Cutter Nos. Cuts Spur Gear Teeth from 1 135 Teeth to Rack 2 55-134 3 35
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1. A spur gear with 46 teeth, 2.5 module has to be cut on a column and knee type horizontal milling machine with a rotary disc type form gear milling cutter. The 2.5 module cutter no. 3 is used on a blank with outside diameter of 120 mm. Index crank rotation is 20/23 rotations. However, after cutting, it is found that the chordal thickness of the gear teeth is less than the correct value. A possible reason could be a. Incorrect choice of cutter b. Incorrect choice of outside diameter c. Incorrect choice of depth of cut d. Incorrect choice of index crank rotation Ans : If the table of cutter no and gear teeth number is consulted - it would be seen that choice of cutter is correct. Hence a is not correct. The outside diameter by calculation is module × number of teeth + 2 × module = 2.5 × 46 + 2 × 2.5 = ф 120, so the outside diameter is also correct. The index crank rotation should be 40/(number of teeth) = 40/46 = 20/23, so that is also correct. So the possible reason for wrong chordal thickness could be incorrect choice of depth of cut. Hence c is correct. Cutter Cuts Spur Gear Nos. teeth from 1 135 teeth to rack 2 55-134 3 35-54 4 26-34 5 21-25 6 17-20 7 14-16 8 12-13 2. A horizontal column and knee type universal milling machine is to be set up for cutting the teeth of a straight spur gear by differential indexing, using a rotary, disc type, form gear milling cutter. In that case, a. The machine table longitudinal feed screw will need to be connected by gear train to the index plate of the indexing head b. The table will have to be rotated about a vertical axis by the angle 2π/Z, where Z is the number of teeth to be cut c. The index plate of the indexing head will not be allowed to rotate. d. None of the others Ans: Machine table lead screw is connected with the index plate in helical milling and hence it is not applicable here. So a is not correct. Table rotation is necessary in case of helical milling – so b is not correct. In case of differential indexing – index plate has to rotate – so c is also not correct. Hence d is correct. 3. During the milling of straight spur gear teeth by rotary disc type form gear milling cutter on a column and knee type milling machine, the following attachment/equipment/ mechanism is necessary a. Taper turning attachment b. Grinding attachment c. Indexing head d. Whitworth mechanism e. None of the others Ans : Only indexing head is necessary. Other attachment and mechanisms are not. Hence c is correct. 4. A straight spur gear of 35 teeth needs to be cut on a milling machine with the help of indexing head. The number of rotations to be provided at the index crank for the purpose of indexing, is nearest to a. 35/40 b. 40/35 c. 1/35 d. 40 e. None of the others Ans : In case of indexing head, 40 turns of index crank produces 1 turn of blank or job. Hence, as we require 1/35th turn of the blank for indexing, the index crank will have to rotate 40/35 turns. Hence b is correct. 5. During milling of one straight spur gear tooth, total length of travel of involute cutter depends on 1. Diameter of the cutter 2. Only on depth of cut 3. Only on width of the gear blank 4. none of these Ans : If the cutter diameter is very large, it would have to travel a larger distance to move from one side of the blank to the other side. Hence 1 is definitely correct. Hence, it does not only depend on the depth of cut, so 2 is not correct. By same logic, it does not only depend on the width of the gear blank, hence 3 is also not correct. So 1 is the only correct answer. 6. A right hand helical gear of 4 module and 30 teeth has to be cut with helix angle 150 on a column and knee type universal milling machine with a rotary disc type form gear milling cutter. The longitudinal feed screw has a pitch of 5 mm. for this case, the gear ratio of the change gears leading from the longitudinal screw to the index plate is nearest to a. 4/3 b. 0.1373 c. 0.1456 d. 1.159 e. None of the others 휋 3.14159 Ans : The pitch circumference of the helical gear is × 푚 × 푧 = × 4 × 30 =390.289 푐표푠 훼 푐표푠 150 mm 휋×퐷 390.289 Hence, the lead is = = =1456.578 mm 푡푎푛 훼 푡푎푛 150 Feed screw rotations = 1456.578/5 because feed screw pitch is 5 mm Change gear output = 1456.578 ×U/5 In helical milling, this passes to the index plate index crank blank, hence 1/40 ratio of worm – worm gear is incorporated in the path Hence, by the time longitudinal motion of lead is executed, work pc should rotate once Hence (1456.578 ×U/5)×(1/40) = 1 So, U = 0.1373 Hence b is correct 7. A right hand helical gear of normal module = 4and 125 teeth has to be cut with helix angle 150 on a column and knee type universal milling machine with a rotary disc type form gear milling cutter. The choice of the cutter number will be Cutter Cuts Spur Gear Nos. teeth from 1 135 teeth to rack 2 55-134 3 35-54 4 26-34 5 21-25 6 17-20 7 14-16 8 12-13 a. 1 b. 2 c. 3 d. 4 Ans : For helical gear cutting, the choice of cutter corresponds to number of teeth decided by Z1= Z/cos3α = 138 and hence, 1 should be correct. Z1 = number of teeth which should be considered for choice of cutter Z = actual number of teeth on gear to be cut = 125 Α = helix angle As Z1 = 138, cutter no 1 should be used and hence, a is the correct answer 8. A set-up (Fig. 2) comprising of an indexing head H with differential change gears U is used to change the angular inclination of an anti-aircraft gun barrel as shown. There is only 1 circular row of holes on the index plate with 83 holes having equal angular separation from each other. The differential change gear U connects up the worm gear shaft WGS (of indexing head) to the rotatable index plate with the help of shaft OS and bevel gears (not shown). U has a ratio (output rpm/input rpm) of 4×40/83. A clockwise rotation of index crank causes a clockwise rotation of index plate and vice versa. In such a configuration, the smallest measurable angle by which the anti aircraft gun barrel inclination can be changed is Differential H Change WGS Anti aircraft gears, ratio Index gun barrel = U plate Index crank OS a. (1/83) of 1 rotation b. (1/79) of 1 rotation c. (1/87) of 1 rotation d. (40/83) of 1 rotation e. None of the others Ans : Smallest measurable angle = index crank rotation from 1 hole to next hole while index plate moves = 휃 amount of rotation (say) 1 1 Hence 휃 × × 푈 = 휃 − 40 83 1 4×40 1 Putting in the value of U, we have 휃 × × = 휃 − 40 83 83 4 1 1 Or 휃 × − 1 = − Or 휃 = 83 83 79 Hence b is the correct answer 9. An accurately machined straight tooth spur gear with 34 teeth and 4 module is measured by a Gear tooth vernier calliper Vertical 6.081mm Y mm gear tooth vernier caliper. The horizontal scale of the caliper is set with an opening of 6.081 mm and put against the tooth of the gear, as shown in figure. A reading of Y mm is registered in the vertical scale in this setting. The value of Y is a. Greater than 4.073 mm b. Exactly equal to 4.073 mm c. Less than 4.073 mm Ans : The chordal thickness of this gear tooth is 6.28 mm. Hence, as it is registering less value of 6.081 on the horizontal scale, it is set above the pitch diameter. Hence, it will read less than the chordal addendum. As chordal addendum is 4.073, the reading Y must be less than that. Hence c is correct. 10. One of the distinctive features of gear milling practice with rotary disc type form gear milling cutter is a. Indexing process is continuous and takes place while cutting is going on b. A single cutter can cut the teeth on all gears with a particular module c. A certain amount of geometrical error is inherent on most of the gears cut by this process d. None of the others Ans : Ideally – for each number of teeth, there should be a unique cutter, because the involutes for different base circles (that means, different numbers of teeth) are different. However, this would raise the costs of production, if so many cutters are kept in stock. As a compromise, only 8 cutters handle all the numbers of teeth. Hence, a certain amount of error is tolerated on most of the gears cut by this process. Hence c is correct. .