STUPIDSID
Solve any one question from Q.1(a,b) & Q.2(a,b)
1(a)
State & explain different types of gear tooth failures, their causes & remedies.
4 M
1(b)
The following data is given for pair of parallel helical gear made up of steel.
Power transmitted = 20kW
Speed of pinion = 720 rpm
No. Of teeth on pinion = 35
No.of teeth on Gear = 70
Center distance = 285mm
Normal module = 5mm
Face width = 50mm
Normal pressure angle = 20°
Ultimate tensile strength = 600N/mm2
Surface Hardness = 300BHN
Grade of Machining = Grade 6.
Service Factor = 1.25
Calculate:-
i) Helix Angle
ii) The Beam strength
iii) The wear strength \[Y'=0.487-\frac{2.87}{Z'};K_v=\sqrt{\frac{5.6}{5.6+\sqrt{V}}}.\]
Power transmitted = 20kW
Speed of pinion = 720 rpm
No. Of teeth on pinion = 35
No.of teeth on Gear = 70
Center distance = 285mm
Normal module = 5mm
Face width = 50mm
Normal pressure angle = 20°
Ultimate tensile strength = 600N/mm2
Surface Hardness = 300BHN
Grade of Machining = Grade 6.
Service Factor = 1.25
Calculate:-
i) Helix Angle
ii) The Beam strength
iii) The wear strength \[Y'=0.487-\frac{2.87}{Z'};K_v=\sqrt{\frac{5.6}{5.6+\sqrt{V}}}.\]
6 M
2(a)
A steel pinion with 20° full depth involute teeth is transmitting 7.5KW power at 1000 rpm from an electric otor. The starting torque of motor is twice the rated torque. The number of teeth on the pinion is 25, while the module is 4mm. The face width is 45mm. Assuming that the velocity Factor account for dyanamic load, calculate:
i) The efffective load on gear tooth \( Y=0.487-\frac{2.87}{Z}. \)/
ii) The bending stress in gear tooth \( K_v=\frac{3}{3+V}. \)/
i) The efffective load on gear tooth \( Y=0.487-\frac{2.87}{Z}. \)/
ii) The bending stress in gear tooth \( K_v=\frac{3}{3+V}. \)/
6 M
2(b)
With neat sketch, discuss the Force Analysis of Bevel Gear.
4 M
Solve any one question from Q.3(a,b) & Q.4(a,b)
3(a)
A pair of straight Bevel gear with 20° pressure angle consists of 20 teeth pinion meshing with 30 teeth gear. The module is 4 mm while the face width is 20 mm. The pinion & gear material has surface hardness of 400 BHN. The pinion rotate at 720 rpm & receives 3kW power from a motor. The service Factor is 1.5 & Barth Factor Dynamic Loading. Determine the factor of safety in pitting.
6 M
3(b)
What is preloading? Explain preloading of taper roller bearing with sketch.
4 M
4(a)
Classify Bevel Gear. State the advantages & limitations of each type.
4 M
4(b)
The following data is given for a belt drive.
Diameter of pulley = 250mm
Shaft Diameter = 20mm
Power transmitted =5kW
Speed = 720rpm
Ration of Belt tensions = 3:1
Load Factor = 3
Assume the pulley to be placed centrally with the belt tensions acting vertically downward the required reliability of the bearing is 95% with the life of 10,000Hrs.
Find the Dynamic Load carrying capacity of bearing so that bearing are selected from manufacturer's catalogue which list dynamic load carrying capacity at 90% reliability.
Diameter of pulley = 250mm
Shaft Diameter = 20mm
Power transmitted =5kW
Speed = 720rpm
Ration of Belt tensions = 3:1
Load Factor = 3
Assume the pulley to be placed centrally with the belt tensions acting vertically downward the required reliability of the bearing is 95% with the life of 10,000Hrs.
Find the Dynamic Load carrying capacity of bearing so that bearing are selected from manufacturer's catalogue which list dynamic load carrying capacity at 90% reliability.
6 M
Solve any one question from Q.5(a,b) & Q.6(a,b)
5(a)
Deduce an expression for efficiency of warm & worm gear pair.
6 M
5(b)
The following Data is given for worm gear pair
Pitch circle diameter of worm = 48mm
Pitch circle diameter of worm gear = 192mm
Axial pitch of worm = 18.85
Pressure Angle in axial plane of worm = 20.10°
Lead of worm = 18.85mm
Effective width of worm gear teeth = 36mm
Worm speed = 3500rpm.
Permissible bending stress for worm gear = 90N/mm2
Worm gear wear factor = 830kN/m2 Coeficient of friction between worm & worm gear tooth= 0.025 Overall heat transfer co-efficient without fan = 16W/m2°C overall heat transfer coefficient with Fan=15.2+8.25×10-3 hw, W/m2°C Effective area of housing = 9×10-5×(a)1.88, m2 Frictional losses in bearing = 4.5% of total input power. Where nw=worm speed, rpm.
a= center distance in mm
Determine:-
i) The dimensions of worm gear pair
ii) The input power rating on the basis of strength.
iii) The temperature of lubricating oil with fan. Is fan necessary. Comment.
Pitch circle diameter of worm = 48mm
Pitch circle diameter of worm gear = 192mm
Axial pitch of worm = 18.85
Pressure Angle in axial plane of worm = 20.10°
Lead of worm = 18.85mm
Effective width of worm gear teeth = 36mm
Worm speed = 3500rpm.
Permissible bending stress for worm gear = 90N/mm2
Worm gear wear factor = 830kN/m2 Coeficient of friction between worm & worm gear tooth= 0.025 Overall heat transfer co-efficient without fan = 16W/m2°C overall heat transfer coefficient with Fan=15.2+8.25×10-3 hw, W/m2°C Effective area of housing = 9×10-5×(a)1.88, m2 Frictional losses in bearing = 4.5% of total input power. Where nw=worm speed, rpm.
a= center distance in mm
Determine:-
i) The dimensions of worm gear pair
ii) The input power rating on the basis of strength.
iii) The temperature of lubricating oil with fan. Is fan necessary. Comment.
12 M
6(a)
In a design of worm gear pair why worm gear governs the design.
4 M
6(b)
The following data refers to worm gear drive used for transmitting 20 kW power from input speed of 1450 rpm. The reduction ratio is 20:5 while the material for worm gear is bronze K= 2.4MPa.
Diameter Factor = 10
Service Factor = 2
Number of starts on worm =2
Permissible bending strength of worm gear material = 275MPa
Wear load factor = 2.4MPa.
Coefficient of friction = 0.026.
Velocity factor \[\frac{6}{6+V9}\]
Form factor for normal pressure angle of 14.5° = 0.314
Norm gear width = 0.73×worm PCD.
Factor of safety = 1.
Standard first preference values of module are as follows:- 1, 1.025, 1.6, 2, 2.5, 3.15, 4.5, 6.8, 10, 12, 16, 20 mm. Design worm & worm gear Derive. Would you recommend blower for gear box? If it is not possible to fit blower then what will be new value of module for worm gear.
Diameter Factor = 10
Service Factor = 2
Number of starts on worm =2
Permissible bending strength of worm gear material = 275MPa
Wear load factor = 2.4MPa.
Coefficient of friction = 0.026.
Velocity factor \[\frac{6}{6+V9}\]
Form factor for normal pressure angle of 14.5° = 0.314
Norm gear width = 0.73×worm PCD.
Factor of safety = 1.
Standard first preference values of module are as follows:- 1, 1.025, 1.6, 2, 2.5, 3.15, 4.5, 6.8, 10, 12, 16, 20 mm. Design worm & worm gear Derive. Would you recommend blower for gear box? If it is not possible to fit blower then what will be new value of module for worm gear.
14 M
Solve any one question from Q.7(a,b) & Q.8(a,b,c)
7(a)
A Horiziontal Flat belt derive is used to transmit 25kW power form a electric motor running at 1440 rpm to a centrifugal water pump expected to run at 720 rpm. The required center distance is 4.5 m. Select the flat belt for the derive. Using following data:
Recommended Range of belt speed = 17.8m/s≤V≤22.9 m/sec.
Load correction factor = Fa= 1.2
Power rating per ply per mm width at 180° arc of contact & V = 10m/sec.
For HI speed belt - 0.023 kW/ply/ mm.
Standard pulley dia:- 90, 100, 112, 125, 140, 160, 180, 200, 224, 250, 280, 315, 355, 400, 450, 500, 560, 630, 710, 80, 90, 1000, 1120, 1250mm. Arc of contact correction factor (Fa).
Standard Belt width
Recommended Range of belt speed = 17.8m/s≤V≤22.9 m/sec.
Load correction factor = Fa= 1.2
Power rating per ply per mm width at 180° arc of contact & V = 10m/sec.
For HI speed belt - 0.023 kW/ply/ mm.
Standard pulley dia:- 90, 100, 112, 125, 140, 160, 180, 200, 224, 250, 280, 315, 355, 400, 450, 500, 560, 630, 710, 80, 90, 1000, 1120, 1250mm. Arc of contact correction factor (Fa).
| Arc of contact Q | 120° | 130° | 140° | 150° | 160° | 170° | 180° |
| Fa | 1.33 | 1.26 | 1.19 | 1.13 | 1.08 | 1.04 | 1.00 |
Standard Belt width
| Num.ply | Standard belt width b, mm |
| 4 | 40, 44, 50, 63, 76, 90, 100, 112, 125, 152 |
| 5 | 76, 100, 112, 125, 152 |
12 M
7(b)
What are different belt tensioning methods, Explain any one with neat sketch?
4 M
8(a)
Derive the condition for maximum power transmission capacity of belt drive based on belt strength & friction capacity.
6 M
8(b)
Show that maximum power transmission capacity of belt occurs at velocity of belt \( V=\sqrt{\frac{Fi}{3m}};\)/ where Fi is the initial tension in the belt & m is mass per unit length of belt.
6 M
8(c)
What are the different modes of roller chain failure.
4 M
Solve any one question from Q.9(a,b) & Q.10(a,b)
9(a)
The following data gives of 360°C hydrodynamic Bearing
Radial load = 10kN
Journal sped= 1450 rpm.
I/d ratio = 1
Bearing length = 50mm
Radial clearance - 20 microns
eccentricity = 15 microns
Speed gravity of lubricant = 2.09kJ/kg°C.
Calculate:
i) The minimum oil flim thickness.
ii) The coefficient of friction.
iii) The power lost in friction.
iv) The Viscocity of lubricant in Cp.
v) The total flow rate of lubricant in 1/ min.
vi) The side leakage &
vii) The average temperature if make up oil is supplied at 30°C.
Radial load = 10kN
Journal sped= 1450 rpm.
I/d ratio = 1
Bearing length = 50mm
Radial clearance - 20 microns
eccentricity = 15 microns
Speed gravity of lubricant = 2.09kJ/kg°C.
Calculate:
i) The minimum oil flim thickness.
ii) The coefficient of friction.
iii) The power lost in friction.
iv) The Viscocity of lubricant in Cp.
v) The total flow rate of lubricant in 1/ min.
vi) The side leakage &
vii) The average temperature if make up oil is supplied at 30°C.
13 M
9(b)
Write short notes on Additives for mineral oil.
3 M
10(a)
A 50 mm dia. Hardended & ground steel Journal rotate at 1440 rpm. In a lathe turned bronze bushing of 50 mm length. For Hydrodynamic Lubrication the minimum oil film thickness should be five times the sum of surface roughness of Journal &bearing. If the class of fit is H8 d8 & the Viscocity of lubricant is 18Cp Determine:
i) The maximum radial load that the journal can carry & still operate under hydrodynamic condition.
ii) The quantity of lubricating oil required.
iii) The side leakage
iv) The temperature rise considering side leakage surface Roughness
Tolerance Take ρ=860 Kg/m3& Cp=2.09kJ/kg°C
i) The maximum radial load that the journal can carry & still operate under hydrodynamic condition.
ii) The quantity of lubricating oil required.
iii) The side leakage
iv) The temperature rise considering side leakage surface Roughness
| Element | Machining method | Surface Roughness (LA) |
| Shaft | Grinding | 1.6 microns |
| Bearing | Turning/Boaring | 0.8 microns |
Tolerance Take ρ=860 Kg/m3& Cp=2.09kJ/kg°C
| Diameter(mm) | Tolerance(mm) |
| 50 | H8 d8 |
| T0.046 -0.100 | |
| 50 | 0.000 -0.196 |
11 M
10(b)
Derive the Petroff's equation for hydrodynamic bearing. Also state its limitation.
5 M
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