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Week 4 Assignment 4 – Detailed Answers

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Week 4 Assignment 4 – Detailed Answers 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.
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