Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2015, Article ID 371342, 8 pages http://dx.doi.org/10.1155/2015/371342

Research Article A Novel Analytical Solution Method for Constraint Forces of the Kinematic Pair and Its Applications

Changjian Zhi, Sanmin Wang, Yuantao Sun, and Bo Li

School of Mechanical Engineering, Northwestern Polytechnical University, Xi’an 710072, China

Correspondence should be addressed to Changjian Zhi; [email protected]

Received 23 December 2014; Revised 11 March 2015; Accepted 12 March 2015

Academic Editor: Vladimir Turetsky

Copyright © 2015 Changjian Zhi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Constraint forces of the kinematic pair are the basis of the kinematics and dynamics analysis of mechanisms. Exploring the solution method for constraint forces is a hot issue in the mechanism theory fields. Based on the observation method and the theory of reciprocal screw system, the solution method of reciprocal screw system is improved and its solution procedures become easier. This method is also applied to the solution procedure of the constraint force. The specific expressions of the constraint forceare represented by the reciprocal screw system of twist. The transformation formula of twist under different coordinates is given and it make the expression of the twist of kinematic pair more facility. A slider-crank mechanism and a single loop spatial RUSR mechanism are taken as examples. It confirms that this method can be used to solve the constraint force of the planar and spatial mechanism.

1. Introduction deformation compatibility method. Russo et al. [7]using the counterweight method and the springs method analyzed The constraint force analysis of kinematic pair is not only the static balancing of spatial parallel manipulator. Lu [8] thekeyofusingmechanismsreasonablyandcreatingnew used virtual work theory and CAD functionalities for solving mechanisms but also the important factor of kinematics active force and passive force of spatial parallel manipulators. and dynamics analysis and the base of structure design of Deepak and Ananthasuresh [9]usednonlinearspringsto mechanism. Traditional methods of constraint force analysis generate minimum torques which keep the cables taut and of kinematic pair include the graphic method and analytical analyzed the static balancing of the parallel cable-driven method [1]. Since they need drawing and the force figures, mechanisms. which go against programming, seeking new analysis meth- The reciprocal screw system represents the constraints ods is very necessary. Many scholars have a try in this filed and have got some achievements. Zhou et al. [2]utilizingthe and constraint forces acting on kinematic pairs. Since its dismantle-bar method and the microdeformation and super- solution method is complex and lacks commonality, based position principle analyzed the static force of the parallel on Huang et al. [10] and Dai and Jones [11, 12]researches, mechanism. Zhao and Huang [3]usingtheanalyticalmethod we improve the solution method of the reciprocal screw analyzed the force of the lower-mobility parallel mechanism system and make the solving process easier. In order to with overconstrained couple. Based on Zhao’s research and express constraint forces of the kinematic pair, the solu- screw theory, Liu et al. [4]analyzedtheforceofsingleloop tion coefficient is introduced. Not only can this method spatial mechanism. Wang et al. [5] combining the traditional solve constraint forces of the kinematic pair of mechanism dismantle-bar method with screw theory analyzed the force but also it can analyze the static and dynamic force by of the spatial parallel mechanism. Jiang et al. [6]using combining it with the foregoing methods. We verify the Newton Euler method and D’Alembert principle established method by solving constraint forces of the kinematic pair the force analysis equations, and they also put forward the of a slider-crank mechanism and a single loop spatial RUSR dynamic analysis model of parallel mechanism based on the mechanism. 2 Mathematical Problems in Engineering

z S0 S

S

hS r S0 −hS

x

y Figure 2: The axis of a screw. Figure 1: A line vector.

where 𝜔 is the instantaneous angular velocity of the rigid 0 2. Theoretic Foundation body and k is the instantaneous velocity a point in the rigid body coincident with the origin. 󸀠 Alinevector(Figure 1)canbedenotedbyadualvector[10] According to formula (4), S canbewrittenas 󸀠 S =(S; S )=(S; r × S) , (1) 󸀠 0 0 S =(𝜔; k −ℎ𝜔)+(0;ℎ𝜔) , (6) (S; S ) S where 0 is called the Plucker¨ coordinates, is the real 0 unit, which is the direction ratios of the line and is not origin- where v −ℎ𝜔 = r × 𝜔; ℎ canbeobtainedbyformula(3). (𝜔; v0 −ℎ𝜔) dependent, S0 is the real unit, which is the moment of the line Therefore, canbeseenasapurerotation, about the origin and is origin-dependent, and r is the vector and (0;ℎ𝜔) can be seen as a pure translation. A twist can be of a point on the line. S and S0 are three-dimensional vectors. decomposed into a rotation and a translation. S and S0 satisfy the orthogonal condition, S⋅S0 =0.When Generally, all the spatial forces acting on a rigid body can 0 S =0, the line passes through the origin. While S =0,theline be reduced to a force (F; F0) and a couple (0; T ).Theforce lies in a plane at infinity, and it becomes a couple. and the couple may have different directions. According to If S and S0 do not satisfy the orthogonal condition, S⋅S0 ≠ thescrewalgebra,thesumoftheforceandthecoupleisa 󸀠 0. S is known as a screw. Ball [13] described that a screw is a new screw, which is called a wrench. Consider straight line with the pitch. It can be expressed as 󸀠 0 0 S =(F; F0 + T )=(F; C ). (7) 󸀠 0 S =(S; S ), (2) 󸀠 According to formula (4), S canbewrittenas where S is the real unit, which is the direction ratios of 0 the screw axis and is not origin-dependent, S is the dual S󸀠 =(F; C0 −ℎF)+(0;ℎF) , 0 (8) unit, which is not origin-dependent. S and S are three- 0 dimensional vectors. where C −ℎF = r × F; ℎ canbeobtainedbyformula(3). S =0̸ ℎ 0 If ,thepitchofascrewis : Therefore, (F; C −ℎF) canbeseenasaforce,and(0;ℎF) S ⋅ S0 canbeseenasacouple.Theyhavethesamedirection. ℎ= . (3) S ⋅ S 0 3. Expressions of Constraint Forces of the In order to decide the position of the axis of a screw, S Kinematic Pair of Mechanism by Screw canbedecomposedintotwoparts(Figure 2). One is parallel to S and the other is perpendicular to S. Consider The specific Plucker¨ coordinate of twist and wrench can be 󸀠 0 written as [10] S =(S; S −ℎS)+(0;ℎS) . (4) S󸀠 =(𝐿,𝑀,𝑁;𝑃∗,𝑄∗,𝑅∗), 0 0 (9) Since S −ℎS is perpendicular to S, S −ℎS = S0. 0 0 ∗ ∗ ∗ (S; S −ℎS) represents a line vector. (S;ℎS) represents a couple. where S =(𝐿,𝑀,𝑁), S =(𝐿 ,𝑀 ,𝑁 ). 󸀠 Obviously, a screw can be decomposed into a line vector and If S represents a twist, (𝐿, 𝑀,)isitsangularvelocity, 𝑁 ∗ ∗ ∗ a couple. According to this formula, a line vector is a special and (𝐿 ,𝑀 ,𝑁 ) is its linear velocity. 󸀠 screw. If S represents a wrench, (𝐿, 𝑀,) 𝑁 is its force part, and ∗ ∗ ∗ The screw can be used to describe motions and forces. (𝐿 ,𝑀 ,𝑁 )isitscouplepart. They are, respectively, called twist and wrench. The instan- In order to calculate the reciprocal screw, a screw can be taneous twist of a rigid body can be written as expressed by a row vector: 󸀠 0 󸀠 ∗ ∗ ∗ S =(𝜔; k ), (5) S =[𝐿𝑀𝑁𝑃 𝑄 𝑅 ]. (10) Mathematical Problems in Engineering 3

A kinematic pair is a combination of two kinematic where G is a (6−𝑛)by(𝑛+1)matrix;C𝑗 is a new bodies which have relative motion with respect to each other. matrix which is formed by deleting the 𝑗th column of Its motion can be described by screw system. The order of the C matrix. The element of V in the 𝑖th row and 𝑘th screw system is the same as the mobility. If the mobility of a column is V𝑖𝑘. It satisfies the following relationship: kinematic pair is 𝑛 (1≤𝑛≤5, 𝑛 is an integer), its twist is expressed as V𝑖𝑘 = G𝑖𝑗 , (16)

∗ ∗ ∗ 𝐿1 𝑀1 𝑁1 𝑃 𝑄 𝑅 𝑘=O [ 1 1 1 ] where 𝑖𝑗 . 󸀠 [ . . ] S = [ . . ] . The rest of elements of the 𝑖th column of V equal 0. [ . . ] (11) C 𝐿 𝑀 𝑁 𝑃∗ 𝑄∗ 𝑅∗ If there is a zero row vector in the matrix ,itis [ 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 ] removed forming a new matrix M. 𝑛 columns are M 𝑛−1 𝑛 N S󸀠 S󸀠r chose from which forms a ( )by matrix .The is a screw system, and its reciprocal screw system is . numerical order of the chosen columns in the matrix 󸀠 They satisfy the following relationship: S are saved in the 𝑖th row of the matrix H which is a 󸀠 󸀠r (6−𝑛)by𝑛 matrix. Consider S ∘ S =0, (12) 𝑛+𝑖+𝑝 󵄨 󵄨 G = (−1) 󵄨N 󵄨 (1≤𝑝≤𝑛), where ∘ is the reciprocal product. 𝑖𝑝 󵄨 𝑝󵄨 (17) 󸀠r Let V = S Δ,whereΔ is transformation operator between the real unit vector and the dual unit vector. The where G is a (6−𝑛)by(𝑛+1)matrix;N𝑝 is a new expression of Δ is matrix which is formed by deleting the 𝑝th column of the N matrix. The element of V in the 𝑖th row and 𝑘th 0I column is V𝑖𝑘. It satisfies the following relationship: Δ=[ ], (13) I0 V𝑖𝑘 = G𝑖𝑝, (18) 󸀠 where I is a 3 by 3 unit matrix. 󸀠 𝑘=H S and V satisfy the following relationship: where 𝑖𝑗 . 𝑖 V 󸀠 The rest of elements of the th column of equal 0. S V =0. (14) (iii) If V can satisfy the requirements, the solution pro- S󸀠 V canberegardedasthenullspaceofthescrewsystems cedures of the reciprocal screw system of are 󸀠 V S . Based on Huang’s observation method and Dai’s linear finished. If cannot satisfy the requirements, the dissatisfactory row is removed and step (2) is repeated algebraic method obtaining reciprocal screw systems, we V improve the solution method of V. The explicit procedures until can satisfy the requirements. are as follows. The reciprocal screw system of twist system is expressed (1) V The solution procedures for the row vectors of as which correspond to the zero column vectors of screw 󸀠 󸀠𝑟 systems S can be described as follows. S = VΔ. (19) There are 𝑒(0≤𝑒≤5)zero column vectors in the screw 󸀠 systems S . D is a submatrix which is the preceding 𝑒 rows The constraint wrench of the kinematic pair is written as of V. D𝑟𝑠 is the element of D in the 𝑟th row and 𝑠th column, 󸀠𝑟 󸀠𝑟 󸀠𝑟 1≤𝑟≤𝑒, 1≤𝑠≤6. 𝑠 is the numerical order of the 𝑟th zero S𝐹 =𝑓1S1 +𝑓2S2 +⋅⋅⋅+𝑓6−𝑛S6−𝑛, (20) 󸀠 column vector in the S .Inthismethod,D𝑟𝑠 equals 1, and the rest of elements of D equal 0. where 𝑓𝑖 (𝑖= 1,...,6− 𝑛) is the solution coefficient of the (2) The solution procedures of the rest of row vectors of constraint wrench. V are as follows. (i) The solution of the 𝑖th row vector of V, 𝑒+1 ≤ 𝑖≤ 4. The Transformation of Twist of Kinematic 6−𝑛. 𝑛+1columnswherethezerocolumnvectoris Pair between Different Coordinate Frames 󸀠 excluded are chosen from S forming the matrix C.If Expressions of twist of kinematic pair of spatial mechanism 𝑖 is different, the chosen column vectors are different. are hard to be obtained. It is necessary to set up a new coor- The numerical orders of the chosen columns are saved dinate frame. The rotation and displacement transformation in the 𝑖th row of the matrix O in order which is a (6−𝑛) matrix from the coordinate frames 2 to 1 are, respectively, R by (𝑛+1)matrix. and d. (ii) If there are no zero rows in the matrix C,thefollowing The twist of a kinematic pair in the coordinate frame 2 is equation is established: written as 󵄨 󵄨 G = (−1)𝑛+𝑖+𝑗 󵄨C 󵄨 (1 ≤ 𝑗 ≤ 𝑛 + 1) , S󸀠 =(S ; S0). 𝑖𝑗 󵄨 𝑗󵄨 (15) 2 2 2 (21) 4 Mathematical Problems in Engineering

. 󸀠 . revolute pair and the prismatic pair are, respectively, S𝑅 and . . 󸀠 . S . 𝑀: S . Si−1 i−2 󸀠 S S =[001𝑦−𝑥0], Linki−1 i−n 𝑅 Link (27) i−2 󸀠 S S Linki−n S𝑀 =[000cos (𝜏) sin (𝜏) 0], (i−1)i (i−2)i ··· S(i−n)i (i − 1)i (i − 2)i where 𝜏 istheincludedanglebetweenthedisplacement (i − n)i direction of the prismatic pair and the 𝑥-axis direction. Linki The reciprocal screw system of the rotation pair of planar mechanism is given by

10000−𝑦 Si [ ] [01000 𝑥] [ ] Figure 3: The force figure of link 𝑖. [ ] S󸀠𝑟 = [00100 0] . 𝑅 [ ] (28) [ ] [00010 0] Itcanbedividedintotwoparts: [00001 0] S󸀠 =(S ; r × S ), 21 2 2 2 According to formulas (20) and (28),theconstraint (22) 󸀠 wrench of the rotation pair in the planar mechanism is given S =(0;ℎS2). 22 by The twist in the coordinate frame 1 is expressed as S𝐹 =[𝑓1 𝑓2 𝑓3 𝑓4 𝑓5 −𝑓1𝑦+𝑓2𝑥]. (29) 󸀠 0 S1 =(S1; S1). (23) The component force of the 𝑧 direction of the rotation 𝑥𝑜𝑦 𝑥 𝑦 Itcanbedividedintotwoparts: pair in the plane and the moment about -and -axis equals zero. Therefore, 𝑓3 =𝑓4 =𝑓5 =0.Thereciprocal 󸀠 S11 =(S1; r1 × S1), screw system of the rotation pair can be simplified as (24) 󸀠 10000−𝑦 S12 =(0;ℎS1), 󸀠𝑟 S =[ ]. (30) 01000 𝑥 where S1 = RS2, r1 = Rr2 + d. The twist in the coordinate frame 1 also can be expressed According to formulas (20) and (28),theconstraint as wrench of the rotation pair in the planar mechanism is 󸀠 expressed as S1 =(RS2;(Rr2 + d)×(RS2)+ℎ(RS2)) . (25) 󸀠 󸀠𝑟 󸀠𝑟 S𝐹 =𝑓1S1 +𝑓2S2 . (31) 5. Equilibrium Equations of Links Similarly, the reciprocal screw system of the prismatic In the mechanism, wrenches acting on links can be classified pair can be simplified as into two types. One is the constraint wrenches of kinematic pair and the other is the external wrench. In Figure 3,link − (𝜏) (𝜏) 0000 󸀠𝑟 sin cos 𝑖 which is a part of a mechanism has 𝑛 kinematic pairs. S =[ ], (32) When it is in an equilibrium state, the following equation is 0 0 0001 established [14, 15]: where 𝜏 istheincludedanglebetweenthedisplacement S(𝑖−1)𝑖 + S(𝑖−2)𝑖 +⋅⋅⋅+S(𝑖−𝑛)𝑖 + S𝑖 =0, (26) direction of the prismatic pair and the 𝑥-axis direction. According to formulas (20) and (28),theconstraint where S(𝑖−𝑘)𝑖 is the constraint wrench acted by link (𝑖−𝑘); S𝑖 wrench of the planar prismatic pair is shown as is the resultant external wrench. 󸀠 󸀠𝑟 󸀠𝑟 S𝐹 =𝑓1S1 +𝑓2S2 . (33) 6. Numerical Examples In order to verify the validity of the foregoing method, a 6.1. The Solution for Constraint Wrenches of Kinematic Pairs slider-crank mechanism (Figure 4) is taken as an example. in the Planar Mechanism. The common kinematic pairs in The known geometry conditions are as follows: 𝜏 = ∘ the planar mechanism mainly have the revolute pair and the 0 , 𝑂𝐴 = 100 mm, and 𝐴𝐵 = 400 mm. In this paper, the prismatic pair. In the coordinate frame 𝑜-𝑥𝑦𝑧, the twist of the unitsofforceandmomentare,respectively,NandN⋅mm. Mathematical Problems in Engineering 5

y 𝛿 A l1 S2 M S b 1 l2 S3 3 C O 𝜃 B x l1 B 4

l Figure 4: The slider-crank mechanism. z1 3 z0 x0 𝛽 x y x1 z 2 1 𝛼 2 𝜃 y0 D y2 The resultant external wrenches acting on each link and the A trimming moment acting on link 1 are, respectively, given by

S1 =[30 40 0 0 0 −270], Figure 5: The single loop spatial RUSR mechanism.

S2 =[25 54 0 0 0 440], (34) S3 =[−24670000], Based on the formula (26) and Figure 3,theanalysismodelof the constraint wrenches of kinematic pairs obtained S𝑏 =[00000𝑀𝑑]. S𝐹𝑂 + S1 + S𝑏 − S𝐹𝐴 =0, Twists of kinematic pairs in the mechanism are written as S𝐹𝐴 + S2 − S𝐹𝐵 =0, (38)

󸀠 S𝐹𝐵 + S3 − S𝐹𝐶 =0. S𝑂 =[001𝑦𝑂 −𝑥𝑂 0], 󸀠 S𝐴 =[001𝑦𝐴 −𝑥𝐴 0], Each wrench equilibrium equation can expand to 3 equa- (35) tions. Therefore, 9 equations can be obtained in total. Since 󸀠 S𝐵 =[001𝑦𝐵 −𝑥𝐵 0], there are 8 unknown solution coefficients and a trimming moment, the number of unknowns equals the number of 󸀠 000𝑦 −𝑥 0 S𝐶 =[ 𝐶 𝐶 ]. equations. The analysis model can be solved. The constraint wrenches of kinematic pairs and the trimming moment of the The reciprocal screw systems of kinematic pairs in the slider-crank mechanism can be obtained: mechanism are given by S𝐹𝑂 =[−31.0 −104.6 0 0 0 0],

10000−𝑦 S𝐹𝐴 =[−1.0 −64.6 0 0 0 −5547.5], S󸀠𝑟 =[ 𝑂], 𝑂 01000 𝑥 𝑂 S𝐹𝐵 =[24 −10.6 0 0 0 −5107.5], (39)

10000−𝑦𝐴 S =[0 56.4 0 0 0 −5107.5], S󸀠𝑟 =[ ], 𝐹𝐶 𝐴 01000 𝑥 𝐴 S =[0 0 0 0 0 −5277.5]. (36) 𝑏 10000−𝑦 S󸀠𝑟 =[ 𝐵], 𝐵 6.2. The Solution for the Constraint Wrenches of the Universal 01000 𝑥𝐵 Spatial Kinematic Pairs. The universal spatial kinematic pairs 010000 include the revolute pair, universal joint, spherical pair, and S󸀠𝑟 =[ ]. 𝐶 000001 prismatic pair. In order to learn the solution method of the constraint wrenches in the spatial kinematic pairs, the single loop spatial RUSR mechanism is taken as example. According to formulas (29), (31),and(33),theconstraint Its schematic figure is shown in Figure 5. 𝐴 and 𝐷 are wrenches of kinematic pairs are as follows: the revolute pairs. 𝐵 and 𝐶 are, respectively, the universal joint and the spherical pair. In order to express twists of 𝐴 𝑥 𝑦 𝑧 S𝐹𝑂 =[𝑓1 𝑓2 0000], kinematic pairs easily, the coordinate frame - 0 0 0,the local coordinate frame 𝐴-𝑥1𝑦1𝑧1,andthelocalcoordinate S𝐹𝐴 =[𝑓3 𝑓4 0 0 0 86.6𝑓4 − 50𝑓3], frame 𝐷-𝑥2𝑦2𝑧2 aresetup.Theaxes𝑥0 and 𝑥1 are coincident (37) with the axis direction of 𝐴,andtheaxis𝑧1 is coincident with S =[𝑓 𝑓 0 0 0 480.3𝑓 ], 𝐹𝐵 5 6 6 link 1. The included angle between 𝑧1 and 𝑧0 is 𝛼.Theincluded angle between 𝑥2 and 𝑥0 is 𝜃.Theincludedanglebetween𝑧2 S =[0𝑓7 000𝑓8]. 𝐹𝐶 and 𝑧0 is 𝛽. 6 Mathematical Problems in Engineering

The known conditions are as follows: 𝑙1 = 250 mm, 𝑙2 = Twists of kinematic pairs in the different coordinate ∘ ∘ 350 mm, 𝑙3 = 150 mm, and 𝑙4 =500mmand𝜃 =30, 𝛼 =60, frames are shown as ∘ ∘ 󸀠 𝛽 =60,and𝛿 =30. The resultant external wrenches acting S𝐴0 =[100000], on link 2 and link 3 are, respectively, 10 0 0𝑙0 󸀠 1 S𝐵1 =[ ], 0−𝑍1 𝑍2 𝑙1𝑍1 00 S2 =[−60 52 −42 230 320 440], (40) 100 0 𝑙𝑍 𝑙 𝑍 𝑍 +𝑙 S3 =[−54 67 21 268 −270 132]. 3 4 3 3 6 4 󸀠 [ ] S𝐶0 = [010 −𝑙3𝑍4 0𝑙3𝑍3𝑍5 ] ,

The coordinates of each kinematic pairs in the different [001−𝑙3𝑍3𝑍6 −𝑙4 −𝑙3𝑍3𝑍5 0 ] coordinate frames are given by 󸀠 S𝐷2 =[100000]. (43) 𝐴0 =(000),

𝐵1 =(00𝑙1), The rotation transformation matrix from coordinate (41) frames 2 to 1 are, respectively, shown as follows: 𝐶2 =(00𝑙3), 10 0 𝑍6 −𝑍4𝑍5 𝑍3𝑍5 𝐷0 =(0−𝑙4 0), [ ] [ ] T10 = [0𝑍8 −𝑍7] , T20 = [𝑍5 𝑍4𝑍6 −𝑍3𝑍6] . [0𝑍 𝑍 ] [ 0𝑍 𝑍 ] where the subscripts represent the numerical order of coor- 7 8 3 4 dinate frames. (44) In the following parts, notations are defined as The displacement transformation matrix from coordinate frames 2 to 1 are, respectively, shown as 𝑍1 = sin (𝛿) , 0 0 𝑍2 = cos (𝛿) , [ ] [ ] P10 = [0] , P20 = [−𝑙4] . (45) 𝑍3 = sin (𝛽) , [0] [ 0 ] 𝑍 = (𝛽) , 4 cos According to formula (25),thecoordinatesandthetwist system of each kinematic pair are obtained: 𝑍5 = sin (𝜃) ,

𝐵0 =(−𝑙1𝑍7 𝑙1𝑍8 0), 𝑍6 = cos (𝜃) , (42)

𝐶0 =(𝑙3𝑍3𝑍5 −𝑙3𝑍3𝑍6 −𝑙4 𝑙3𝑍4), 𝑍7 = sin (𝛼) , 10 0 0𝑙𝑍 𝑙 𝑍 󸀠 1 8 1 7 (46) 𝑍8 = cos (𝛼) , S𝐵0 =[ ], 0−𝑍9 𝑍10 𝑙1𝑍1 00 𝑍9 = sin (𝛼+𝛿) , 󸀠 S𝐷0 =[𝑍6 𝑍5 000𝑙4𝑍6]. 𝑍10 = cos (𝛼+𝛿) , Based on the improved solution method of V,therecip- 𝑍 =𝑙 (𝛽) (𝜃) +𝑙 . 11 3 sin cos 4 rocal screw systems of kinematic pairs are shown as

100000 [ ] [010000] [ ] [ ] S󸀠𝑟 = [001000] , 𝐴0 [ ] [ ] [000010] [000001]

0 000−𝑍10 −𝑍9 [ ] [ 2 ] [ 0𝑙1𝑍1 0−𝑙1𝑍1𝑍8 00] S󸀠𝑟 = [ ] , 𝐵0 [ 2 ] [𝑙1𝑍10𝑍8 0000−𝑙1𝑍1𝑍8] 2 2 [ 0−𝑙1𝑍1𝑍7 𝑙1𝑍1𝑍8 000] Mathematical Problems in Engineering 7

1000𝑙3𝑍4 𝑍11 [ ] 󸀠𝑟 [ 2 ] S𝐶0 = [ −𝑙3𝑍3𝑍5 𝑍11 0𝑙3𝑍4𝑍11 −𝑙3𝑍4𝑍3𝑍5 0 ] , 2 2 2 [𝑙3𝑍4𝑍3𝑍5 −𝑙3𝑍4𝑍11 𝑙3𝑍4 000] 10 0 0 0 0 [ ] [01 0 0 0 0] [ ] [ ] S󸀠𝑟 = [00 0 0 0 1] . 𝐷0 [ ] [ ] [000𝑍5 −𝑍6 0]

[00−𝑍5 0𝑙4𝑍6 0] (47)

According to formula (20), the constraint wrenches of all kinematic pairs in the spatial mechanism are obtained:

S𝐴𝐹 =[𝑓1 𝑓2 𝑓3 0𝑓4 𝑓5],

2 2 2 2 S𝐵𝐹 =[𝑓8𝑙1𝑍10𝑍8 𝑓7𝑙1𝑍1 −𝑓9𝑙1𝑍1𝑍7 𝑓9𝑙1𝑍1𝑍8 −𝑓6𝑙1𝑍1𝑍8 −𝑓5𝑍10 −𝑓5𝑍9 −𝑓7𝑙1𝑍1𝑍], (48) 2 2 2 2 S𝐶𝐹 =[𝑓10 −𝑓11𝑙3𝑍3𝑍5 +𝑓12𝑙3𝑍3𝑍4𝑍5 𝑓11𝑍11 −𝑓12𝑙3𝑍11𝑍4 𝑓12𝑙3𝑍4 𝑓11𝑙3𝑍4𝑍11 𝑓10𝑙3𝑍4 −𝑓11𝑙3𝑍4𝑍3𝑍5 𝑓10𝑍11],

S𝐷𝐹 =[𝑓13 𝑓14 −𝑓17𝑍5 𝑓16𝑍5 −𝑓16𝑍6 +𝑓17𝑙4𝑍6 𝑓15].

According to formula (26) and Figure 3,theanalysis (2) The improved solution can solve the reciprocal screw model of the constraint wrenches of the single loop RUSR system by the programming conveniently and swiftly. mechanism is obtained: It can enhance the solution efficiency of constraint wrenches. S𝐴𝐹 + S1 + S𝑏 − S𝐵𝐹 =0, (3) The constraint wrench has an important significance S𝐵𝐹 + S2 − S𝐶𝐹 =0, (49) for the analysis and application of the mechanism. According to the examples mentioned in this paper, S + S − S =0. 𝐶𝐹 3 𝐷𝐹 the solution method of the constraint wrenches of kinematic pairs, which is based on the reciprocal Equations (49) can expand into 18 equations. Since screw, can solve the constraint wrenches of the planar thereare17unknownsolutioncoefficientsandatrimming and spatial mechanism. moment, the number of unknowns equals the number of equations. The analysis model is solvable. The constraint wrenches of kinematic pairs and the e trimming moment of Conflict of Interests the slider-crank mechanism are shown as follows: The authors declare that there is no conflict of interests S𝐴𝐹 =[2251.5 5040.0 −682.3 0 −8.2 989.1], regarding the publication of this paper. S =[2251.5 5040.0 −682.3 −482.3 −8.2 989.1], 𝐵𝐹 Acknowledgment S =[2191.5 5092.0 −724.3 −252.3 311.8 1429.1], 𝐶𝐹 The authors gratefully acknowledge the financial support of S =[2137.5 5159.0 −703.3 15.7 581.8 1561.1]. the National Natural Science Foundation of China (Grant no. 𝐷𝐹 51175422). (50) References 7. Conclusion [1] J. J. Uicker, G. R. Pennock, and J. E. Shigley, Theory of Machines (1) The constraint wrenches of kinematic pairs can be and Mechanisms, Oxford University Press, Oxford, UK, 2011. expressed by the reciprocal screw system. Based on [2] Y. Zhou, L. Liu, and F. Gao, “Static full-solutions of spherical this fact, the solution method of constraint wrenches parallel mechanism 3-RRR with 3-DOF,” Chinese Journal of is formed. Mechanical Engineering, vol. 44, no. 6, pp. 169–176, 2008. 8 Mathematical Problems in Engineering

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