This time of the year is called spring. It put sparkles in the eyes of lovers and make the hearts beat faster. The days became longer and the beaches call.
But!
Please!
Remember that this time of the year is also the time when students must apply their new found knowledge to impress the lecturers in order to be promoted to the next level or, to earn a Certificate that will determine their futures.
You are one of those students. You and me both know that you have worked hard during the year in order to pass your tests. Completed your tasks in order to obtain good ISAT marks.
Please do not throw away all your hard work during the year by falling into the Spring-trap.
Now is the time for you to have self-discipline. To stay off the beaches and study.
A little bit of discipline and hard work now can prevent tears and a bad holiday later.
Please guys, make use of the Question papers on the log to study. Remember that, although I expect a lot from you, you are still doing this for yourself and YOUR OWN future.
Monday 14 October 2013
Wednesday 9 October 2013
Milling Level 4 Indexing
Fitting and Turning Level 4
Milling machine – Indexing calculations
INDEXING EQUIPMENT:
Indexing is done on the milling machine using a dividing head. The dividing head is a very important tool used to divide the circumference of a work piece into equally spaced divisions. This is important when milling gears, splines, hexagons and so on. The dividing headset consists of the headstock with footstock, chuck, center rest, index plates and change gears.
The chuck clamps the work piece when milling while the footstock supports the work piece on the opposite side of the dividing head when milling between centers.
The center rest supports long work pieces when milling between centers.
SIMPLE INDEXING:
Simple indexing is performed in the following manner.
Have the worm and worm gear engaged.
Calculate the indexing
Place the adjustable plunger into the required hole on the chosen index plate.
Adjust the sector arms to count the required number of holes.
Turn the index crank handle the number of full turns and part of a turn determined by the sector arms and put into the set hole
Move the sector arms and repeat the procedure until the required number of divisions or sides have been cut.
Two types of dividing heads have been developed namely Brown and Sharp as well as the Cincinnati.
Brown and Sharp: The dividing head id supplied with three index plates:-
Plate 1: 15, 16, 17, 18, 19 and 20 holes
Plate 2: 21, 23, 27, 29, 31 and 33 holes
Plate 3: 37, 39, 41, 43, 47 and 49 holes
The Cincinnati dividing head is supplied with only one indexing plate that is reversible:
Side 1: 24, 25, 28, 30, 34, 37, 38, 39, 41, 42 and 43 holes
Side 2: 46, 47, 49, 51, 53, 54, 57, 58, 59, 62 and 66 holes
It is also important to note that the crank handle must turn through 40 full revolutions before the work piece and worm wheel is turned through one revolution.
Out of this information we use the following formula for indexing.
Formula =40/N where N = the number of divisions or sides to be machined.
Let us do two examples. In one example the divisions is more than 40 (in which case there will be no full turns of the handle) and in the next the divisions is less than forty (in which case there will be some full turns of the handle.)
Example 1
We want to cut a gear with 44 teeth. In our formula N will be equal to 44 and we are using the Cincinnati dividing head.
Formula =40/N
=40/44 We can simplify this
=10/11 × 66/1 (on the Cincinnati plate 66 is the first number of holes dividable by 11)
= 60
Our full answer will now be: 0 full turns of the crank handle, 60 holes on a 66-hole circle plate.
Example 2
We want to cut a gear with 12 teeth on a Brown and Sharp dividing head.
Formula =40/N
=40/12 We can simplify this
=3 1/3 × 15/1 (on the Brown and Sharp plates 15 is the first number of holes dividable by 3)
=1/3 × 15/1 (we ignore the three and calculate only the fraction)
Type equation here.
= 5
Our full answer will now be: 3 full turns of the crank handle, 5 holes on a 15-hole circle plate.
ANGULAR INDEXING:
If we want to cut grooves or slots at an angle in a work piece. If we are working with angles it is important to remember that 40 full turns of the crank handle will0 rotate the work piece through only one full turn. (360°) To work out how many degrees the crank handle will turn the work piece in one full turn of the handle, we must divide 360° by 40. This will give us 9°.
The formula for indexing when working with degrees will therefore be N/(9°)
For example:- Using the Cincinnati head, calculate the indexing to cut a gear with 38° teeth.
Our Formula =N/(9°)
=38/9
=4 2/9
=2/9 × 54/1 (we ignore the 4 and 54 is the first number of holes dividable by 9)
= 12
Our full answer will now be: 4 full turns of the crank handle, 12 holes on a 54-hole circle plate.
DIFFERENTIAL INDEXING:
Sometimes it is impossible to calculate the indexing because the plates do not provide a number of holes that are divisible by the number of teeth or divisions. To overcome this the headstock can be attached to a number of interchangeable gears. The gears will in fact slightly change the number of turns the crank handle make for one full revolution of the work piece.
How do we do these calculations?
Step 1: Round of the divisions either up or down.
Step 2: make a note that, if you have rounded up, the plate must turn the same direction as the crank handle and if you have rounded down, the plate must turn in the opposite direction as the crank handle. (make use of an idler gear to change direction.)
Step 3: Use the following formulae: Indexing =40/n and Gear Ratio (n-N) × 40/n (Where N = Real number of divisions and n = rounded number of divisions.
The change gears supplied are the following: 2 of each 24, 28, 32, 40, 44, 48, 56, 64, 72, 86, 100
EXAMPLE:
Using the Cincinnati dividing head calculate the indexing for 99 divisions.
Round the divisions to 100
Indexing =40/n
=40/100
=4/10 × 20/1 (20 is the first divisible number of holes on Cincinnati plate)
= 8
Gear Ratio =(n-N) ×40/n
=(100-99) ×40/100
=1×40/100
=40/100
=4/10 ×10/10
=40/100
Our full answer will be: Indexing = 0 full turns of the crank handle, eight holes on a 20-hole circle plate with a gear ratio of 40/100 and the index plate turning in the same direction as the crank handle.
RAPIT INDEXING:
No difficult calculations needed. You want to cut a hexagon or a square
EXAMPLE:
Determine the indexing to complete six sides on you work piece.
For rapid indexing follow these steps:
Using the handle on the dividing head, disengage the worm from the worm wheel.
On the 12 slot plate, mark every second slot with chalk.
Place the plunger in the first marked slot and cut the first division.
Remove the plunger from the slot, turn the spindle by hand to the next marked
slot.
Engage the plunger and cut the next division.
Repeat the proses until all six divisions have been cut.
Tuesday 8 October 2013
AET Level 4 Beams
Applied Engineering Technology – Level 4
A look at simple beams
A possible question on beams can be as follows:
Use the following information and draw the beam to the scale 1 meter = 1 cm.
A beam of 12 meter is supported at L and R. Point L is 3 meter from the left end of the beam and point R is 2 meter from the right end of the beam. There is a point load of 40N on the left end of the beam and a point load of 20N on the right end of the beam. There is another point load of 50N, 3 meter from the right end of the beam. A uniformly distributed load of 10N/meter starts at L and end at the point load of 5N. Ignore the mass of the beam.
Draw a simplified diagram to scale to solve the reactions at L and R
At first glance this problem seem to be massive because of all the confusing information, but remember that you can eat a big elephant one bite at a time. So let us attack this problem one step at a time.
Step 1: A beam of 12 meter – We stop here and draw a line 12 cm long (scale 1meter = 1 cm.)
________________________________________________________________________________
Step 2: Point L is 3 meter from the left end of the beam. We measure 3 cm from the left end to find L
Step 3: Point R is 2 meter from the right end of the beam. We measure 2 cm from the right end to find R
Step 4: There is a point load of 40N on the left end of the beam.
Step 5: There is a point load of 20N on the right end of the beam.
Step 6: There is another point load of 50N, 3 meter from the right end of the beam.
Step 7: A uniformly distributed load of 10N/meter starts at L and end at the point load of 5N.
And that is your final drawing, your first elephant has been eaten. Now we must start to work out the forces at L and R.
First we need to transform the UDL (uniformly distributed load) of 10N/m to a point load. It is important to work out the length over which the load is distributed. The load starts 3 meter from the one end of the 12 meter beam and ends 3 meter from the other end. Therefore we are going to take the length of the beam and subtrack the the two 3 meter ends. The calculation will look like this. 12-(3+3) = 12-6 = 6 meters.
Now we know that the UDL is distributed over 6 meters at a load of 10N/m (given) The pointload will thus be 6 X 10 = 60N and it will be in the centre of the UDL. Let’s make it part of our drawing:
Let us start with the value of L. – To work out the value of L we must determine the movements around the centre point R. remember that the whole beam can now rotate around the point of R. If we count all the forces we find the following from left to right: 40N , L , 60N , 50N and 20N. Note that we do not count R. So there are 5 forces and we must make sure that we take all of them into account.
Now we must decide which of these five forces will work clockwise around R and which of them will work counter-clockwise. This is easy if we follow the following rule of thumb:
All downward forces to the left of the point R will move counter-clockwise.
All downward forces to the right of the point R will move clockwise.
The opposite is true for the upward forces.
All upward forces to the left of the point R will move clockwise.
All upward forces to the right of the point R will move counter clockwise.
Let us start with the force at L because we want L to be on the left hand side of our equation.
You will find that L is an upward force on the lefthand side of the centre at R and moves the beam in a clockwise direction around R. To find the downward forces that will move the beam clockwise around R (the same direction as L) we must look for downward forces to the right of R. There is only one downward force to the right of R and that is the 20N force ar the end of the beam. Logic now dictate that the remaining three forces wil be counter-clocwice around R because they are downward and to the left of R.
Each force stand in relation to the distance from R and the clokcwise forces must be equel to the counter-clockwise forces. We have two clockwise forces = three counter-clockwise forces. The Equation wil be:
(force X distance)+ (force X distance) = (force X distance)+ (force X distance)+ (force X distance)
(L X 7m)+(20N X 2m) = (40N X 10m)+(60N X 4m)+(50N X 1m) Make sure your distances from R is correct.
(L X 7m)+ 40 = 400 + 240 + 50
(L X 7) = 690 – 40
L = 650 ÷ 7
L = 92.86N
We worked out L. Now we must do the same for R. To work out the value of R we must determine the movements around the centre point L. Remember that the whole beam can now rotate around the point of L. If we count all the forces we find the following from left to right: 40N , 60N , 50N , R and 20N. Note that we do not count L. So there are 5 forces and we must make sure that we take all of them into account.
Now we must decide which of these five forces will work clockwise around L and which of them will work counter-clockwise. This is easy if we follow the following rule of thumb:
All downward forces to the left of the point L will move counter-clockwise.
All downward forces to the right of the point L will move clockwise.
The opposite is true for the upward forces.
All upward forces to the left of the point L will move clockwise.
All upward forces to the right of the point L will move counter clockwise.
Let us start with the force at R because we now want R to be on the left hand side of our equation.
You will find that R is an upward force on the righthand side of the centre at L and moves the beam in a counter-clockwise direction around L. To find the downward forces that will move the beam Counter-clockwise around L (the same direction as R) we must look for downward forces to the left of L. There is only one downward force to the left of L and that is the 40N force ar the end of the beam. Logic now dictate that the remaining three forces wil be clocwice around L because they are downward and to the right of L.
Each force stand in relation to the distance from L and the clokcwise forces must be equel to the counter-clockwise forces. We have two counter-clockwise forces = three clockwise forces. The Equation wil be:
(force X distance)+ (force X distance) = (force X distance)+ (force X distance)+ (force X distance)
(R X 7m)+(40N X 3m) = (60N X 3m)+(50N X 6m)+(20N X 9m) Make sure your distances from L is correct.
(R X 7m)+ 120 = 180 + 300 + 180
(R X 7) = 660 – 120
R = 540 ÷ 7
R = 77.14N
Well done! We have calculated the forces at L and R. Now to make sure thet we have done so correctly. If the forces at L and R is correct the sum of the two upward forces should be equal to the sum of all the downward forces.
L + R = F1 + F2 + F3 + F4
92.86 + 77.14 = 40 + 60 + 50 + 20
170 = 170 And they balance so our calculations were correct.
Now, let us work out the Shear force diagram: We do this directly underneathour drawing that we started with. Draw very light construction lines down from every force (up or down force) and then draw a zero line.
Our Shear force diagram must now be build around the zero line. We will work from left to right and start on the left side of the zero line. Our first force is 40N downward. Strarting from zero we go four centimeter down to A.
For the next three meters there are no force on the beam. We can therefore draw a line straight across to B.
At B we find an upward force of 92.86N (Force at L) Starting at B we draw a line of 9.3cm straight up to C.
From here we find a downward UDL (Unifomed Distributed Load) of 60N over the next 6 meters. We measure 6cm across to D (6 meters) and 6cm down to E (60N). Now we draw a diagonal line from C to E. A UDL will always form a diagonal line.
At point E there is another downward force of 5N so from E we draw a 5cm line down to F.
From F to G there is no force. We draw a 1cm line from F to G.
At G we find that the upward force at R apply. That tells to draw a line of 7.7cm straight up from G to H.
From H to I there is no force. We draw a 2cm line from H to I.
The downward force of 20N at I turn into a 2 cm line that will end on the zero l
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