderately stubby beam structures. This element is based on the Timoshenko beam theory. Shear deformation is included. The results of the comprehensive elliptic integral solution agree well with NFEA.Figure 6
Plots of momentM vs angle Δθ.Figure 7
Deflected shape of positive pivot subject toFx=0N, Fy=0N, and M=0.49N⋅m.Figure 8
Plots of momentM vs horizontal displacement ΔX.Figure 9
Plots of momentM vs vertical displacement ΔY.Figure 10
Plots of momentM vs angle Δθ.For the LITF pivot subject to pure moment loaded at point C, the relationship between the rotation angleΔθ and the moment M is shown in Figure 6. The LITF pivot reveals fine linearity for Δθ less than 10∘ (the dashed line in Figure 6 expresses the linear approximation of the LITF pivot with small deformation). However, when Δθ is larger than 10∘, the nonlinearity of the stiffness for this kind of pivot becomes remarkable. For M=0.49N⋅m, the maximum curvature of the pivot occurring at point A is equal to 48.5519·m−1, which is substituted into equation (16) to obtain the maximum stress, σmax=33.986MPa, close to the yield strength. Meanwhile, Δθ attains 27.04∘, for which the deflected shape of the pivot is shown in Figure 7.IfFx=5N and M are loaded at point C simultaneously, the relationships of ΔX, ΔY, and Δθ with M before yield failure have slight nonlinearities, as shown in Figures 8–10. Otherwise, for Fx=−5N, Fy=0N, and different M, the nonlinearities of the pivot become obvious. For M from 0 to 0.5N·m, ΔX varies from −0.005m m to 0.011 m and ΔY increases from −0.0007m to 0 m and then decreases to −0.004m again and Δθ changes from −7.6∘ to 18.9∘, as shown in Figures 8–10. The curve of Fx=−5N and Fy=0N is intersected at M=0.2 with that of Fx=−5 and Fy=−5, where the resultant moments for the rotation center O′ are equal to zero, so that the pivot returns to the original position. The corresponding deflected shapes of the pivot for Fx=−5N, Fy=0N, and M=0∼0.5N⋅m, as shown in Figure 11, incline to the left and then to the right.Figure 11
Deflected shapes of the positive pivot subject toFx=−5N, Fy=0N, and M=0∼0.5N⋅m.


### 3.1.2. Workspace Evaluation

The stress of the deflected pivot solved by the kinetostatic model is checked by equation (16), and then the load-bearing capacity in different load cases and the motion range of the pivot are obtained.(1) Horizontal Force and Moment. Figure 12 shows that the pivot subject to different M and Fx can bear a range of horizontal force. The arrows drawn in Figure 12 roughly mark the descending direction of the stress, and the covered area is the safe working region.Figure 12
Allowable maximum horizontal force of the pivot subject to different moments.The pivot only subjected to horizontal force, i.e.,M=0, can bear the maximum horizontal force reaching Fx=±13.6N, for which the corresponding rotation angles are Δθ=±16.9∘, as shown in Figure 13. With the incremental moment, the maximum positive horizontal force of the pivot gradually decreases and the anticlockwise rotation angle of link AB shows the increasing tendency.Figure 13
Corresponding rotation angle of the pivot subject to different moments and maximum horizontal force depicted in Figure12.ForM=0.49N⋅m, the maximum stress of the pivot without horizontal force reaches the yield strength. When negative horizontal force and moment act on the pivot, the negative allowable horizontal force increases gradually and the corresponding angle decreases slightly with the increasing moment, as shown in Figures 12 and 13. For M=0.5N⋅m, the allowable negative force is Fx=−24.12N and Δθ=−13.06∘, as shown in Figure 14, and the maximum curvature also happens at point A.Figure 14
Deformed shape of the positive pivot subject toFx=−24.21N, Fy=0N, and M=0.5N⋅m.The relative errors of the rotation angles between the comprehensive solutionΔθCS and the nonlinear finite element ΔθFEA results are expressed as(17)Error=ΔθCSΔθFEA−1.The errors of the positive rotation angles depicted in Figure13 between the comprehensive solution and the nonlinear finite element results are less than 1.5%, which is shown in Figure 15.Figure 15
The errors of the rotation angles between the comprehensive solution and the nonlinear finite element results.(2) Vertical Force and Moment. For the pivot subject to M and Fy, Figure 16 draws the maximum vertical force that the pivot subject to different moments can bear. Similarly, the declining direction of the stress is masked roughly by the arrows in Figure 16. Positive vertical force can counteract the rotation angle of the pivot caused by the moment. It should be noted that the tensile stress might lead to the failure of the pivot when positive vertical force reaches a certain value because the Bernoulli–Euler beam theory neglects the effect of axial elongation and the maximum positive vertical force cannot be predicted, the discussion of which is outside the scope of this paper.The corresponding rotation angles subject to different moments and the maximum vertical forces depicted in Figure16 are shown in Figure 17. The rotation angle slightly increases for M<0.2N⋅m and then decreases for M>0.2N⋅m. When M=0.2N⋅m, the allowable negative vertical force attains Fy=−11.42N and the rotation angle is Δθ=31.86∘, for which the deflected shape agrees well with the result calculated by NFEA, as shown in Figure 18. The maximum stress of the pivot subject to Fy=−11.42N and M=0.2N⋅m is σmax=33.995MPa and occurs in the deflected spring O2B shown by the diamond shape in Figure 18. When M is greater than 0.49N⋅m, the pivot subject to the negative force directly leads to the failure of the spring, so in this case, the pivot can only withstand the positive vertical force.Figure 16
Allowable maximum vertical force of the pivot subject to different moments.Figure 17
Corresponding rotation angle of the pivot subject to different moments and maximum vertical force depicted in Figure16.Figure 18
Deformed shape of the positive pivot subject toFx=0N, Fy=−11.42N, and M=0.2N⋅m.ForM=0N⋅m, the pivot only subject to vertical force and the buckling of the spring may take place. For the buckled LITF pivot, the maximum bending stress σmax may be less than the yield strength of the material used, but the LITF pivot has been invalidated, so the maximum negative vertical force for M=0N⋅m is equal to the critical buckling force.The buckled springs can be seemed as the fixed-guided beams with two inflection pointsm=2 that perhaps have two deformed shapes (I) and (II), as shown in Figure 19. The vertical displacement of the freedom is δ, and the end slope of the buckled spring remains constant, i.e., θo≡0.Figure 19
Buckling deformation of the LITF pivot.For the buckled springs, the coordinates of the free end are given as(18)a=L−δcosβb=−δsinβ.The vertical forceFy can be solved as(19)Fy=2Psinβ−ncosβ.Substitutingm=2, θo≡0, and equation (8) into equation (9) yields(20)P=16EI1+n2L2F2t,(21)aL=−n1+n21−2EtFt,(22)bL=11+n21−2EtFt.From equations (18), (21), and (22), n is(23)n=−ab=L−δcosβδsinβ.Substituting equations (20) and (23) into equation (19) yields(24)Fy=32EIL2δ−Lcosβδ2+L2−2LδcosβF2t.WhenFy reaches the critical buckling force, we have n⟶∞ and a/L⟶0, and then equation (21) reduces to Ft/Et⟶1 and has(25)t=0.We haveFt=π/2; then, the critical buckling force Fcr from equation (24) is(26)Fcr=−32EIcosβL2F20=−8π2EIcosβL2.Thus, forM=0, the maximum negative vertical force of the pivot Fy is determined by the critical buckling force solved by equation (26) and equal to −41.0272N, as shown in Figure 16. When the pivot is loaded only by the vertical force, the leaf with two inflection points includes two deflection paths, which are shown in the left-hand leaf and right-hand leaf of Figure 19. The choice of the two solutions is decided by the processing factor of the leaf.


## 3.1.1. Deflected Shapes under Different Loads

The deflected results of the pivot subject to different loads are obtained separately by using the comprehensive elliptic integral solution and a nonlinear finite element analysis (NFEA) model, as shown in Figures6–10. For the NFEA model built with the ANSYS software, springs O1A and O2B are meshed into 100 elements with BEAM188, respectively, and the large displacement analysis option is turned on. BEAM188 is suitable for analyzing slender to moderately stubby beam structures. This element is based on the Timoshenko beam theory. Shear deformation is included. The results of the comprehensive elliptic integral solution agree well with NFEA.Figure 6
Plots of momentM vs angle Δθ.Figure 7
Deflected shape of positive pivot subject toFx=0N, Fy=0N, and M=0.49N⋅m.Figure 8
Plots of momentM vs horizontal displacement ΔX.Figure 9
Plots of momentM vs vertical displacement ΔY.Figure 10
Plots of momentM vs angle Δθ.For the LITF pivot subject to pure moment loaded at point C, the relationship between the rotation angleΔθ and the moment M is shown in Figure 6. The LITF pivot reveals fine linearity for Δθ less than 10∘ (the dashed line in Figure 6 expresses the linear approximation of the LITF pivot with small deformation). However, when Δθ is larger than 10∘, the nonlinearity of the stiffness for this kind of pivot becomes remarkable. For M=0.49N⋅m, the maximum curvature of the pivot occurring at point A is equal to 48.5519·m−1, which is substituted into equation (16) to obtain the maximum stress, σmax=33.986MPa, close to the yield strength. Meanwhile, Δθ attains 27.04∘, for which the deflected shape of the pivot is shown in Figure 7.IfFx=5N and M are loaded at point C simultaneously, the relationships of ΔX, ΔY, and Δθ with M before yield failure have slight nonlinearities, as shown in Figures 8–10. Otherwise, for Fx=−5N, Fy=0N, and different M, the nonlinearities of the pivot become obvious. For M from 0 to 0.5N·m, ΔX varies from −0.005m m to 0.011 m and ΔY increases from −0.0007m to 0 m and then decreases to −0.004m again and Δθ changes from −7.6∘ to 18.9∘, as shown in Figures 8–10. The curve of Fx=−5N and Fy=0N is intersected at M=0.2 with that of Fx=−5 and Fy=−5, where the resultant moments for the rotation center O′ are equal to zero, so that the pivot returns to the original position. The corresponding deflected shapes of the pivot for Fx=−5N, Fy=0N, and M=0∼0.5N⋅m, as shown in Figure 11, incline to the left and then to the right.Figure 11
Deflected shapes of the positive pivot subject toFx=−5N, Fy=0N, and M=0∼0.5N⋅m.


## 3.1.2. Workspace Evaluation

The stress of the deflected pivot solved by the kinetostatic model is checked by equation (16), and then the load-bearing capacity in different load cases and the motion range of the pivot are obtained.(1) Horizontal Force and Moment. Figure 12 shows that the pivot subject to different M and Fx can bear a range of horizontal force. The arrows drawn in Figure 12 roughly mark the descending direction of the stress, and the covered area is the safe working region.Figure 12
Allowable maximum horizontal force of the pivot subject to different moments.The pivot only subjected to horizontal force, i.e.,M=0, can bear the maximum horizontal force reaching Fx=±13.6N, for which the corresponding rotation angles are Δθ=±16.9∘, as shown in Figure 13. With the incremental moment, the maximum positive horizontal force of the pivot gradually decreases and the anticlockwise rotation angle of link AB shows the increasing tendency.Figure 13
Corresponding rotation angle of the pivot subject to different moments and maximum horizontal force depicted in Figure12.ForM=0.49N⋅m, the maximum stress of the pivot without horizontal force reaches the yield strength. When negative horizontal force and moment act on the pivot, the negative allowable horizontal force increases gradually and the corresponding angle decreases slightly with the increasing moment, as shown in Figures 12 and 13. For M=0.5N⋅m, the allowable negative force is Fx=−24.12N and Δθ=−13.06∘, as shown in Figure 14, and the maximum curvature also happens at point A.Figure 14
Deformed shape of the positive pivot subject toFx=−24.21N, Fy=0N, and M=0.5N⋅m.The relative errors of the rotation angles between the comprehensive solutionΔθCS and the nonlinear finite element ΔθFEA results are expressed as(17)Error=ΔθCSΔθFEA−1.The errors of the positive rotation angles depicted in Figure13 between the comprehensive solution and the nonlinear finite element results are less than 1.5%, which is shown in Figure 15.Figure 15
The errors of the rotation angles between the comprehensive solution and the nonlinear finite element results.(2) Vertical Force and Moment. For the pivot subject to M and Fy, Figure 16 draws the maximum vertical force that the pivot subject to different moments can bear. Similarly, the declining direction of the stress is masked roughly by the arrows in Figure 16. Positive vertical force can counteract the rotation angle of the pivot caused by the moment. It should be noted that the tensile stress might lead to the failure of the pivot when positive vertical force reaches a certain value because the Bernoulli–Euler beam theory neglects the effect of axial elongation and the maximum positive vertical force cannot be predicted, the discussion of which is outside the scope of this paper.The corresponding rotation angles subject to different moments and the maximum vertical forces depicted in Figure16 are shown in Figure 17. The rotation angle slightly increases for M<0.2N⋅m and then decreases for M>0.2N⋅m. When M=0.2N⋅m, the allowable negative vertical force attains Fy=−11.42N and the rotation angle is Δθ=31.86∘, for which the deflected shape agrees well with the result calculated by NFEA, as shown in Figure 18. The maximum stress of the pivot subject to Fy=−11.42N and M=0.2N⋅m is σmax=33.995MPa and occurs in the deflected spring O2B shown by the diamond shape in Figure 18. When M is greater than 0.49N⋅m, the pivot subject to the negative force directly leads to the failure of the spring, so in this case, the pivot can only withstand the positive vertical force.Figure 16
Allowable maximum vertical force of the pivot subject to different moments.Figure 17
Corresponding rotation angle of the pivot subject to different moments and maximum vertical force depicted in Figure16.Figure 18
Deformed shape of the positive pivot subject toFx=0N, Fy=−11.42N, and M=0.2N⋅m.ForM=0N⋅m, the pivot only subject to vertical force and the buckling of the spring may take place. For the buckled LITF pivot, the maximum bending stress σmax may be less than the yield strength of the material used, but the LITF pivot has been invalidated, so the maximum negative vertical force for M=0N⋅m is equal to the critical buckling force.The buckled springs can be seemed as the fixed-guided beams with two inflection pointsm=2 that perhaps have two deformed shapes (I) and (II), as shown in Figure 19. The vertical displacement of the freedom is δ, and the end slope of the buckled spring remains constant, i.e., θo≡0.Figure 19
Buckling deformation of the LITF pivot.For the buckled springs, the coordinates of the free end are given as(18)a=L−δcosβb=−δsinβ.The vertical forceFy can be solved as(19)Fy=2Psinβ−ncosβ.Substitutingm=2, θo≡0, and equation (8) into equation (9) yields(20)P=16EI1+n2L2F2t,(21)aL=−n1+n21−2EtFt,(22)bL=11+n21−2EtFt.From equations (18), (21), and (22), n is(23)n=−ab=L−δcosβδsinβ.Substituting equations (20) and (23) into equation (19) yields(24)Fy=32EIL2δ−Lcosβδ2+L2−2LδcosβF2t.WhenFy reaches the critical buckling force, we have n⟶∞ and a/L⟶0, and then equation (21) reduces to Ft/Et⟶1 and has(25)t=0.We haveFt=π/2; then, the critical buckling force Fcr from equation (24) is(26)Fcr=−32EIcosβL2F20=−8π2EIcosβL2.Thus, forM=0, the maximum negative vertical force of the pivot Fy is determined by the critical buckling force solved by equation (26) and equal to −41.0272N, as shown in Figure 16. When the pivot is loaded only by the vertical force, the leaf with two inflection points includes two deflection paths, which are shown in the left-hand leaf and right-hand leaf of Figure 19. The choice of the two solutions is decided by the processing factor of the leaf.


## 3.2. Parallelogram Flexure

A parallelogram flexure is a one-degree-of-freedom device that obtains accurate motion by the bending of the springs. Many authors have contributed to this problem; for example, Awtar et al. [18] proposed a beam constraint model and Dibiasio et al. [19] presented a pseudorigid-body model to simplify the derivation and calculation. In the paper, the kinetostatic model is also suitable to analyze the parallelogram flexure.For the parameters of the mechanism given in Table1, the leaf springs O1A and O2B with one inflection point guide the motion of link AB with minimal rotation. When Fx is applied at point C, the horizontal displacement ΔX is obtained to arrive at a static equilibrium state, as shown in Figure 20. With increasing horizontal force, the nonlinear characteristics of the curve are gradually obvious and the rotation angle Δθ is slowly increasing, as shown in Figure 21. When Fx=8.5N is loaded at point C, ΔX=0.032m and Δθ=−0.145∘, for which the maximum curvature occurring at point O2 is Kmax=46.9396m−1 and the maximum bending stress σmax solved by equation (16) is slightly less than Sy.Figure 20
Plots of horizontal forceFx vs horizontal displacement ΔX.Figure 21
Plots of rotation angleΔθ of parallelogram flexure subject to horizontal force Fx.The rotation angleΔθ of link AB is a parasitic error motion that is undesirable in response to the horizontal force Fx, which may be eliminated by an appropriate combination of moment M or vertical force Fy [18]. When Fx and M are loaded simultaneously at point C to ensure Δθ≡0, we have, from equations (4)–(6),(27)MFx=−M1oP1.For a parallelogram flexure because each deflected leaf spring carries one inflection point, where the resultant moment is equal to zero and the rotation angles at the fixed and free ends of each deflected leaf spring are both equal to zero, the inflection point occurs at the middle of the deflected spring, i.e.,x=a1/2. The moment at the inflection point is(28)a12P1+M1o=0.Substituting equation (28) into equation (27) yields(29)MFx=a12=a22=L+ΔY2.ForFy=0, the ratios in equation (29) during the intermediate stage are approximatively constant, which agree well with the results of Ref. [18] equal to 0.5L, as shown in Figures 22 and 23. Then, the ratios between M and Fx are less than 0.5L with increasing horizontal force Fx and the corresponding transverse stiffness gradually increases for protecting Δθ≡0. As listed in Table 2, the corresponding moments and displacements for Fx=2∼10N solved by the elliptic integral solution agrees well with the load-deflection relationship expressed in equation (29) and the deflected shapes of the pivot subject to Fx and M are shown in Figure 24.Figure 22
MomentM vs Fx for end angle Δθ≡0 of parallelogram flexure.Figure 23
Horizontal displacementΔx vs Fx for end angle Δθ≡0 of parallelogram flexure.Table 2
Loads and deflections of parallelogram flexure.
F
x(N)
F
y(N)
M (N⋅m)
Δ
X (m)
Δ
Y (m)
2
0
0.0692
0.0095
−
7.84
×
10
−
4
4
0
0.1343
0.0181
−
2.87
×
10
−
3
6
0
0.1929
0.0252
−
5.71
×
10
−
3
8
0
0.2448
0.0308
−
0.0088
10
0
0.2908
0.0354
−
0.0118Figure 24
Deflected shapes of parallelogram flexure loaded byFx and M for Δθ≡0.WhenFx and Fy are loaded at point C simultaneously, M is needed to ensure that Δθ≡0. In this case, the inflection points also appear in the middle of the deflected leaf springs. However, if Fy is a tensile force, the ratios between M and Fx are less than a1/2 because of n being less than zero. On the contrary, for Fy as a pressure, the ratios between M and Fx are greater than a1/2. The more obvious nonlinearity of the pivot appears with increasing pressure Fy, as shown in Figure 23. Until the pressure reaches the critical buckling force calculated by equation (26) Fy=−47.3741N, the buckling of the pivot leads directly to failure.


## 4. Conclusions

The comprehensive elliptic integral solution was used for building the generalized model of LITF pivots and solving nonlinear deflection problems. For the LITF pivot, the accurate deflected shapes are described subject to different horizontal forces, vertical forces, and moments. Furthermore, based on the strength check and the analysis of the critical buckling force, motion range and load-bearing capability for the pivot are evaluated. For the parallelogram flexure, two cases for free rotation angle and constant rotation angle are discussed. The more accurate ratio between horizontal force and moment is proposed to ensure that the rotation angle remains constant. The analytical results for the maximum rotation angle of the LITF pivot subject to horizontal force and moment solved by the comprehensive elliptic integral solution are within 1.5 percent error compared to the finite element analysis results.



---

*Source: 1390692-2020-02-12.xml*