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

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

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I.N. Matsyuk, E. . Shlyakhov, N.V Zyma. Study of Planar Mechanisms Kinetostatics Using the Theory of Complex Numbers with MathCAD PTC. Mechanics, Materials Science & Engineering Journal, 2017, 8, 10.2412/mmse.40.52.685. [hal-01508541](https://hal.science/hal-01508541)

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

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

1. State Higher Educational Institution “National Mining University, Dnipro, Ukraine

a. shlyahove@nmu.org.ua

<figure><img src="image_2.png" /></figure>

DOI 10.2412/mmse.40.52.685 provided by [Seo4U.link](https://link.springer.com/10.2412/mmse.40.52.685)

**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].

The research has been provided using MathCAD PTC that is powerful program for operations with complex numbers. All calculations are performed in the international standard system (NIST).

Starting with the group, which consists of connected rod 4 and slider 5 (Fig. 1). Assume, that link 4 is connected to link 2 by a revolute joint *E*, and slider 5 is connected with a frame 0 by a prismatic joint. The mass centre *s*4 is in the middle of the link *EF*.
The position of the slider $s_5$ mass center coincides with point F.

<figure><img src="image_3.png" /><figcaption>Fig. 1. Second class Assur group (connecting rod – slider): a – group skeleton; b – calculation scheme.</figcaption></figure>

The initial data for the calculation are as follows:

* geometrical parameters – $l_4 = l_{EF} = 0,15$ m, $\alpha = 45^{\circ}$;

* links masses – $m_4 = m_{EF} = 6$ kg; $m_5 = m_{sl} = 20$ kg;

* links inertia moments – $I_{s4} = I_{EF} = 11,2 \cdot 10^{-3} \text{kgm}^2$;

* resistance force – $P = 500$ N.

As it is known, from previous kinematical analysis:

* connecting rod EF, is interpreted by the vector $l_4 = 0.109 + 0.103i$;

* vectors of acceleration of links mass centres $a_{s4} = -59.32 - 39.14i$ and $a_{s5} = -52.69 - 52.69i$;

* connecting rod angular acceleration – $\varepsilon_4 = -192 \text{ s}^{-2}$.

The acting loads on the links:

* gravity forces (hereinafter use approximate value of the gravitational acceleration $g = 10 \text{ m/s}^2$)
  $G_4 = -60i$ and $G_5 = -400i$;

* resistance force – $P = -353.6 - 353.6i$;

* inertial loads – $F_{i4} = 355.9 + 234.8i$, $F_{i5} = 2105 + 2105i$, $M_{i4} = 2.15$.

Note, that the vectors $\varepsilon_4$ and $M_{i4}$ can not be expressed in complex numbers because they are directed perpendicular to the complex plane. Operations with them will be described below. Following a fragment from MathCAD 11, which shows the input of initial data, is provided (fig. 2).
$$ \alpha := \frac{\pi}{4} \qquad l_4 := 0.109 + 0.103 \cdot i $$

$$ a_{s4} := -7.84 - 5.18 \cdot i \qquad a_{s5} := -6.959 - 6.959 \cdot i $$

$$ m_4 := 6 \qquad m_5 := 10 \qquad G_4 := -60 \cdot i \qquad G_5 := -100 \cdot i \qquad I_{s4} := 11.2 \cdot 10^{-3} $$

$$ F_{i4} := -m_4 \cdot a_{s4} \qquad F_{i4} = 47.04 + 31.08i $$

$$ F_{i5} := -m_5 \cdot a_{s5} \qquad F_{i5} = 69.59 + 69.59i $$

$$ \varepsilon_4 := -25.4 \qquad M_{i4} := -I_{s4} \cdot \varepsilon_4 \qquad M_{i4} = 0.284 $$

$$ P := -565.7 - 565.7 \cdot i $$

<caption>Fig. 2. Input of initial data.</caption>

Let's proceed directly to the forces analysis of group, in which necessary to determine the forces in external kinematic pairs of groups $R_{24}$ (action, for instance, of link 2 on link 4) and $R_{05}$ (action of frame 0 on slider 5), and the force $R_{54}$ in 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:

$$ \text{link } 4 - \mathbf{R}_{24} + \mathbf{G}_{4} + \mathbf{F}_{i4} + \mathbf{R}_{54} = 0 \qquad (1) $$

$$ \text{link } 5 - \mathbf{R}_{54} + \mathbf{G}_{5} + \mathbf{F}_{i5} + \mathbf{P} + \mathbf{R}_{05} = 0. \qquad (2) $$

The equilibrium equation of moments acting on the connection rod 4 around its mass centre:

$$ -0.5l_4 \times \mathbf{R}_{24} + \mathbf{M}_{i4} + 0.5l_4 \times \mathbf{R}_{05} = 0. \qquad (3) $$

In formulas (1) and (2) two vectors are unknown $R_{24}$ and $R_{05}$. The last vector's magnitude is 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 $R_{05}$ to line, which tilted to X axis on angle $\alpha$ will be equal to:

$$ \mathbf{R}_{05} \cdot \overline{e^{\alpha i}} = -\overline{\mathbf{R}_{05}} \cdot e^{\alpha i}, \qquad (4) $$

where the top underline denote the conjugate vectors.
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:

$$M(h, F) := Im(\bar{h} \cdot F). \qquad (5)$$

The resulting system of equations in MathCAD solved by a block Given-Find.

The corresponding fragment is given below (fig. 3).

$$M(h, F) := Im(\bar{h} \cdot F)$$

$$R_{24} := 100 + 100 \cdot i$$

$$R_{54} := 100 + 100 \cdot i$$

$$R_{05} := 100 + 100 \cdot i$$

Given

$$R_{24} + G_{4} + F_{i4} + R_{54} = 0$$

$$-R_{54} + G_{5} + F_{i5} + P + R_{05} = 0$$

$$R_{05} \cdot \exp(\alpha \cdot i) = -R_{05} \cdot \exp(\alpha \cdot i)$$

$$M(-0.5l_{4}, R_{24}) + M_{i4} + M(0.5l_{4}, R_{54}) = 0$$

$$(R_{24} \quad R_{54} \quad R_{05}) := \text{Find}(R_{24} \quad R_{54} \quad R_{05})$$

$$R_{24} = 532 + 542i$$

$$R_{05} = -83 + 83i$$

$$R_{54} = -579 - 513i$$

<caption>Fig. 3. The resulting system of equations.</caption>

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 – $l_2 = l_{BC} = 0,15$ m, $l_{21} = l_{BE} = 0,09$ m, $l_{BE} = l_{CE}$;

* links masses – $m_2 = 4$ kg; the rocker mass is small and it has not been taken into account;

* link's moment of inertia – $I_{s2} = 7,5 \cdot 10^{-3}$ kgm².
<figure><img src="image_4.png" /><figcaption>Fig. 4. Second class Assur group (connecting rod – rocker): a – group skeleton; b – calculation scheme</figcaption></figure>

In addition, we have previously obtained kinematics analysis:

* the sides of triangular connecting rod is interpreted by vectors: $l_2 = 0.145 + 0.037i$, $l_{21} = 0.061 + 0.067i$ and link 3 – by vector $l_3 = 0.021 + 0.072i$;

* the acceleration vector of the mass centre for the link 2: $\boldsymbol{a}_{s2} = -7.415 - 2.871i$;

* rocker angular acceleration $2 - \varepsilon_2 = 42.25 \text{ s}^{-2}$.

The loads acting on the links in the group:

* gravity forces (hereinafter use approximate value of the gravity constant $g = 10 \text{ m/s}^2$) $G_2 = -40i$;

* a reaction of link 4 on link 1 (considered above) $R_{42} = -532 - 542 \cdot i$, (N);

* inertia loads – $F_{i2} = 29.66 + 11.48i$, $M_{i2} = -0.317$.

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

$$
\begin{align*}
l_2 &:= 0.145 + 0.037 \cdot i & l_{21} &:= 0.061 + 0.067 \cdot i & l_3 &:= 0.021 + 0.072 \cdot i \\
a_{s2} &:= -7.415 - 2.871 \cdot i & m_2 &:= 4 & G_2 &:= -40 \cdot i & I_{s2} &:= 7.5 \cdot 10^{-3} \\
F_{i2} &:= -m_2 \cdot a_{s2} & F_{i2} &:= 29.66 + 11.48i & \varepsilon_2 &:= 42.25 & & \\
M_{i2} &:= -I_{s2} \cdot \varepsilon_2 & M_{i2} &:= -0.317 & R_{42} &:= -532 - 542 \cdot i & &
\end{align*}
$$

<caption>Fig. 5. Initial data input.</caption>

In the force analysis necessary to determine the forces in the external joints of group $R_{12}$ (action of link 1 on link 2) and $R_{03}$ (action of frame 0 on rocker 3), and the force in the internal joint $R_{32}$ (action of rocker 3 on connecting rod 2).
The link 3 does not affected by the external loads, so a force $\mathbf{R}_{32}$ is equal to a force $\mathbf{R}_{03}$ and both 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 $\mathbf{R}_{32}$ to vector $\mathbf{l}_3$.

$$ \mathbf{R}_{32} \cdot \overline{\mathbf{l}_3} = \overline{\mathbf{R}_{32}} \cdot \mathbf{l}_3 $$

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 arm $\mathbf{l}_{BS_2}$ of forces $G_2$ and $F_{i2}$.

$$ \mathbf{l}_{BS_2} = \frac{2}{3} \cdot \frac{\mathbf{l}_2 + \mathbf{l}_{21}}{2} $$

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

$$ M(h,F) := \operatorname{Im}(\bar{h} \cdot F) \qquad l_{Bs2} := \frac{2}{3} \cdot \frac{l_2 + l_{21}}{2} \qquad l_{Bs2} = 0.069 + 0.035i $$

$$ R_{12} := 100 + 100 \cdot i \qquad R_{32} := 100 + 100 \cdot i $$

Given

$$ R_{12} + G_2 + F_{i2} + R_{42} + R_{32} = 0 \qquad R_{32} \cdot \overline{\mathbf{l}_3} = \overline{\mathbf{R}_{32}} \cdot \mathbf{l}_3 $$

$$ M[l_{Bs2}, (G_2 + F_{i2})] + M_{i2} + M(l_{21}, R_{42}) + M(l_2, R_{32}) = 0 $$

$$ (R_{12} \quad R_{32}) := \operatorname{Find}(R_{12} \quad R_{32}) $$

$$ R_{12} = 501 + 565i \qquad R_{32} = 1.57 + 5.37i $$

$$ R_{03} := R_{32} \qquad R_{03} = 1.57 + 5.37i $$

<caption>Fig. 6. MathCAD PTC fragment.</caption>

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 – $l_3 = l_{CD} = 0,49$ m, $l_{31} = l_{CB} = 0,329$ m;

* mass of links – $m_3 = 18$ kg; the slider mass is small and it does not take into account;

* link inertia moment – $I_{s3} = 0,43$ kgm².
<figure><img src="image_5.png" /><figcaption>Fig. 7. A second class Assur group ( slotted link-slider block): a – skeleton of group; b – diagram for a forces calculation</figcaption></figure>

In addition, we know from previous kinematic analysis:

* the slotted link CD and its part CB were interpreted by vectors: $\mathbf{l}_3 = 0.17 + 0.46i$ и $\mathbf{l}_{31} = 0.114 + 0.308i$;

* the acceleration vector of the mass centre for the link: $\mathbf{a}_{s3} = -2.25 + 0.056i$;

* angular acceleration of the slotted link – $\boldsymbol{\epsilon}_3 = 8.52 \ \text{s}^{-2}$;

The loads acting on the links in the group:

* gravity forces (hereinafter are using approximate value of the acceleration of free fall $g = 10 \ \text{m/s}^2$) $\mathbf{G}_3 = -180i$;

* a reaction of link 4 on link 3 (of considered above) $\mathbf{R}_{43} = 1640 - 2i, (\text{N})$;

* inertial loads – $\mathbf{F}_{i3} = 40.47 - 1,01i, \mathbf{M}_{i3} = -3.66$.

Following is a document fragment from MathCAD, which shows the input of initial data (fig. 8).
$$l_3 := 0.17 + 0.46 \cdot i \qquad l_{31} := 0.114 + 0.308 \cdot i$$

$$a_{s3} := -2.25 + 0.056 \cdot i$$

$$m_3 := 18 \qquad G_3 := -180 \cdot i \qquad I_{s3} := 0.43$$

$$F_{i3} := -m_3 \cdot a_{s3} \qquad F_{i3} = 40.5 - 1.01i$$

$$\varepsilon_3 := 8.52 \qquad M_{i3} := -I_{s3} \cdot \varepsilon_3 \qquad M_{i3} = -3.66$$

$$R_{43} := 1640 - 2 \cdot i$$

<caption>Fig. 8. Input of initial data.</caption>

It is necessary to determine the forces in the external joints B and C $R_{12}$ (action of link 1 on link 2) and $R_{03}$ (action of frame 0 on slotted link 3) and the reaction of slider 2 on slotted link 3 – $R_{32}$.

On the first stage of forces analysis, we do not include friction, so force $R_{23}$ is equal to the force $R_{12}$, 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 $R_{23}$ to the vector $l_3$.

$$R_{23} \cdot \bar{l}_3 = -\overline{R_{23}} \cdot l_3$$

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

$$M(h, F) := Im(\bar{h} \cdot F)$$

$$R_{23} := 100 + 100 \cdot i \qquad R_{03} := 100 + 100 \cdot i$$

Given

$$R_{23} + G_3 + F_{i3} + R_{43} + R_{03} = 0 \qquad R_{23} \cdot \bar{l}_3 = -\overline{R_{23}} \cdot l_3$$

$$M(l_3, R_{43}) + M_{i3} + M(l_{31}, R_{23}) + M[0.5l_3, (G_3 + F_{i3})] = 0$$

$$(R_{23} \quad R_{03}) := \text{Find}(R_{23} \quad R_{03})$$

$$R_{23} = 2075 - 767i \qquad R_{03} = -475.2 + 945.5i$$

$$R_{12} := R_{23} \qquad R_{12} = 2075 - 767i$$

<caption>Fig. 9. MathCAD document fragment.</caption>
In conclusion, consider the equilibrium of input crank AB (fig. 10). The mass center of the crank $s_1$ is located on its axis of rotation.

<figure><img src="image_6.png" /><figcaption>Fig. 10. The input crank: a – skeleton of group; b – diagram for a forces calculation.</figcaption></figure>

Take the following initial data:

* crank length $l_1 = 0,49$ m;

* mass of crank $m_1 = 3$ kg.

Previously we identified:

* the crank AB was interpreted by vectors: $l_1 = 0.114 + 0.038i$;

* angular acceleration of the crank $\varepsilon_1 = 0$.

The loads acting on the crank:

* gravity forces $G_1 = -30i$;

* a reaction value which acts from link 2 to the crank 1 – $R_{21} = -2072 + 764i$, (N);

* inertial loads – $F_{i1} = 0, M_{i1} = 0$.

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

$$l_1 := 0.114 + 0.038 \cdot i$$

$$m_1 := 3 \quad G_1 := -30 \cdot i$$

$$R_{21} := -2072 + 764 \cdot i$$

<caption>Fig. 11. MathCAD document fragment.</caption>
From the equilibrium conditions of the crank ($R_{21} + G_1 + R_{01} = 0$; $M_1 = l_1 \times R_{21}$) easily find the force $R_{01}$ and the torque required to drive the input link 1 $M_1$ (fig. 12).

$$ M(h,F) := \operatorname{Im}(\bar{h} \cdot F) \qquad R_{01} := -G_1 - R_{21} \qquad R_{01} = 2072 - 734i \\ M_1 := -M(l_1, R_{21}) \qquad M_1 = -165.8 $$

<caption>Fig. 12. MathCAD document fragment.</caption>

**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.**

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[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

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[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.

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