Week 4 Assignment 4 – detailed answers

Due date for this assignment: 2017-08-23, 23:59 IST. Click only the correct answers. There is no negative marking.

In all cases of gear Hobbing – assume

(i) that the Hob is getting fed past the blank parallel to the axis of rotation of the blank.

(ii) And in Helical Hobbing, in addition to the feed of the Hob past the blank along axis of blank, there is additional rotational motion of blank provided through differential mechanism.

1. On the gear hobbing machine, the speed ratio of the Hob and the blank is decided by Option (a) : The number of starts on the Hob and the number of teeth on the Hob Option (b) : The number of teeth on the Hob and the number of teeth to be cut on the part Option (c) : The number of teeth to be cut on the part and the number of starts on the Hob Option (d) : None of the others 1 point

Ans : The hob and the gear blank are in reality having motion relation as the worm and worm gear. Hence their speed ratio is k/Z, where k = number of starts on the Hob and Z = number of teeth to be cut on the blank. Hence, option (c ) is the correct answer.

2. There is a right hand helical gear to be cut with helix angle of 150. The left handed hob is having an angle of 30 (Fig. 1). In order to cut the gear on the gear hobbing machine with this Hob, the following configuration is correct

Option (a) : A

Option (b) : B

Option (c) : C

Option (d) : D

Option (e) : None of the others 1 point

Ans : In order to have correct orientation for cutting, the helical thread of the Hob facing the blank should be aligned with (that is, in line with) the potential teeth on the blank. That is only happening in case of option (c ). Hence, option (c ) is correct. 3. One advantage of hobbing over shaping of gears is that

Option (a) : Hobbing can produce internal gears far more easily than shaping

Option (b) : Hobbing can cut helical gears more easily than shaping

Option (c) : Hobbing employs continuous indexing while Shaping uses discontinuous indexing

Option (d) : None of the others 1 point

Ans : Actually – it is shaping which can cut internal gears, hence option (a) is not correct.

Hobbing can cut helical gears more easily than shaping, as it does not require any special guideway (Helical cutting guides) nor any type of special cutter. Hobbing simply requires a differential and a lead change gear box.

Set-up time for changeover from cutting straight spur to Helical gear in case of shaping is comparatively much higher than that in case of Hobbing.

It is true that CNC controls can replace helical cutting guides in shaping process, but let us compare between equals – CNC shaping and CNC Hobbing. CNC controls can replace differential in Hobbing. So that makes them even, but the helical shaping cutter is still required while in Hobbing, the same Hob can be used for cutting straight spur as well as helical gears. Hence option (b) is correct.

Option (c) is not correct as both the methods shaping and hobbing employ continuous indexing.

Hence option (b) is the correct answer.

4. The differential mechanism inside a gear Hobbing machine is useful for

Option (a) : Developing very high or very low values of cutting speed

Option (b) : Cutting of helical gears

Option (c) : Cutting of gears with employment of differential indexing

Option (d) : None of the others 1 point

Ans : The differential mechanism is used for cutting of helical gears. Hence option (b) is correct. The differential mechanism has no connection to the cutting speed, hence option (a) is not correct. Further, differential mechanism as employed in Hobbing and differential indexing are not related. Hence option (c) is not correct.

Hence option (b) is correct.

5. In the figure of a gear hobbing machine (Fig. 2) , the correct position of the index change gear box is

Option (a) : 1

Option (b) : 2

Option (c) : 3

Option (d) : 4 1 point

Ans : The task of the index gear box is to control the number of teeth to be cut. If it is placed at the location 1, any change in the index gear setting would change the cutting speed of the Hob – which is not desired. Hence option (a) is not correct. If it is placed at 2, it is in the line of Hob and blank and will be able to control their rpm ratio (and hence number of teeth on blank). Hence option (b) is correct. If it is placed at location 3, any change in the index gear box setting would change the feed of the Hob in mm of movement parallel to the axis of the blank per rotation of the blank. This is not desired and hence option (c ) is not correct. Hence option (b) is the correct answer.

6. In the following gear hobbing machine configuration as shown in Fig. 3, there will be a problem as follows

Option (a) : No problem at all – this configuration is correct

Option (b) : The index change gear Ui is not at the correct position Option (c) : During helical Hobbing, change in vertical feed will change the helix angle being cut

Option (d) : Every time lead change gear box (Ud) setting is changed, the vertical feed is going to get affected

Option (e) : None of the others 1 point

X

Ans : The feed gear box is not at the correct location in the Fig. 3. It should have been in location X (the location X is added to the figure for clarity of this detailed explanation). Due to its wrong location in Fig. 3, the feed gear box is placed in the kinematic chain between vertical feed and rotation of blank for helical cutting. Hence, every time feed gear box setting is changed for getting a different vertical feed, helix angle being cut on the blank will get changed. This is not desired. Hence Option (c ) is correct.

Hence option (a) is wrong. Also, option (d) is wrong as change in lead change gear box does not affect vertical feed. 7. A mechanical engineer buys a junked Hobbing machine, takes out the differential mechanism and fits it into an Electric rickshaw. All the bevel gears of the differential are identical.

Motors are fixed to the rickshaw bodies. If Z1 = 50, Z2= 150, Z3 =

100, Z4= 100, Z5 = 40, Z6= 160, motor rpm = 300 with direction of rotation as shown, the model that will give him speed higher than 30 km/hr is

Option (a) : 1

Option (b) : 2

Option (c) : 3

Option (d) : 4 1 point

Ans : In these mechanisms, we have to find out the rpm of the wheels by application of Willich’s equation.

Let p = left hand bevel gear and q = right hand bevel gear

Willich’s equation states

푵풑 − 푵풂풓풎 = 풆 푵풒 − 푵풂풓풎

Here, Ni represents the rpm of the element i. Here, the “arm” is the wheel shaft. So, Narm is the rpm of the wheel. The machine elements p and q are the two bevel gears on the two sides of the bevel differential.

The value e = epicyclic gear train value of the differential = speed ratio of element p and q = -1 (in this case)

This “e” value is obtained by keeping the arm fixed (that is, Narm = 0) and noting the ratio of RPMs of p and q (=Np/Nq) in that case.

You will find that in that case, they have equal and opposite rpm (as arm is not rotating). To be more clear  Bevel q rotates, Bevel on arm rotates with it as arm is held steady, Bevel p rotates so that Bevel on arm is rendered an idler. Hence Np = - Nq

Let p = left hand bevel gear and q = right hand bevel gear

For the first case, obviously, Np = -Nq (because they are obtained from motors with same RPM and through same gear pair)

In that case

푵풑 − 푵풂풓풎 = −ퟏ −푵풑 − 푵풂풓풎

Hence Np – Narm = Np + Narm

So Narm = 0

In the second case

Np = Nq = 100 rpm

Hence, in the Willich eqn

ퟏퟎퟎ − 푵풂풓풎 = −ퟏ ퟏퟎퟎ − 푵풂풓풎

Which gives us

100 – Narm = -100 + Narm  Narm = 100 rpm

In the third case

Np = 300 rpm

Nq = 300/4 = 75 rpm

Hence,

푵풑 − 푵풂풓풎 ퟑퟎퟎ − 푵 = 풂풓풎 = −ퟏ 푵풒 − 푵풂풓풎 ퟕퟓ − 푵풂풓풎

Hence, 300 – Narm = -75 + Narm

So that Narm = 375/2 rpm

In the fourth case

Np = 300 rpm

Nq = -75 rpm

Hence

푵풑 − 푵풂풓풎 ퟑퟎퟎ − 푵풂풓풎 = = −ퟏ 푵풒 − 푵풂풓풎 −ퟕퟓ − 푵풂풓풎

Hence, 300 – Narm = 75 + Narm

So that Narm = 225/2 rpm

So, let us check whether the highest rpm of 375/2 is able to achieve a linear velocity of 30 km/hr or not

Linear velocity in the third case

ퟔퟎ ퟔퟎ = 흅 × 푫 × 푵 × = ퟑ. ퟏퟒퟏퟓퟗ × ퟏ × ퟑퟕퟓ × = ퟑퟓ. ퟑퟒퟒ km/hr ퟏퟎퟎퟎ ퟐퟎퟎퟎ

Lets check the next higher value

ퟔퟎ ퟔퟎ 흅 × 푫 × 푵 × = ퟑ. ퟏퟒퟏퟓퟗ × ퟏ × ퟐퟐퟓ × = ퟐퟏ. ퟐ km/hr hence does not qualify ퟏퟎퟎퟎ ퟐퟎퟎퟎ

Therefore the correct answer is option (c )

8. In a Hobbing machine, the operator has just finished the cutting of one Left Hand helical gear. The blank rotational axis is vertical. The Hob started from the top and has reached the lowest point of its vertical motion. The worker now takes out the finished gear and puts another blank for the same Left Hand gear and without any other change, starts the machine in reverse, which reverses ALL motions of the machine. In that case

Option (a) : Cutting will be done properly

Option (b) : Cutting will not be done and accident may occur

Option (c) : Cutting will be done properly but Right Hand helical gear will get cut

Option (d) : None of the others 1 point

Ans : If the direction of rotation of the Hob (the cutter) is reversed, it would not be able to cut. Moreover, the rear faces of the teeth of the cutter would impact the work piece and may result in an accident like cutter breakage / blank damage / machine damage.

Hence, option (b) is correct

9. Two straight tooth spur gears of the same module, same width have 25 (Gear A) and 100 (Gear B) numbers of teeth respectively. They are cut on the same gear hobbing machine with same speed (m/min), feed (mm of Hob travel past blank per rotation of blank) and depth of cut values. In that case

Option (a) : Gear A will necessarily require more Hobbing time for completion compared to that for Gear B

Option (b) : Gear B will necessarily require more Hobbing time for completion compared to that for Gear A

Option (c) : The two gears will necessarily require equal Hobbing time for completion 1 point

Ans : The same cutting speed is employed in both the cases, so Hob speed and rpm are the same in the two cases. Hence, the blanks, which are going to have different numbers of teeth, are going to have different rotations per min (RPMs). This is because ratio of rotations per minute of Hob and blank is k/Z, Z being the number of teeth being cut.

Hence, as Nblank/NHob = k/Z, Nblank = NHob X (k/Z)

Hence the rpm of blank with higher number of teeth will be lower

If the RPMs of the blanks are different for the two cases, in order to achieve the same feed (mm of Hob movement per rotation of blank), the time taken by the Hob to cut the same width will be more in case of larger number of teeth compared to that in case of lower number of teeth.

Hence option (b) will be correct

10. The main problem in Gear Hobbing is that, since blank and cutter rotation are related, increase in cutting speed necessarily results in

Option (a) : Increase of feed motion in mm of Hob travel past blank parallel to the axis of blank/rev of blank

Option (b) : Increase in depth of cut

Option (c) : Increase in the number of teeth being cut

Option (d) : None of the others

Ans : None of the others are related to the RPM of the Hob. Feed motion in mm/rev of blank is not dependent on cutting speed

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