STUPIDSID
Solve any one question from Q1 and Q2
1 (a)
What is the beam strength of spur gear? Derive the expression for it.
6 M
1 (b)
What is virtual number of teeth for a helical gear?
4 M
2 (a)
spur gear pair with 20 o full depth involute tooth profile, consist of 19
teeth pinion meshing with 40 teeth gear. The pinion is mounted on a
crank shaft of 7.5 kW single cylinder diesel engine running at 1500 rpm,
the driven shaft is connected to a machine. Take service factor as 1.5.
The pinion & gear are made of steel with ultimate tensile strength 600
N/mm 2 , the module is 4 mm while the face width is 10 x module. The
gears are ordinarily cut. Take Lewis form factor as 0.314 for 19 teeth.
Calculate the factor of safety based on beam strength.
6 M
2 (b)
Explain the following terms related to helical gears.
i) Tooth advance
ii) Leading edge
iii) Trailing edge
iv) Minimum face width
i) Tooth advance
ii) Leading edge
iii) Trailing edge
iv) Minimum face width
4 M
Solve any one question from Q3 and Q4
3 (a)
For a bevel gear explain the force analysis by considering the total load
is shared by one pair of teeth.
4 M
3 (b)
Suggest suitable bearing for the following applications with Justification
i) Lathe spindle
ii) Table fan shaft
iii) Wind turbine shaft
iv) Railway wheels and axle
v) Hand drill spindle
vi) Household mixer grinder
i) Lathe spindle
ii) Table fan shaft
iii) Wind turbine shaft
iv) Railway wheels and axle
v) Hand drill spindle
vi) Household mixer grinder
6 M
4 (a)
What is addendum modification in gears? How it is done?
4 M
4 (b)
A ball bearing, subjected to a radial load of 5000 N is expected to have a life of 8000 hours with a reliability of 99% at 1450 RPM. Calculate the dynamic load carrying capacity of the bearing, so that the bearing can be
selected from the manufacturer's catalogue. Use the following relation. [ \dfrac {L}{L_{10}} = \left ( \dfrac {ln (1/R)} {ln (1/R_{90})} \right )^{frac {1}{1.17}} ]
6 M
Solve any one question from Q5 and Q6
5 (a)
A Vee belt is used to connect an electric motor with an agitator. Determine
the number of belts required and the pitch length of the belt using following
Data.
| Power capacity | 20 kW |
| Motor capacity | 1440 rpm |
| The pitch diameter of the motor pulley | 300 mm |
| The pitch diameter of the agitator pulley | 900 mm |
| Coefficient of friction for the belt and pulleys as | 0.2 |
| Centre distance of | 1 m |
| Mass density of the belt material | 0.97 g/cc |
| Maximum tension in the belt | 850 N |
| Maximum width at top | 22 mm |
| Minimum width at bottom | 12 mm |
| Depth | 14 mm |
| Groove angle | 40° |
12 M
5 (b)
Explain the various methods used for the belt tensioning.
6 M
6 (a)
Derive an expression for the length of the open flat belt drive.
6 M
6 (b)
Explain the procedure for the selection of flat belt from manufacturer's
catalogue.
6 M
6 (c)
What is polygonal action in roller chain drive? How to control it?
6 M
Solve any one question from Q7 and Q8
7 (a)
Design a worm gear pair based on wear strength and suggest the minimum
surface area to be provided for the gear box if it has to work with natural
circulation. Use following data.
| Number of starts on worm | single |
| Motor power | 3 kW |
| Motor speed | 1500 rpm |
| Required reduction | 30:1 |
| Wear factor | 0.6 N/mm2 |
| Gear tooth system | 20deg; full depth involute |
| Service factor | 1.2 |
| Factor of safety | 1.4 |
| Permissible temperature rise | 50° C |
| Coefficient of friction | 0.03 |
| Overall heat transfer coefficient | 18 W/m2 °C |
| Standard modules: 1, 1.25, 1.5, 2, 2.5, 3, 4, 6, 8, 10, 12, 16 mm | |
12 M
7 (b)
Why the worm gear is always weaker than the worm?
4 M
8 (a)
Compare crossed helical gear drive with worm and worm gear drive.
4 M
8 (b)
A worm gear box with an effective surface area of 1.5 m 2 is working in
still air with a heat transfer coefficient of 15 W/m2 °C. The permissible
temperature rise of the lubricant is 50°C. The worm gear drive is
designated as 1/30/10/8. The motor speed is 1440 rpm and the normal
pressure angle is 20°. Calculate the power rating of the motor using a
coefficient of friction 0.024. Consider a wear factor of 0.6 N/mm2 the
material strength as 180 N/mm2 and a Lewis form factor as 0.46. Take
required factor of safety as 1.5 & service factor 1.2.
12 M
Solve any one question from Q9 and Q10
9 (a)
The following data refers to a 360 o hydrodynamic journal bearing.
Assuming that the total heat generated in the bearing is carried away by the total oil flow in the bearing, Determine
i) Dimensions of the bearing
ii) Coefficient of friction
iii) Power lost in friction
iv) Total oil flow
v) Side leakage
vi) Temperature rise
Refer the following table.
| Radial load | 10 kN |
| Journal speed | 1440 rpm |
| 1/d | 1 |
| Unit bearing pressure | 1000 kPa |
| Clearance ratio | 800 |
| Viscosity of the lubricant | 30 Mpa - S |
Assuming that the total heat generated in the bearing is carried away by the total oil flow in the bearing, Determine
i) Dimensions of the bearing
ii) Coefficient of friction
iii) Power lost in friction
iv) Total oil flow
v) Side leakage
vi) Temperature rise
Refer the following table.
| 1/d | h0/c | ε | S | (r/c)f | Q/(rcns1) | Qs/Q | P/Pmax |
| 0 | 1.0 | 0 | 0 | 0 | 1 | 0 | |
| 0.3 | 0.97 | 0.00474 | 0.514 | 4.82 | 0.973 | 0.152 | |
| 0.1 | 0.9 | 0.0188 | 1.05 | 4.74 | 0.919 | 0.247 | |
| 1 | 0.2 | 0.8 | 0.0446 | 1.7 | 4.62 | 0.842 | 0.313 |
| 0.4 | 0.6 | 0.121 | 3.22 | 4.33 | 0.68 | 0.415 | |
| 0.6 | 0.4 | 0.264 | 5.79 | 3.99 | 0.497 | 0.484 | |
| 0.8 | 0.2 | 0.631 | 12.8 | 3.59 | 0.28 | 0.529 |
12 M
9 (b)
Derive an expression for friction loss in hydrodynamic journal bearing.
4 M
10 (a)
Explain the desirable properties of the material used for the sliding contact
bearings. Also suggest the suitable materials mapped with the desirable
Properties.
8 M
10 (b)
Write the Reynolds' equation for 2D flow and explain the significance
of each term in it.
4 M
10 (c)
Compare the sliding and rolling contact bearings.
4 M
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