Study of Planar Mechanisms Kinetostatics Using the Theory of Complex Numbers with MathCAD PTC

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Study of Planar Mechanisms Kinetostatics Using the

Theory of Complex Numbers with MathCAD PTC

I.N. Matsyuk, E.�. Shlyakhov, N.V Zyma

To cite this version:

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Numbers with MathCAD PTC

Matsyuk I.N. 1,a, Shlyakhov E.М. 1, Zyma N.V.1

1 – State Higher Educational Institution “National Mining University, Dnipro, Ukraine a – [email protected]

DOI 10.2412/mmse.40.52.685 provided by Seo4U.link

Keywords: MathCAD, planar mechanism, vector, complex number, second class Assur group, kinetostatic.

ABSTRACT. The paper describes the convenient method to study the kinematics of planar mechanisms. Very convenient to study the kinematics of planar mechanisms with the help of complex numbers. The current article proposed the kinetostatic analysis also be carried out by operating in the field of complex numbers.

It is recommended to use the MathCAD PTC, having great potential for operations with complex numbers. The basic idea is presented by the example of the three second class Assur groups found most frequently in modern planar mechanisms.

Introduction. As it is well known, complex numbers are used to solve geometric problems [1] and

for research purposes of the motion of planar bodies and mechanisms [2, 3, 4]. Information about complex numbers usage to solve kinetostatics problems of planar mechanisms has been described insufficiently in known literature.

In case, when inertia forces of the links are taking into account for kinetostatics analysis of mechanism, to perform kinematic analysis is necessary first of all. This analysis made with representing vectors as complex numbers and has a definite advantage over other methods. In this case, it is logical to provide force analysis using complex numbers too. This is facilitated by the fact, that the popular modern mathematical software (Maple, Wolfram Mathematica, MathCAD) have appropriate calculation algorithm to operate with such numbers.

Analysis of the recent research. The research of the force analysis of mechanisms does not represent any difficulties today. Historically, in the beginning it has been solved only using the graphical-analytical methods [5]. Gradually graphical-analytical methods are substituted by purely analytical methods, that became a prior due to development of computer technology [6 - 9].

Determination of forces in kinematic pairs can be carried out separating the mechanism on the Assur’s groups [10] which are kinetostatically determined [11]. Another approach is to partition mechanism on links, for each of which the equilibrium equation should be written. Solving the obtained system of equations defines all unknown reactions [12].

These reactions may be represented by two components (normal and tangential [11] or horizontal and vertical [12]), i.e. vectors with known directions, but unknown magnitudes. Representation of required reactions as unknown vectors is more compact. In this case, the number of equations is minimal [7].

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The position of the slider

s

₅mass center coincides with point F.

a)

b)

Fig. 1. Second class Assur group (connecting rod – slider): a – group skeleton; b – calculation scheme.

The initial data for the calculation are as follows: • geometrical parameters – l₄l_EF0,15 m,  45 ; • links masses – m₄m_EF 6 kg; m₅m_sl 20 kg; • links inertia moments – 3

4 11,2 10 

s EF

I I kgm2;

• resistance force – P500 N.

As it is known, from previous kinematical analysis:

connecting rod EF, is interpreted by the vector l₄0.109 0.103i ;

vectors of acceleration of links mass centres a_s₄ 59.32 39.14 i and a_s₅ 52.69 52.69 i;

connecting rod angular acceleration – ₄ 192 s-2_. The acting loads on the links:

gravity forces (hereinafter use approximate value of the gravitational accelerationg 10 m/s2) 4 60i

G and G₅400i;

resistance force – P 353.6 353.6i ;

inertial loads – F_i₄355.9234.8i, F_i₅2105 2105 i, M_i₄2.15.

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P 565.7565.7 i Mi4 0.284 Mi4 Is44 ₄  25.4 Fi5 69.59 69.59i  Fi5 m5as5 Fi4 47.04 31.08i  Fi4 m4as4 Is4 11.2 10 3 G5  100i G4  60i m5  10 m4  6 as5 6.9596.959 i as4 7.845.18 i l4  0.109 0.103 i   4  . P  565.7565.7 i Mi4 0.284 Mi4 Is44 ₄  25.4 Fi5 69.59 69.59i  Fi5 m5as5 Fi4 47.04 31.08i  Fi4 m4as4 Is4 11.2 10 3 G5  100i G4  60i m5  10 m4  6 as5 6.9596.959 i as4 7.845.18 i l4  0.109 0.103 i   4  .

Fig. 2. Input of initial data.

Let’s proceed directly to the forces analysis of group, in which necessary to determine the forces in external kinematic pairs of groups R₂₄ (action, for instance, of link 2 on link 4) and R (action of ₀₅ frame 0 on slider 5), and the force R₅₄in the internal revolute joint (action of slider 5 on connecting rod 4).

Generally, a system of four vector equations can be written: two for each of the links in the group. In the first equation the sum of all the forces acting on the link is zero. In the second one, the sum of the moments of these forces around any centre is zero. In our case, the equation of moments will be one - only for the connecting rod [7].

Vector of equilibrium equations of forces acting on the links:

link 4 – R₂₄G₄F_i₄R₅₄0 (1) link 5 – R₅₄G₅F_i₅ P R₀₅0. (2) The equilibrium equation of moments acting on the connection rod 4 around its mass centre:

4 24 4 4 05

0.5 _i 0.5 0

 l R M  l R  . (3)

In formulas (1) and (2) two vectors are unknown R and ₂₄ R . The last vector’s magnitude is ₀₅ unknown. Its direction is known: it is perpendicular to the trajectory of the slider. From the theory of complex numbers known, that the condition of the perpendicular vector R to line, which tilted ₀₅ to X axis on angle  will be equal to:

05 05

i i

e e

   

R R , (4)

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Condition (4) must be added to equations (1) and (2).

The vector product represented by complex numbers finds in the following way. If to multiply the conjugate to one of the two complex numbers to the second, then the imaginary part of the multiplication considering the sign is the vector product of these vectors.

In MathCAD a small subroutine to determine the moment of force can be arranged in case, if a lever arm h and the force F are known:

( , ) : ( )

M h F Im h F .  (5)

The resulting system of equations in MathCAD solved by a block Given-Find. The corresponding fragment is given below (fig. 3).

M h F(  )  Im h

 

F R24 100 100 i _R54  100 100 i _R05  100 100 i Given R24 G4  Fi4R54 0 _R54 _G5 _Fi5P _R05 0 R05 exp

 

i _R05exp

 

i M



0.5_{l4 R24}



 _Mi4 _{M 0.5l4 R54}







0 R24 R54 R05





 Find R24 R54 R05





R24  532 542i _R05  83 83i _R54  579513i

Fig. 3. The resulting system of equations.

Next, consider the Assur group consisting of a connecting rod 2 (triangle BCE), and the rocker 3 (Fig. 4). The rocker and frame form revolute joint D and the connecting rod is connected by joints B and E to links 1 and 4. The mass centre of connecting rod is located at the intersection of medians of the triangle BCE.

Consider the following initial data:

• geometrical parameters – l₂ l_BC 0,15 m, l₂₁l_BE0,09 m, l_BE l_CE;

• links masses – m₂4 kg; the rocker mass is small and it has not been took into account; • link’s moment of inertia – 3

2 7,5 10 s

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a) b)

Fig. 4. Second class Assur group (connecting rod – rocker): a – group skeleton; b – calculation scheme

In addition, we have previously obtained kinematics analysis:

the sides of triangular connecting rod is interpreted by vectors: l₂ 0.145 0.037i , 210.061 0.067i

l and link 3 – by vector l₃0.021 0.072i ;

the acceleration vector of the mass centre for the link 2: a_s₂  7.415 2.871 i;

rocker angular acceleration 2–  ₂ 42.25 s-2. The loads acting on the links in the group:

gravity forces (hereinafter use approximate value of the gravity constantg10 m/s2) G₂ 40i;

a reaction of link 4 on link 1 (considered above) R₄₂ 532 542 i  , (N);

inertia loads – F_i₂ 29.66 11.48 i, M_i₂  0.317.

Below a MathCAD fragment is given, which shows the initial data input (fig. 5).

l2  0.145 0.037 i _l21  0.0610.067 i _l3 0.021 0.072 i as2 7.4152.871 i _m2  4 _G2 40i _Is2 7.5 10 3 Fi2 m2as2 Fi2 29.66 11.48i  2  42.25

Mi2 Is22 Mi2  0.317 R42  532542 i

Fig. 5. Initial data input.

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The link 3 does not affected by the external loads, so a force R₃₂ is equal to a force R and both ₀₃ these forces are directed parallel to the longitudinal axis of link 3.

Therefore, in this case it is enough to write a system of two vector equations (forces and moments) for link 2 and to add a condition of parallelism of vector R to vector ₃₂ l . ₃

32 3 32 3 R l R l

For writing the equilibrium equation of moments of forces for link 2 around a point B, first, we have to find the vector, that expresses a lever armlBS₂of forces G₂and F . i2

2 2 21 2 3 2 BS   l l l .

The respective fragment of MathCAD is given below (fig. 6).

M h F(  )  Im h

 

F _lBs2 2 3 l2 l21 2   _{lBs2 0.069 0.035i}  R12  100 100 i _R32  100 100 i Given R12 G2  Fi2 R42R32 0 _{R32 l3} _R32_l3 M lBs2 G2 Fi2













 Mi2 M l21 R42







 M l2 R32







0 R12 R32





 Find R12 R32





R12  501 565i _R32  1.57 5.37i R03  R32 R03  1.57 5.37i M h F(  )  Im h

 

F _lBs2 2 3 l2 l21 2   _{lBs2 0.069 0.035i}  R12  100100 i _R32  100100 i Given R12 G2  Fi2 R42 R32 0 _{R32 l3} _R32_l3 M lBs2 G2 Fi2













 Mi2M l21 R42







 M l2 R32







0 R12 R32





 Find R12 R32





R12  501 565i _R32  1.575.37i R03  R32 R03  1.575.37i .

Fig. 6. MathCAD PTC fragment.

At last, consider the Assur group consisting of slotted link 3 and slider block 2 (Fig. 7). The slotted link is rotates about fixed point C, and is connected to link 4 by a joint D. The mass centre of slotted link

s

₃ is located at the middle of it.

So, take the following initial data:

• geometrical parameters – l₃l_CD 0,49m, l₃₁l_CB0,329 m;

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a) b)

Fig. 7. A second class Assur group ( slotted link-slider block): a – skeleton of group; b – diagram for a forces calculation

In addition, we know from previous kinematic analysis:

the slotted link CD and its part CB were interpreted by vectors: l₃0.17 0.46i и 310.114 0.308i

l ;

the acceleration vector of the mass centre for the link: a_s₃ 2.25 0.056 i;

angular acceleration of the slotted link –  ₃ 8.52 s-2; The loads acting on the links in the group:

gravity forces (hereinafter are using approximate value of the acceleration of free fall g10 m/s2₎

3 180i

G ;

a reaction of link 4 on link 3 (of considered above) R₄₃1640 2i ,(N);

inertial loads – F_i₃40.47 1,01 i, M_i₃ 3.66.

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l3  0.17 0.46 i _l31  0.114 0.308 i as3 2.250.056 i

m3  18 _G3  180i _Is3 0.43 Fi3 m3as3 Fi3 40.5 1.01i 

₃  8.52 _Mi3 _Is3₃ _Mi3  3.66 R43  16402 i

Fig. 8. Input of initial data.

It is necessary to determine the forces in the external joints B and C R (action of link 1 on link 2) ₁₂ and R (action of frame 0 on slotted link 3) and the reaction of slider 2 on slotted link 3 – ₀₃ R . ₃₂ On the first stage of forces analysis, we do not include friction, so force R is equal to the force ₂₃

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R , and both these forces are directed perpendicular to the longitudinal axis of slotted link 3. Therefore, in this case it is enough to write a system of two vector equilibrium equations (of forces and moments about a point C) for link 3 and add a condition of perpendicularity of the vector R₂₃

to the vector l . ₃

23  3 23 3

R l R l

The respective fragment of MathCAD is given below (fig. 9).

M h F(  )  Im h

 

F R23  100 100 i _R03  100 100 i Given R23 G3 Fi3 R43 R03 0 _{R23 l3} _R23_l3 M l3 R43







 Mi3 M l31 R23







 M 0.5l3 G3 Fi3













0 R23 R03





 Find R23 R03





R23  2075767i _R03  475.2 945.5i R12  R23 R12  2075767i .

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1

is located on its axis of rotation.

a) b)

Fig. 10. The input crank: a – skeleton of group; b – diagram for a forces calculation.

Take the following initial data: • crank length – l₁0,49 m; • mass of crank – m₁3 kg. Previously we identified:

the crank AB was interpreted by vectors: l₁0.114 0.038i ;

angular acceleration of the crank – ₁0. The loads acting on the crank:

gravity forces G₁ 30i;

a reaction value which acts from link 2 to the crank 1 – R₂₁ 2072 764i ,(N);

inertial loads – F_i₁0, M_i₁0.

The following is a fragment MathCAD, which shows the input of initial data (fig. 11).

l1  0.114 0.038 i m1  3 _G1  30i R21  2072 764 i

.

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From the equilibrium conditions of the crank (R₂₁G₁R₀₁0; M₁ l₁ R ) easily find the ₂₁ force R and the torque required to drive the input link 1₀₁ M₁ (fig. 12).

M h F(  )  Im h

 

F _R01  _G1_R21 _R01  2072734i M1  M l1 R21







M1  165.8

.

Fig. 12. MathCAD document fragment.

Summary. Thus, kinematics analysis and force analysis of planar linkage can be performed in a

field of complex numbers. MathCAD PTC has enough powerful apparatus for work with them. The vector product of vectors in complex form does not lie in the field of complex numbers and can not be expressed by a complex number. It is proposed to find module of vector product using the properties of complex numbers, which allow including in the equilibrium equations of the moments of force to determine unknown reactions in the kinematics pairs.

References.

[1] Podolskiy M.E. (1954). O primenenii kompleksnyih chisel k izucheniyu ploskogo dvizheniya tverdogo tela, Trudyi Leningradskogo korablestroitelnogo institute, pp. 213-218.

[2] Ponarin Ya. P. (2004) Algebra kompleksnyih chisel v geometricheskih zadachah, MTsNMO, 160 p.

[ 3 ] Kinematics and dynamics of machines by Martin, George Henry (1969), New York, McGraw-Hill.

[4] Fundamentals of Kinematics and Dynamics of Machines and Mechanisms (2000). Oleg Vinogradov, CRC Press Edition, 306 p

[5] Yudin V.A. Kinetostatika ploskih mehanizmov (1939). Moscow, Izdanie voenno-inzhenernoy akademii RKKA, 205 p.

[6] F.Y. Zlatopolskiy, G.B. FIlImonIhIn, V.V. Kovalenko, O.B. Chaykovskiy (2000). Rozrahunok ploskih mehanizmiv z vikoristannyam PEOM. Navchalniy posibnik, Kirovograd, PP «KOD», 124 p.

[7] Matsyuk I.N., Shlyahov E.M. (2015). The research of plane link mechanisms of a complicated structure with vector algebra methods. Eastern-European Journal of Enterprise Technologies, 3 (7 (75)), 34–38. doi: 10.15587/1729-4061.2015.44236.

[8] Matsyuk I.N., Zyma N.V., Shlyahov E.M (2014). Kinematika ploskih mehanizmov v programme MathCAD s ispolzovaniem teorii kompleksnyih chisel. Sbornik nauchnyih trudov mezhdunarodnoy konferentsii «Sovremennyie innovantsionnyie tehnologii podgotovki inzhenernyih kadrov dlya gornoy promyishlennosti i transporta 2014», NGU, pp. 514-520.

[9] Matsyuk I.N., Shlyahov E.M. (2013) Opredelenie kinematicheskih i kinetostaticheskih parametrov ploskih sterzhnevyih mehanizmov slozhnoy strukturyi, Sovremennoe mashinostroenie. Nauka i obrazovanie: Materialyi 3-y Mezhdunar. nauch.-prakt. Konferentsii, Saint Petersburg, pp. 788-796.

[10] Kolovsky M.Z., Evgrafov A.N., Semenov Yu.A., Slousch A.V. (2000) Advanced Theory of Mechanisms and Machines, Berlin, New York, London, Paris, Tokyo: Springer, 394 р.

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