Theory of machines Velocity and Acceleration Polygons PDF

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Velocity and Acceleration Polygons summaries from the subcourse theory of machines. Which is in the second term of modelling....

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Snelheid en Versnelling-veelhoek / Velocity and Acceleration Polygons Prosedure om ‘n Versnellings-veelhoek te skets: Dit is baie belangrik om die reëls van vektoroptelling en vektor aftrekking te onthou. Vir vektoroptelling, moet vektore A en B so geplaas word dat die stert van een vektor aan die pylpunt van die ander een raak. Vir vektor aftrekking, moet vektore A en B so gerangskik word dat óf die punte van beide in aanraking wees op 'n punt, of die sterte van die twee. Procedure to sketch an Acceleration polygon: It is very important to remember the rules of vector addition and vector subtraction. For vector addition, vectors A and B must be so placed that tail of one vector touches tip of arrowhead of the other. For vector subtraction however, vectors A and B must be so arranged that either the tips of arrowhead of both touch at a point or the tails of the two vectors touch at a common point. The procedure to draw acceleration polygon can be summarised in the following steps: 1. Draw mechanism to some scale in the given configuration. 2. Write down velocity equation to connect known velocity of one point 𝐴 with unknown velocity of second point 𝐵 on the same link through their relative velocity. Keep unknown velocity vector 𝑉𝐵 on the left hand side of the equal to sign. 3. Select suitable location for velocity pole 𝑂𝑣 . Choose suitable scale. From 𝑂𝑣 , draw a line in a direction parallel to the direction of the known absolute velocity. From the tip of this vector, draw a line parallel to the direction of (partially known) relative velocity 𝑉 𝐵𝐴 on R.H.S. Similarly from 𝑂𝑣 draw a line parallel to the line of action of (partially known) absolute velocity vector 𝑉𝐵 on the L.H.S. The two lines intersect at a point completing the polygon. 4. Measure the magnitudes of velocity vectors which were known partially. Fix their directions of arrow heads so as to satisfy velocity equation. 5. Write down acceleration equation to connect unknown acceleration 𝐴𝐵 of one point with known acceleration 𝐴𝐴 of the other point on the same link. Write down all acceleration quantities in terms of their radial and tangential components. 6. Using information of velocities from velocity polygon, identify acceleration components that are partially or completely known. Usually normal components of acceleration, which depend on velocity (𝑎𝐴𝑛 = 𝜔2 𝑅), are known. Also tangential components of acceleration (𝑎𝐴𝑡 = 𝛼𝑅) which depend on angular acceleration of a link are not known in magnitude, but are known in direction – note that the tangential component is always perpendicular to normal, which means it is perpendicular to the link.

7. Select location of acceleration pole 𝑂𝑎 suitably, and choose a convenient scale for the acceleration polygon. Starting from 𝑂𝑎 plot all ‘completely known’ absolute acceleration vectors. From the tip of these vectors, draw other ‘completely known’ vectors one after the other, in such a way as to satisfy acceleration equation. When all the known (known in magnitude and direction) vectors are exhausted on R.H.S., draw lines parallel to the direction of the remaining (partially known) acceleration components. The two lines meet at a point defining solution. Assign arrowheads on each acceleration vector so as to ensure vector addition and subtraction as per acceleration equation. 8. With this, construction of acceleration polygon is complete. Using the scale of acceleration diagram already chosen, unknown accelerations can be measured from the corresponding side of the polygon. Example Assume 𝝎 = 𝟏𝟖𝟎 𝒓𝒑𝒎, 𝜶 = 𝟏𝟎𝟎

𝒓𝒂𝒅 𝒔𝟐

, 𝑪𝑷 = 𝟑𝟎𝟎

Determine 𝑉𝑃 , 𝑉𝐶 , 𝑉𝑃𝐶 , all the acceleration components of 𝑃 and 𝐶 , as well as 𝜶𝑷𝑪

Solution: Angular speed w of crank 𝑂𝐶, is 𝜔 = 180 𝑟𝑝𝑚 = 18.85

𝑟𝑎𝑑 𝑠

Velocity of point C is: 𝑉𝐶 = 𝑂𝐶 ∗ 𝜔 = 25 ∗ 18.85 = 471.2 𝑐𝑚/𝑠 We also know 𝑉𝑃 would be to the right. Next write down the velocity equation to help you to draw the velocity polygon: 𝑽𝑷 = 𝑽𝑪 + 𝑽𝑷𝑪

where 𝑉𝐶 is known completely, its magnitude is 𝟒𝟕𝟏. 𝟐 𝒄𝒎/𝒔 and the direction is tangent to crank circle as shown. Then basically you just follow the first line: For vector subtraction, vectors A and B must be so arranged that either the tips of arrowhead of both touch at a point or the tails of the two vectors touch at a common point. ∴ 𝑽𝑷 − 𝑽𝑪 = 𝑽𝑷𝑪

𝑪

𝒂

𝑉𝐶 = 471.2 𝑉𝑃𝐶 ∅

𝑂𝑣

    

𝒃 𝑉𝑃

𝑷

𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝐶

The angle is easy to obtain here, since the direction of 𝑽𝑪 is given. By inspection, ∅ = 60°. Since the velocity of 𝑃 w.r. to 𝐶 is perpendicular to 𝑃𝐶 and, as per velocity equation, is to be added to vector 𝑉𝐶 , a line is drawn from point 𝒂, perpendicular to direction 𝑃𝐶. Since magnitudes of 𝑽𝑷 and 𝑽𝑷𝑪, are not known, intersection of lines parallel to direction of 𝑽𝑷𝑪 and 𝑽𝑷 define point 𝒂. Note that 𝑽𝑷 can exist only along line of stroke. Since line 𝒂𝒃 was drawn parallel to 𝑽𝑷𝑪 and is additive to vector 𝑽𝑪 (see first velocity equation), the arrowhead on line 𝒂𝒃 is from 𝒂 to 𝒃 as shown. Similarly, as velocity 𝑽𝑷 is subtractive from 𝑽𝑪 (see again first velocity equation), the arrowhead on line 𝑶𝒗 𝒂 should indicate direction 𝑶𝒗 to 𝒂 as shown. Thus 𝑶𝒗 𝒂 gives 𝑽𝑷 in magnitude and direction, while 𝒂𝒃 gives velocity 𝑽𝑷𝑪 in magnitude and direction.

By measurement from velocity polygon, 𝑉𝑃 = 195 𝑐𝑚/𝑠 and 𝑉𝑃𝐶 = 405 𝑐𝑚/𝑠 Now the acceleration: 𝒂𝑷 = 𝒂𝑪 + 𝒂𝑷𝑪 𝑛 𝑡 + 𝑎𝑃𝐶 = 𝑎𝑡𝐶 + 𝑎𝐶𝑛 + 𝑎𝑃𝐶

Note that as slider reciprocates, it has only an acceleration component 𝒂𝑷 along the line of stroke. Also note that because of the indication of 𝛼, 𝒂𝑪𝒕 occurs in the direction opposite to 𝑽𝑪 . The three completely known acceleration components are: 𝑎𝑛𝐶 = 𝑂𝐶 ∗ 𝜔2 = 8883 𝑐𝑚/𝑠 2 𝑎𝐶𝑡 = 𝛼 ∗ 𝑂𝐶 = 2500 𝑐𝑚/𝑠2 𝑛 𝑎𝑃𝐶 =

2 𝑣 2 𝑉𝑃𝐶 = 1640.2 𝑐𝑚/𝑠 2 = 𝜌 𝑃𝐶

Now we need to draw the polygon: 𝒕 The other two accelerations 𝒂𝑷 and 𝒂𝑷𝑪 are known only in directions.

𝒂𝑷 acts along line of stroke 𝒕 𝒂𝑷𝑪 acts in a direction perpendicular to PC (this is always the case – the relative acceleration’s tangential component is perpendicular to the link).

Step 1 – Start off by making the origin – let us call it 𝒐𝒂 𝒐𝒂

Step 2 – From this origin, draw known direction and magnitude accelerations first (in this case, 𝒂𝑷 , which we assume occurs to the right). 𝒐𝒂

𝑎𝑃

Step 3 – Also from the origin, draw the normal component of the second acceleration, namely 𝒂𝒄𝒏 , which we know acts in the direction from 𝐶 to 𝑂 (always in line with the crank / link to the origin) 𝑎𝑃

𝒐𝒂 𝒂𝒏𝒄

Step 4 – Next, you always complete the vector of Step 3 by drawing its tangential component, namely 𝒂𝒕𝒄, which we know is always perpendicular to 𝒂𝒄𝒏 and in the direction of the rotation, which is CCW. 𝑎𝑃

𝒐𝒂

𝒂𝒏𝒄

𝒂𝒄𝒕

Step 5 – Now we know the direction of 𝑎𝐶 , since the equation of vector addition shows us that it needs to start from the origin. Also we know that the arrowhead needs to end up at 𝑎𝐶𝑡 . 𝒂𝑷 𝑷

𝒐𝒂

𝒂𝒏𝒄 𝒂𝑪 𝒂𝒄𝒕

𝑪 Step 6 – From the equation we can now complete the first part of the polygon by adding the relative acceleration arrow. Remember that because we are working with 𝑎𝑃𝐶 , the arrow is from 𝐶 to 𝑃. 𝒂𝑷 𝑷

𝒐𝒂

𝒂𝒏𝒄

𝒂𝑷𝑪

𝒂𝑪 𝒂𝒄𝒕

𝑪 Step 7 – Now you add the two components of the relative acceleration. Let’s have a look at the acceleration equation again: 𝒂𝑷 = 𝒂𝑪 + 𝒂𝑷𝑪 where 𝑛 𝑡 + 𝑎𝑃𝐶 𝒂𝑷𝑪 = 𝑎𝑃𝐶 𝒕 acts in a direction perpendicular to PC, so we know the line should lie here: 𝒂𝑷𝑪

𝒂𝑷 𝑷

𝒐𝒂

𝒂𝒏𝒄

𝒂𝑷𝑪

𝒂𝑪 𝒂𝒄𝒕

𝑪 However, what should the direction be?

𝒏 To choose the correct one, focus on the other relative component - 𝒂𝑷𝑪

From the equation: 𝑛 𝑡 + 𝑎𝑃𝐶 𝑎𝑃 = 𝑎𝐶𝑡 + 𝑎𝐶𝑛 + 𝑎𝑃𝐶 𝑛 𝒕 𝒏 Note that 𝒂𝑷𝑪 is away from 𝑎𝐶𝑡 (to the left), and perpendicular to 𝒂𝑷𝑪 : is additive to 𝒂𝑪𝒕 . Hence, 𝑎𝑃𝐶

𝒂𝑷 𝑷

𝒐𝒂

𝒂𝒏𝒄

𝒂𝑷𝑪

𝒂𝑪 𝒂𝒄𝒕 𝒂𝒏𝑷𝑪

𝑪

𝒕 Now we know the direction of 𝒂𝑷𝑪 , which is away from 𝒂𝒏𝑷𝑪

𝒂𝑷 𝑷

𝒐𝒂 𝒕 𝒂𝑷𝑪

𝒂𝒏𝒄

𝒂𝑷𝑪

𝒂𝑪 𝒂𝒄𝒕 𝒂𝒏𝑷𝑪

𝑪

By measurement: 𝑨𝑷 = 𝟓𝟒𝟒𝟎 𝒄𝒎/𝒔𝟐 𝑨𝒕𝑷𝑪 = 𝟔𝟓𝟔𝟎 𝒄𝒎/𝒔𝟐 𝜶𝑷𝑪 =



𝑨𝑷𝑪 𝒓𝒂𝒅 = 𝟔𝟓. 𝟔 𝟐 𝒄𝒄𝒘 𝑷𝑪 𝒔

For questions regarding gravitational accelerations, the acceleration of G is calculated as follow:

𝒂𝑷𝑪 represents the acceleration image of Link 𝑃𝐶. Hence, if acceleration of any point 𝐺 on 𝑃𝐶 is required, the link is to be divided at 𝐺 , such that:

𝑪𝑮 𝟏𝟎𝟎 𝟏 = = 𝑪𝑷 𝟑𝟎𝟎 𝟑 Then 𝐺 lies a third of the length of 𝑃𝐶 on the polygon.

𝒂𝑷 𝑷

𝒐𝒂 𝒕 𝒂𝑷𝑪

𝒂𝒏𝒄

𝑮 𝒂𝑷𝑪

𝒂𝑪

𝒂𝒄𝒕 𝒂𝒏𝑷𝑪

𝑪

This is the extent of what I can explain with regards to how you should draw the polygons. Furthermore, the use of the trig-identities and rules is what will allow you to solve the magnitudes and directions.:...