position analysis of mechanisms

7/23/2019 Position Analysis of Mechanisms

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3.4 Vector Loops of a Mechanism

The main difference between freely moving bodies and the moving links ina mechanism is that they have a constrained motion due to the joints inbetween the links. The links connected by joints form closed polygons thatwe shall call a loop. The motion analysis of mechanisms is based onexpressing these loops mathematically.

In kinematic analysis we shall assume that all the necessary dimensions of each link is given and link length dimensions (i.e. the distance between the

joints or the angles) can be determined from the given dimensions usingthe geometry of the link.

In Section 2.1.2, we have seen that it is sufficient to represent the positionof each link (rigid body) by describing the position of any two points on thatlink. One way of selecting these two points on a link is to use thepermanently coincident points. It is obvious that in such a procedure, theorigin of a vector will be defined by the previous vector and thus the

number of parameters to define the link positions will be decreased.

Let us consider a four-bar mechanism as shown above as a simpleexample. In this mechanism A0 ,is a permanently coincident point between

links 1 and 2, A is peranently coincident point between links 2 and 3, Bbetween 3 and 4 and B0 between 1 and 4. Let us disconnect joint B. In

such a case we will obtain two open kinematic chains A0 AB (links 2,3) with

two degrees of freedom and A0 B0B (links 1,4) with one degree of freedom

(Fig.2.7b). To determine the positions of the links we must have a

reference frame. One obvious choice is to select the fixed pivots A0, B0 asone of the co-ordinate axes and select A0 or B0 as the origin. Next,in order

to define the position of link 2, we must define angle 12 , which is related

with the degree of freedom of the joint between links 1 and 2. To determinethe position of link 3, since the location of the permanently coincident point

A between 2 and 3 can be determined when 12 defined, we must now

define 13 , which is related to the freedom of the joint between links 2 and

3. Similarly 14 must be defined to determine the position of link 4. Hence

we need 3 parameters ( 12 , 13 and 14 ) which are all related to the




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joint freedoms for the open kinematic chains obtained when we disconnecta joint to eliminate a loop.

Each link can be defined by a vector fixed on that link, let us select thepermanently coincident points between the links as the tips of these vectorsand define vector A0A (for link 2), AB (for link 3), B0B (for link 4) and A0B0

(for link 1). Except A0B0 , the other three vectors will be a function of time

(since the distances between the two points on the same link are fixed, the

magnitudes will remain constant but the directions of these vectors willchange in time). Since the mechanism contains revolute joints only, Themagnitude of the vectors are constant link lengths(A0 A=a2, AB=a3,

A0B0=a1 ve B0B=a4). The angular orientation of these vectors will be

rotation variables ( 12 , 13 ve 14 ). When the joint at B is disconnected,

B3 and B4 may not be coincident. For the open kinematic chain, the

position of point B may be defined in two different forms as:

A0A +AB=A0B3 (1,2,3 open loop)

A0B0 +B0B= A0B4 (1,4 open loop)

However, at every instant the revolute joint between links 3 and 4 mustexist and point B must remain a permanently coincident point for differentvalues of the position variables if the system we are considering is amechanism. Therefore the vector A0B3 and A0B4 obtained from the two

equations using the two open kinematic chains must be equal and thisresults with the vector equation::

A0A +AB=A0B0 +B0B

This vector equation must be valid for all positions due to the permanetlycoincident points. If this vector equation can nott be satisfied for a giveninput angle, then that position cannot exist (mechanism cannot beassembled at that position).

In a four-bar mechanism there is a single loop formed and the vector equation describes the closure of this loop mathematically.The equation(s)that describes the closure of the loop(s) formed in the mechanism areknown as loop closure equation(s). The variables in the loop closure

equations are always related by the joint freedoms and we can solve for two position variables from any loop equation. In plane the vector equationwill correspond to two scalar equations. In the four-bar example there arethree variables ( 12, 13 and 14) which we shall call "position

variables". If one of the position variable ( say 12 ), the other position

variables ( 13 and 14) can be solved from this vector loop equation. The

number of independent parameters that are required will always be equalto the degree-of-freedom of the mechanism. The relation between theposition variables is a nonlinear, trigonometric relation.





One simple and concise form of writing the vector loop equations is to usecomplex numbers. for example, If the length of the vector A0A is a2 and if

the vector makes an angle 12:

A0A = a2cos 12+ ia2sin 12

or, using Euler's equation:

A0A = a2

In a similar fashion if the link lengths are denoted as ai olarak ( a1= A0B0 ,

a2= A0 A, etc.) the vector loop equation in complex numbers can be written

as:

a2 +a3 =a1+a4

If required, the equation can be written in cartesian coordinates as:

x and y components can be equated separetely two yield two scalar equations in the form::

a2cos 12 + a3cosq 13 = a1 + a4cos 14

a2sin 12 + a3sinq 13 = a4sin 14

In case of a four-bar, the vectors in the loop closure equation have fixedmagnitudes. However, the angular inclinations of the three vectorsrepresenting the moving links will change. Hence, there are three positionvariables ( 12, 13 and 14). If one of these variables is defined, the

remaining two variables can be solved from the vector equation. If we refer to the definition of the degree-of-freedom of a mechanism, the variable thatmust be defined is the input variable and for a constrained motion thenumber of input variables must be equal to the degree-of-freedom of the

joints involved. In case of a four-bar, since all the connections are revolute

joints, the variables are all rotation variables. In case of a prismatic joint,the variable will be the magnitude of a vector or a vector component.Consider a slider-crank mechanism as shown in Fig. A.. Let us disconnectthe revolute joint at B (Fig.B). In order to determine the positions of links 2and 3 we must define 12and 13. To locate the position of link 4 its

displacement along the slider axis must be known and the position variables14 must be defined. The resulting loop closure equation is:





AoA + AB = AoB

Again there are 3 variables ( 12, 13 and s14) one of which must be

specified as the input. In this case the vectors AoA and AB have fixed

magnitudes and varying directions. The vector AoB has a fixed y

component (length c) and a changing x component (s14). Depending on the

applications either 12 (i.e. in pumps) or s14 (i.e. internal combustionengines) is the input.In complex numbers the vector loop equation will be:

a2 +a3 = s14 + ic

The vectors defined and the variables used in the loop closure equationsare not unique. For example, for the slider crank mechanism, rather thandisconnecting the revolute joint at B, one can as well disconnect therevolute joint at A between links 2 and 3 (Fig C). We must now define theangle 13'= xBA instead of the angle 13 o determine the position of link

3. Note that the angles. 13 and 13' differ by a constant angle (In this

case by 180o). The resulting loop equation is:

AoA = AoB + AB

or in complex numbers:

a2 = s14 + ic +a3'





Example : The loop closure equations of a six link mechanism

Referring to the four-bar mechanism above, one can write a vector equation in the form::

AoA +AB = AoB

Considering the vector AoB the magnitude and direction of this vector arevariables (or its x and y components) and these variables are not relatedwith the joint freedoms.we can solve for the vector AoB provided that the

magnitudes and the directions of the other two vectors are known. Such anequation will not help us for the solution of position variables. Although it isa valid vector equation, it is not a loop closure equation. One can identifysuch loops by noting the variables involved are not related with the jointfreedoms and these equations are not obtained by disconnecting the joints.

A similar argument can also be made for the vector equations:

AoA + AC = AoC ve AB + BC = AC

In the later case, all three vectors are on the same link and they have afixed orientation with respect to each other. If the origin and the angular orientation of one of the vectors is known, due to rigidity the orientation of the other vectors will be known.

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In certain other cases we may have to use instantaneously coincidentpoints or other points on the links as the tips of the vectors, as shown inFig. 2.14. The variable involved in such a case is the relative displacementof one link with respect to another link. The vector BoA can be separated

into two components: BoC and CA so that one component is of constant

magnitude and the magnitude of the other vector is related to thedisplacement of the prismatic joint between links 3 and 4. The loop closure

equation can than be written in the form:

AoA = AoBo + BoC + CA

or in complex numbers:

Points A2 ve A3 are permanently coincident points. Point A4 is

instantaneously coincident with point A3 when the mechanism is moved

from this position the two points will be displaced by a distance s along

the slider axis relative to each other. Points A2 and A3 will be two other

different points that will be coincident with (A') (Fig. C).

The vector CA changes both its magnitude and direction. However itsorientation with respect to the vector B0C will be fixed, and no new variable





is needed. If the orientation of the vector B0C is whown by the variable 14

measured from the positive x-axis of our reference, the orientation of thevector CA with respect to positive x axis is 14+ 4 and angle 4 is a

constant angle measured on link 4 between two lines BoC and CP (point P

is any point on the slider axis on link 4) The position variables in the loopequation will be 12, 14 and s43 .

Note that the same vector loop equation can be derived for the swingingblock mechanism shown above. Therefore, the inverted slider crankmechanism is the same as the swinging block mechanism although their construction is different. Although it is a different construction, If the linkdimensions (a2, a4, 4) are the same, the motion of the two mechanisms

will be the same.

When writing the vector loop equations one must be sure that theequations are valid for every position of the mechanism. The mechanismmay be at a special position such that one or more of the links are collinear as shown above. (Although links 2 and 1 are collinear, they will havedifferent orientations at some other instant). In such a case you may redrawthe mechanism slightly offset from the critical position or show the variableangle 12 , as shown. In some cases if the constant link angles between

two vectors are of a certain simple value (such as 900), then it is advisableto simplifiy the equations accordingly. For example if the angle 4 of the

inverted slider crank mechanism is a right angle, The loop closure equationmust be written as:





The solution to the loop equations may not exist for every value of theindependent parameter. This will mean that for that particular link lengthsthe mechanism cannot be assembled at the requested position.

-How will you determine the number of independent loops from the

number of joints and links?-

In planar mechanisms we can write vector loop equations for each loop of the mechanism. This corresponds to L (L= number of independent loops)vector equations or 2L scalar equations, if we equate the x and ycomponents of vectors. The number of parameters involved in theseequations will be 2L+ F, where F is the degree of freedom of themechanism. If we now define F number of variables (independent variablesor input parameters (variables)), then theoretically, we must be able tosolve for the other variables (dependent position parameteres). We canchange the input variable within a given range in certain increments andobtain the values for the dependent variables. For example if the inputvariable corresponds to the angle that defines the angular position of an

input crank, we change this angle from 0 to 360o. If the input is themovement of a piston inside a cylinder, then we change this length of thepiston starting from the closed position to the most extended position (thedifference is the stroke of the piston).

When we determine the values of all the position variables correspondingto a certain input variable, then we can determine the position of any pointon any link of the mechanism.

One need not draw the mechanism with dismantled joints to writethe loop equations. After some practice one can conceptuallydisconect the joints, identify the loops and the write the necessaryloop equations. Initially of course, as a visual aid, we have shown the

joints are shown disconnected.

In the following examples the necessary loop equations for somemechanisms are written. The loop equations and the variables defined arenot unique. You are to determine the joint that is disconnected when writing

the given loop equation.

In recent years different package programs are available for the analysis of mechanisms. When using these programs, you must input these loops bytelling the program which link is connected to which link by what kind of a

joint (i.e. what kind of freedom is permitted by that joint). If you are usingmathematical packages such as matlab or mathcad, you must type theseequations in one form or another. A slight mistake in the loop equationsresults with erroneous results. Please keep in mind that these equationsdefine the mathematical model of an existing mechanism. Thismathematical model can be solved in different ways as we shall see in the

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coming sections.

Example I.

(A0 A=a2, AB= a3, BC=a4)

A0A + AB + BC = A0C

A0A + AD = A0D0 + D0D

Example II.





A0A = A0B0 + B0B + BA

A0C = A0B0 + B0B + BC

Example III.




B0B + BC + CD = B0A0 + A0A + AD

B0B + BC = B0C0 + C0C

Due to the gear pair: r 3 13= -r 2( 12- 2) (when 13=0; 12= 2)

©es



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position analysis of mechanisms

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