1. The shaft angle of two helical gears is 60°. The normal module is 6 mm. The larger gear has 50 teeth and the gear ratio is 2. The centre distance is 300 mm. Find the helix angles of the two gears.
a) Ψ1 = 11.68°, Ψ2 = 48.32°
b) Ψ1 = 13.02°, Ψ2 = 46.98°
c) Ψ1 = 10.11°, Ψ2 = 49.89°
d) Ψ1 = 12.21°, Ψ2 = 47.79°
Explanation: Given: Ψ1 + Ψ2 = 60°; Ψ1 = 60° – Ψ2; mn = 6 mm; T2 = 50 and T1 = T2/G = 25. C = 300 mm
C = 300 = (mn/2) ((T1/cos Ψ1) + (T2/cos Ψ2)) = 3((25/cos(60° – Ψ2))+(50/cos Ψ2)
100 = ((25/cos(60° – Ψ2))+(50/cos Ψ2)
By trial and error method, we get Ψ2 = 47.79°, Ψ1 = 12.21°.
2. Find the spiral angles of the two helical gears for maximum efficiency if the shaft angle is 100° and the friction angle is 10°.
a) 55°,55°
b) 50°,50°
c) 55°,45°
d) 65°,45°
Explanation: For maximum efficiency of the helical gears, Ψ1 = (θ + φ)/2 = (100° + 10°)/2 = 55°
Ψ2 = 100° – 55° = 45°.
3. Find the normal circular pitch as well as the normal module if the larger gear has 40 teeth and the gear ratio is 4. The helix angle is 20° and the shaft angle is 50°. Assume the centre distance between the gears to be 200 mm.
a) 22.12 mm, 7.04 mm
b) 21.31 mm, 6.78 mm
c) 20.37 mm, 6.48 mm
d) 21.98 mm, 6.996 mm
Explanation: Given: T2 = 40 and T1 = T2/G = 40/4 = 10, Ψ1 + Ψ2 = 50°; Ψ1 = 20°; Therefore, Ψ2 = 30°; C = 200 mm
C = 200 = (mn/2) ((T1/cos Ψ1) + (T2/cos Ψ2))
400 = mn x (56.83)
mn = 7.04 mm and thus pn = πmn = 22.12 mm.
4. The normal circular pitch of two helical gears in a mesh is 7 mm. If the number of teeth on the smaller gear is 22 and the gear ratio is 2, find the pitch diameters of the gears. Ψ1 = 40° and Ψ2 = 25°.
a) 83.01 mm, 109.93 mm
b) 63.99 mm, 108.17 mm
c) 85.34 mm, 96.32 mm
d) 79.12 mm, 100.13 mm
Explanation: T1 = 22 and T2 = T1 x G = 22 x 2 = 44, pn = 7 mm, Ψ1 = 40° and Ψ2 = 25°
d1 = P1T1/π = pnT1/(cos Ψ1 x π) = 63.99 mm
d2 = P2T2/π = pnT2/(cos Ψ2 x π) = 108.17 mm.
5. For two gears having shaft angle 90° and friction angle 6°, find the maximum efficiency of the helical gears.
a) 89.23%
b) 85.64%
c) 79.02%
d) 81.07%
Explanation: θ = 90°, φ = 6°
ηmax = (cos(θ+φ)+1)/ (cos(θ-φ)+1) = 0.8107 = 81.07%.
6. When large gear reductions are needed _________ gears are used.
a) helical
b) spur
c) worm
d) bevel
Explanation: Worm gears are used where large speed reductions are needed. The horizontal portion of the gear is called as a worm and the assembly of it with the gear is known as a worm gear. Worm can easily turn a gear but a gear cannot turn a worm because of the shallow angle on the worm.
7. The driven gear in the worm gear is a helical gear. True or false?
a) True
b) False
Explanation: The helical gear is driven in the worm gear and the driving element is the screw or the worm. Worm gears are used in transmission of power between two non-parallel and non-intersecting shafts.
8. Which is of these is an advantage of worm gear?
a) It is expensive
b) Has high power losses and low transmission efficiency
c) Produce a lot of heat
d) Used for reducing speed and increasing torque
Explanation: Reduction of speed and increasing the torque is an advantage of worm gear. The rest are the disadvantages of the worm gear. Worm gears are used in gate control mechanisms, hoisting machines, automobile steering mechanisms, lifts, conveyors and presses.
9. The distance between corresponding points on adjacent teeth measured along the direction of the axis is called ____________
a) joint line
b) normal link
c) axial pitch
d) lead
Explanation: Axial pitch is the distance between corresponding points on adjacent teeth measured along the direction of the axis. The axial pitch of the worm gear is the same thing as the circular pitch of the helical gear.
10. The distance by which a helix advances along the axis of the gear for one turn around is called _____________
a) joint line
b) normal link
c) axial pitch
d) lead
Explanation: Lead is the distance by which a helix advances along the axis of the gear for one turn around. The axial pitch is equal to the lead in a single helix and the axial pitch is one half of the lead in a double helix and so on.