Gears

Question 1
A steel spur pinion has a module (m) of 1.25 mm, 20 teeth and 20^{\circ} pressure angle. The pinion rotates at 1200 rpm and transmits power to a 60 teeth gear. The face width (F) is 50 mm, Lewis form factor Y = 0.322 and a dynamic factor K_v = 1.26. The bending stress (\sigma) induced in a tooth can be calculated by using the Lewis formula given below.

If the maximum bending stress experienced by the pinion is 400 MPa. the power transmitted is __________ kW (round off to one decimal place).

Lewis formula: \sigma =\frac{K_vW^t}{FmY}, where W^t is the tangential load acting on the pinion.
A
10
B
20
C
30
D
40
GATE ME 2020 SET-2   Machine Design
Question 1 Explanation: 
\begin{aligned} F_{t} \times c_{v} s &=b m y\left[\sigma_{b}\right]_{\max } \\ c_{v} &=1.26 \\ F_{t} \times 1.26 \times 1 &=50 \times 1.25 \times 0.322 \times 400 \\ F_{t} &=6388.88 \mathrm{N} \\ P_{\text {angle }} &=F_{t} \times v=\frac{F_{t} \times \pi D_{p} N}{60} \\ &=\frac{6388.88 \times \pi \times 1.25 \times 20 \times 1200}{60}=10 \mathrm{kW} \end{aligned}
Question 2
A helical gear with 20^{\circ} pressure angle and 30^{\circ} helix angle mounted at the mid-span of a shaft that is supported between two bearings at the ends. The nature of the stresses induced in the shaft is
A
normal stress due to bending only
B
normal stress due to bending in one plane and axial loading; shear stress due to torsion
C
normal stress due to bending in two planes and axial loading; shear stress due to torsion
D
normal stress due to bending in two planes; shear stress due to torsion
GATE ME 2020 SET-1   Machine Design
Question 3
Which one of the following is used to convert a rotational motion into a translational motion?
A
Bevel gears
B
Double helical gears
C
Worm gears
D
Rack and pinion gears
GATE ME 2014 SET-4   Machine Design
Question 3 Explanation: 
Bevel gears: Rotational motion transfer between axes at right angle.
Worm gears: For large reduction ratio in a single stage.
Double helical gears: Rotational motion transfer between parallel axes.
Rack and Pinion gears: Rotational to linear motion conversion.
Question 4
A spur pinion of pitch diameter 50 mm rotates at 200 rad/s and transmits 3 kW power. The pressure angle of the tooth of the pinion is 20^{\circ}. Assuming that only one pair of the teeth is in contact, the total force (in newton) exerted by a tooth of the pinion on the tooth on a mating gear is _______
A
638.5N
B
548.6N
C
985.6N
D
784.5N
GATE ME 2014 SET-2   Machine Design
Question 4 Explanation: 
Given,
\begin{aligned} d&=50 \mathrm{mm} \\ \omega&=200 \mathrm{rad} / \mathrm{s} \\ P&=3000 \mathrm{W} \\ \phi&=20^{\circ} \\ T&=\frac{P}{\omega}=\frac{3000}{200}=15 \mathrm{N.m} \end{aligned}

\begin{aligned} F_{T} \times r &=T \\ \therefore F_{T} &=\frac{T}{r}=\frac{15}{0.025}=600 \mathrm{N} \\ \therefore F_{T} &=F \cos \phi \\ F &=\frac{600}{\cos 20^{\circ}}=638.5 \mathrm{N} \end{aligned}
Question 5
A pair of spur gears with module 5 mm and a center distance of 450 mm is used for a speed reduction of 5:1. The number of teeth on pinion is _______
A
30
B
40
C
50
D
60
GATE ME 2014 SET-1   Machine Design
Question 5 Explanation: 


\begin{aligned} \frac{N_{P}}{N_{G}} &=5=\frac{T_{G}}{T_{P}} \\ T_{G} &=5 T_{P} \\ C &=\frac{m}{2}\left(T_{P}+T_{G}\right) \\ T_{P}+T_{G} &=\frac{2 C}{m}=\frac{2 \times 450}{5} \\ &=180 \\ \text{or }\quad 6 T_{P}&=180 \\ \text{or }\quad T_{P}&=30 \end{aligned}
Question 6
For the given statements:

I. Mating spur gear teeth is an example of higher pair
II. A revolute joint is an example of lower pair
Indicate the correct answer
A
Both I and II are false
B
I is true and II is false
C
I is false and II is true
D
Both I an d II are true
GATE ME 2014 SET-1   Machine Design
Question 6 Explanation: 
Example of Lower Pair:
Revolute pair, Prismatic joint, Screw pair, cylindrical joint, spherical joint.
Example of higher pair:
Wheel rolling an surface, cam and follower contact, meshing teeth of two gears.
Question 7
Two cutting tools are being compared for a machining operation. The tool life equations are:

Carbide tool: VT^{1.6}=3000
HSS tool: VT^{0.6}=200

where V is the cutting speed in m/min and T is the tool life in min. The carbide tool will provide higher tool life if the cutting speed in m/min exceeds
A
15
B
39.4
C
49.3
D
60
GATE ME 2013   Machine Design
Question 7 Explanation: 
\begin{aligned} T^{0.16} &=3000 N \qquad \cdots(i)\\ T^{0.6} &=200 N \qquad\cdots(ii)\\ \therefore \qquad \frac{T^{1.6}}{T^{0.6}} &=\frac{3000}{200} \\ T^{1.6-0.6} &=15 \\ \Rightarrow \quad T &=15 \mathrm{min}\\ \therefore \quad V(15)^{1.6}&=3000\\ \Rightarrow V&=39.4 \mathrm{m} / \mathrm{min} \end{aligned}
If the cutting speed exceeds 39.4 m/min, the carbide tool will provide higher tool life
Question 8
A 20^{\circ} full depth involute spur pinion of 4mm module and 21 teeth is to transmit 15kW at 960rpm. Its face width is 25mm.

Given that the tooth geometry factor is 0.32 and the combined effect of dynamic load and allied factors intensifying the stress is 1.5; the minimum allowable stress (in MPa) for the gear material is
A
242
B
166.5
C
121
D
74
GATE ME 2009   Machine Design
Question 8 Explanation: 
Tooth geometry factor, Y=0.32
Combined effect of dynamic load and allied factor
intensifying the stress is f_{s}=1.5
F_{t}=\frac{\sigma_{b} b Y_{m}}{f_{s}}
\Rightarrow 3552 N=\frac{\sigma_{b}(25 \mathrm{mm})(0.32)(4 \mathrm{mm})}{1.5}
\Rightarrow \quad \sigma_{b}=166.5 \mathrm{MPa}
Question 9
A 20^{\circ} full depth involute spur pinion of 4mm module and 21 teeth is to transmit 15kW at 960rpm. Its face width is 25mm.

The tangential force transmitted (in N) is
A
3552
B
2611
C
1776
D
1305
GATE ME 2009   Machine Design
Question 9 Explanation: 
Module: m=4 \mathrm{mm}
Number of teeth: \quad T=21
Speed: N=960 \mathrm{rpm}
Pressure angle: \quad \phi=20^{\circ}
Face width: b=25 \mathrm{mm}
Full depth involute spur pinion
Transmission of power
P=15 \mathrm{kW} \text { at } 960 \mathrm{rpm}
Power = Torque x Angular speed
P=T_{w}=F_{t} \times r \times \omega
= Tangential force \times Radius \times \omega
\begin{aligned} \text { Radius: } r&=\frac{D}{2}=\frac{m T}{2}=\frac{4 \times 21}{2}=42 \mathrm{mm} \\ P&=F_{t} \times r \times \omega \\ \Rightarrow 15 \times 1000&=F_{t} \times 42 \times 10^{-3} \times \frac{2 \pi \times 960}{60}\\ \Rightarrow \quad F_{t}&=3552.56 \mathrm{N} \end{aligned}
Question 10
One tooth of a gear having 4 module and 32 teeth is shown in the figure. Assume that the gear tooth and the corresponding tooth space make equal intercepts on the pitch circumference. Thedimension 'a' and 'b', respectively, are closest to
A
6.08 mm, 4 mm
B
6.48 mm, 4.2 mm
C
6.28 mm, 4.3 mm
D
6.28 mm, 4.1 mm
GATE ME 2008   Machine Design
Question 10 Explanation: 
\begin{aligned} Given: \quad m=4 and T=32 So\\ D &=m \times T=4 \times 32 \\ &=128 \mathrm{mm}\\ P_{c}=\frac{\pi D}{T}=\frac{\pi \times 128}{32} \\ P_{c}=12.566 \mathrm{mm}\\ but \quad P_{c}=a+a=2 a \\ \end{aligned}
(because tooth thickness = tooth space)
\begin{aligned} 2 a &=12.566 \\ a &=6.28 \mathrm{mm} \\ \text{and}\quad b &=4.1 \mathrm{mm} \end{aligned}


\begin{aligned} \because \quad \phi &=\frac{360^{\circ}}{32 \times 4}=2.8^{\circ} \\ \therefore \quad A R &=64 \times \cos 2.8^{\circ} \\ &=63.9^{\circ} \\ \Rightarrow \quad b&=m+F R[\because F R=F A-A R] \\ &=m+(64-63.9) \\ \Rightarrow \quad b&=4.1 \mathrm{mm} \\ \text{Addendum}\quad &= \text{module}\\ \text{Dedendum}\quad &=1.1 \times \text{module} \end{aligned}
There are 10 questions to complete.

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