Structural analysis3 of the crank-slider mechanism. Slide-crank mechanism Kinematic scheme synthesis

Given (Figure 2.10): j 1, w 1 \u003d const, l BD, l DC, l AB, l BC, m l [m / mm ] .

Speed V B\u003d w 1 l A Bpoint B is directed perpendicular to link AB in the direction of its rotation.

To determine the speed of point C, we compose a vector equality:

C \u003d B + SV

The direction of the absolute speed of point C is known - parallel to the line xx. The speed of point B is known, and the relative speed V C B is directed perpendicular to the link BC.

We build a plan of speeds (Fig. 2.11) in accordance with the above equation. Moreover, m n \u003d V B / Рв[m / s mm ].

Point B's absolute acceleration is equal to normal acceleration a p VA(since w 1 \u003d const, e 1 \u003d 0 and a t B \u003d 0) a B \u003d a p VA \u003d w 2× l VA[m / s 2]

and is directed along the AB link from point B to point A.

Scale factor of the acceleration plan m a \u003d a B /p in[m / s mm], where p in - a segment of arbitrary length, representing acceleration on the plan a B.

Point C acceleration:

(1 way),

where a p CB \u003d V 2 CB / l CB[m / s 2]

A segment representing this acceleration in the acceleration plan:

n sv \u003d a n sv /m a[mm]

We choose the pole p of the acceleration plan. Draw a line from the pole along which the acceleration is directed a B (// AB) and postpone the selected segment p indepicting this acceleration on the plan (Fig. 2.12). From the end of the resulting vector, draw a line of direction of the normal component a n SVparallel to the SV link and postpone the segment n svdepicting on scale m athis is normal acceleration. From the end of the normal acceleration vector, draw a line of direction of the tangential component a t CB, and from the pole p - direction of absolute acceleration of point C ( ïï xx). At the intersection of these two directions, we get point C; in this case, the vector pС represents the required acceleration.

The module of this acceleration is:

a C \u003d (p from)m a[m / s 2]

The angular acceleration e 2 is determined as:

e 2 = a t CB / l CB= (t CB)m a / l SV[1 / s 2]

Direction e 2 shown in the diagram of the mechanism.

To find the speed of point D, you must use similarity theorem,which is used to determine the speeds and accelerations of the points of one link, when the speeds (accelerations) of the other two points of this link are known: the relative speeds (accelerations) of the points of one link form on the plans of speeds (accelerations) figures similar to the figure of the same name on the mechanism diagram. These figures are similarly located, i.e. when reading letter designations in one direction on the mechanism diagram, letters on the plan of speeds (accelerations) follow in the same direction.

To find the speed of point D, you need to build a triangle, similar to the triangle on the mechanism diagram.

Triangles D cvd (on the plan of speeds) and DCBD (on the plan of the mechanism) are triangles with mutually perpendicular sides. Therefore, to construct a triangle D cvd draw perpendiculars to CD and BD from points with and in respectively. At their intersection, we get point d, which we connect to the pole.

The acceleration of point D is also determined by the similarity theorem, since the accelerations of the other two points of link 2 are known, namely a In and a C. It is required to construct a triangle D on the acceleration plan incd, similar to triangle DBCD in the mechanism diagram.

To do this, we first build it on the mechanism diagram, and then transfer it to the acceleration plan.

Line segment " sun»The plan of accelerations is transferred to the segment of the same name CB on the mechanism diagram, putting it on the CB link from any point (C or B) (Fig. 2.10). Then along the segment “ sun»Triangle D is constructed on the mechanism indс, similar to the triangle DBDC, for which a straight line "dс" is drawn from the point "C", parallel to the straight line DC, until it intersects with the straight line BD. We get D indc ~ DBDC.

The obtained sides of the triangle r 1 and r 2 are equal in size to the sides of the desired

Figure 2.10

Figure 2.11

Figure 2.12

a triangle on the acceleration plan, which can be built using serifs (Figure 2.12). Next, you need to check the similarity of the location of the figures. So, when reading the letter designations of the vertices of the DBDC triangle on the clockwise mechanism diagram, we get the order letters B-D-C; on the plane of accelerations in the same direction, i.e. clockwise, we should get the same letter order in-d-c. Consequently, the solution is satisfied by the left intersection point of the circles r 1 and r 2.

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1. Structural Analysis mechanism

The crank-slider mechanism is presented.

The number of degrees of the mechanism under study is determined by the Chebyshev formula:

(1)

where n - the number of moving links in the studied kinematic chain; p 4 and p 5 - respectively, the number of pairs of the fourth and fifth class.

To determine the value of the coefficient n analyze block diagram mechanism (Figure 1):

Figure 1 - Block diagram of the mechanism

The structural diagram of the mechanism consists of four links:

1 - crank,

2 - connecting rod AB,

3 - slider B,

0 - rack,

wherein links 1 - 3 are movable links, and rack 0 is a fixed link. It is represented as part of a structural diagram with two hinged fixed supports and a slide guide 3.

Hence, n \u003d 3.

To determine the values \u200b\u200bof the coefficients p 4 and p 5 we will find all the kinematic pairs that are part of the considered kinematic chain. We enter the results of the study in table 1.

Table 1 - Kinematic pairs

№	Kinematic pair (KP)	Kinema diagram tic pair	Kinema class tic pair	The degree of mobility
1	0 – 1		rotational
2	1 – 2		rotational	1
3	2 – 3		rotational	1
4	3 – 0	rotational		1

From the analysis of the data in Table 1, it follows that the investigated iCE mechanism with an increased piston stroke consists of seven pairs of the fifth class and forms a closed kinematic chain. Hence, p 5 \u003d 4, a p 4 \u003d 0.

Substituting the found values \u200b\u200bof the coefficients n, p 5 and p 4 into expression (1), we get:

To identify the structural composition of the mechanism, we divide the scheme under consideration into Assur structural groups.

The first group of links 0-3-2 (Figure 2).

Figure 2 - Structural group of Assur

This group consists of two moving links:

connecting rod 2 and slider 3;

two leashes:

and three kinematic pairs:

1-2 - rotational pair of the fifth grade;

2-3 - rotational pair of the fifth class;

3-0 - forward pair of the fifth grade;

then n \u003d 2; p 5 \u003d 3, a p 4 \u003d 0.

Substituting the revealed values \u200b\u200bof the coefficients into expression (1),

Consequently, the group of links 4-5 is an Assur structural group 2 of class 2 of order 2 of the type.

The second group of links 0-1 (Figure 3).

Figure 3 - Primary mechanism

This group of links consists of a movable link - crank 1, rack 0 and one kinematic pair:

0 - 1 - rotational pair of the fifth class;

then n \u003d 1; p 5 \u003d 1, a p 4 \u003d 0.

Substituting the found values \u200b\u200binto expression (1), we get:

Consequently, the group of links 1 - 2 is indeed the primary mechanism with mobility 1.

Structural formula of the mechanism

MECHANISM \u003d PM (W \u003d 1) + SGA (class 2, order 2, type 2)

2. Synthesis kinematic diagram

For the synthesis of the kinematic scheme, it is first necessary to set the scale factor of the lengths μ ℓ. To find μ ℓ, it is necessary to take the natural size of the crank OС and divide it by the size of a segment of arbitrary length │OС│:

After that, using the scale factor of the lengths, we translate all natural dimensions of the links into segments, with the help of which we will build a kinematic diagram:

After calculating the dimensions, we proceed to constructing one position of the mechanism (Figure 4) using the serif method.

To do this, first draw a rack 0 on which the crank is fixed. Then we draw a horizontal line XX through the center of the circle that was drawn to build the rack. It is necessary for the subsequent finding of the center of the slider 3. Further, from the center of the same circle, draw two other radii

and. Then from there we draw a segment of length at an angle to the horizontal line XX. The intersection points of this segment with the constructed circles will be points A and C, respectively. Then from point A we build a circle with a radius.

The point of intersection of this circle with line XX will be point B. Draw a guide for the slider, which will coincide with line XX. We build the slider and all the other details of the drawing are needed. We designate all points. The synthesis of the kinematic scheme has been completed.

3. Kinematic Analysis flat mechanism

We proceed to building a speed plan for the position of the mechanism. To simplify the calculations, you should calculate the speeds and directions for all points of the mechanism position, and then build a speed plan.

Figure 4 - One of the positions of the mechanism

Let us analyze the scheme of the crank-slider mechanism: point O and O 1 are fixed points, therefore, the moduli of the speeds of these points are equal to zero (

The velocity vector of point A is the geometric sum of the velocity vector of point O and the speed of the relative rotational motion of point A around point O:

. (2)

Velocity vector action line

is perpendicular to the axis of the crank 1, and the direction of action of this vector coincides with the direction of rotation of the crank.

Speed \u200b\u200bmodule point A:

, (3) - angular velocity link OA; - length of OS.

Angular velocity

slider mechanism

2.1. Block diagram of the mechanism

Fig 2.1 Block diagram of the crank-slider mechanism

2.2. Identifying complex and spaced kinematic pairs

There are no spaced-apart kinematic pairs in the crank-slider mechanism. Couple INcomplex, so we will consider it as two kinematic pairs.

2.3. Classification of kinematic pairs of the mechanism

Table 2.1

P / p No.	The numbers of the links forming a pair	Symbol	Name	Mobility	Higher / Inferior	Closure (Geometric / Power)	Open / Closed
			Rotational
			Rotational
			Rotational
			Rotational
			Rotational
			Rotational
			Translational

The investigated mechanism consists only of one-movable kinematic pairs ( r 1 = 7, r\u003d 7), where r 1 - the number of one-moving kinematic pairs in the mechanism, r- the total number of kinematic pairs in the mechanism.

2. 4. Classification of links of the mechanism

Table 2.2

P / p No.	Link numbers	Symbol	Name	Traffic	Number of vertices
				Is absent
			Crank	Rotational

				Rotational

				Translational

The mechanism has: four () two-vertex () linear links 1,2,4,5; one (n 3 \u003d 1) three vertex link, which is the base link; five () moving links.

Find the number of connections to the rack. The conveyor mechanism has three () attachments to the rack.

In the studied complex mechanism, one elementary mechanism can be distinguished

Figure: 2.4 Crank-slider mechanism.

There are no mechanisms with open kinematic chains in the investigated crank-slider mechanism.

The mechanism contains only simple stationary mechanisms.

There are no fixing links in the investigated mechanism. Link 3 is simultaneously included in two simple mechanisms - an articulated four-link mechanism and a crank-slider. Hence, for this link

We classify the mechanism. The investigated mechanism has a constant structure, is complex and of the same type. It consists of one elementary mechanism and two stationary simple ones, which contain only closed kinematic chains.

The mechanism exists in a three-movable space.

Formulas for determining the mobility of these mechanisms will take the following form:

Let's define the mobility of the hinged four-link link. This mechanism has: three () movable links 1,2,3; four () one-way kinematic pairs O, A, B, C.

Let's find the mobility of the crank-sliding mechanism. It has: () movable links 3,4,5 and four () kinematic pairs C, B, D, K. Its mobility is determined in the same way:

We determine the mobility of a complex mechanism by the formula:

We analyze the structural model of the machine tool. We check whether the investigated mechanism corresponds to the structure of the mathematical model. The mechanism has: seven () single-moving kinematic pairs; five () movable two-vertex () links, the base is; three attachments to the rack () and no anchor links ().

Mathematical model:

;

Since the equations of the model have turned into identities, the investigated device has the correct structure and is a mechanism.

Let's select and carry out the classification of structural groups. The elementary mechanism is conventionally classified as a class I mechanism.

The class of the structural group is determined by the number of kinematic pairs included in a closed loop formed by internal kinematic pairs. The group order is determined by the number of external kinematic pairs. The type of group is determined depending on the location of the rotational and translational kinematic pairs on it.

2-order

It can be seen that the identified structural groups are completely similar in the species and quantitative composition of links and kinematic pairs. Each of the structural groups has: two movable links (), and the links are two-vertex () and, therefore, the base link also has two vertices (); three () one-way kinematic pairs, of which two are external ().

We check whether the selected structural groups correspond to mathematical models. Since the groups are similar, we check only one group, for example, OAB. Mathematical models of structural groups are as follows:

The crank-slider mechanism belongs to the II class.

3. Kinematic analysis of the mechanism

The kinematic analysis of any mechanism consists in determining: extreme (dead) positions of the machine, including the determination of the trajectories of individual points; the speeds and accelerations of the characteristic points of the links according to the well-known law of motion of the initial link (generalized coordinate).

3.1 Determination of extreme (dead) positions of the mechanism

The extreme (dead) positions of the mechanism can be determined analytically or graphically. Since analytics gives higher accuracy, it is preferred when determining extreme positions.

For a crank-slider and an articulated crank-rocker four-link, the extreme positions will be when the crank and the connecting rod are stretched (), then folded () in one line.

Figure: 3.1 Determination of extreme positions of the mechanism.

3.2 Determination of the positions of the links of the mechanism graphically.

Figure: 3.3 Construction of closed vector contours.

We place the structural diagram of the mechanism in a rectangular coordinate system, the origin of which is placed at point O. We connect the vectors with the links of the mechanism so that their sequence is two closed contours: OABCO and CBDC.

For the OABCO circuit: (3.1)

We represent the equation in projections on the coordinate axes.

Perm State Technical University

DEPARTMENT "Mechanics of Composite Materials and Structures".

COURSE PROJECT

ON THEORYMECHANISMS AND MACHINES

Theme:

The task:

Option:

Completed:group student

Checked:professor

Poezzhaeva E.V.

Perm 2005

Structural analysis of the mechanism ……………………………………………… 3

Kinematic analysis of the mechanism ………………………………………… ..4

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perpendicular OA. Draw a straight line through m. A perpendicular to AB. The point of intersection of the x-axis (selected in the directions of m. C) with this straight line will give m. C, connecting m. C with the pole, we obtain the velocity vector m. C. Determine the magnitude of the speed t. In:

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