Question and answers
| Que.No | Marks | |
|---|---|---|
| Q 4a)(iii) |
4 |
Question:State the effect of keyway on the strength of the shaft.Answer: The keyway is a slot machined either on the shaft or in hub to accommodate the key. It is cut by vertical or horizontal milling cutter. A little consideration will show that the keyway cut into the shaft reduces the load carrying capacity of the shaft. This is due to the stress concentration near the corners of the keyway and reduction in the crosssectional area of the shaft. It other words, the torsional strength of the shaft is reduced. The following relation for the weakening effect of the keyway is based on the experimental results by H.F. Moore.
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| Que.No | Marks | |
|---|---|---|
| Q 1a)(iii) |
8 |
Question:A hollow shaft for a rotary compressor is to be designed to transmit maximum torque of 4750 N-m. The shear stress in the shaft is limited to 50 MPa. Determine the inside outside diameter of the shaft if the ratio of inside to outside diameter of the shaft is 0.4.Answer: Design of Hollow shaft: Given Data: T =4750 N-m = 4750 X 103 N-mm , Ʈ = 50 N/mm2 , K=Di/Do =0.4 The hollow shaft is designed on the basis of strength from the derived torsion equation. 4750 X 103 N-mm ---- Thus Do= 79.18 mm 80 mm ( Say ) Di = 0.4 x Do = 0.4 x 80 = 32 mm ----------------------------------------------------------------------------------------------------- |
| Q 1b)(ii) |
6 |
Question:Design a bushed pin type flexible coupling for connecting a motor shaft to a pump shaft for the following service conditions. Power to be transmitted = 40 KW. Speed of the motor shaft = 1000 RPM. Diameter of the motor shaft = 50 mm Diameter of the pump shaft = 45 mm The bearing pressure in the rubber bush and allowable stress in the pins are to be limited to 0.45 N/mm2 and 25 MPa respectivelyAnswer:
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| Q 2b)(ii) |
8 |
Question:Draw neat sketch of a protected type flanged coupling showing all details.Answer: distance to some degree. c) For effective conjugate action i.e for maintaining a constant velocity ration, in case of involute gearing system, the center distance can be changed without affecting angular velocity ratio. d)In involute gearing as the path of contact is a straight line and the pressure angle is constant . Sketch of Protected type flanged coupling with details : ----------------------------------------------------------------------------------------------------- |
| Q 3 e ) |
4 |
Question:A shaft 30 mm. diameter is transmitting power at a maximum shear stress of 80 MPa. If a pulley is connected to the shaft by means of a key, find the dimension of the key so that stress in the key is not to exceed 50 MPa and length of the key is 4 times the width.Answer: Comparison of welded joints with screwed joint. 1) Welded Joint is rigid & permanent. Screwed joint is temporary. 2) Cost of welded assembly is lower than that of screwed joints. 3) Strength of welded structure is more than screwed joints. 4) For welding joints, highly skilled worker are required 5) Welded joints are tight & leak proof as compared to Screwed joints. 6) Welded joint is very difficult to inspect compared to other joints.
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| Q 4a)(iii) |
4 |
Question:Explain effect of keyways on strength of shaft. Name one type of key which does not affect strength of shaftAnswer: Effect of keyways – when the keyways are cut on the shafts, material is removed at the skin, there by weakening the cross section of the shaft. Stress concentration effect is also serious at the corner of the keyways. Thus the shaft become weak. Type of key- Hollow saddle key or Tangent key ----------------------------------------------------------------------------------------------------- |
| Q 4b)(i) |
6 |
Question:(i) Explain different causes of gear tooth failure and suggest possible remedies to avoid such failuresAnswer: Causes-1) Bending failure-every gear tooth acts as a cantilever. If the total repetitive dynamic load acting on the gear tooth is greater than the beam strength of the gear tooth then the gear tooth will fail in bending remedies-module and face width of the gear is adjusted so that the beam strength is greater than the dynamic load 2)Pitting-surface fatigue failure which occurs due to many repetition of Hertz contact stresses , failure occurs when the surface contact stresses are higher than the endurance limit of the material. It starts with the formation of pits which continue to grow resulting in the rupture of the tooth surface. Remedies- dynamic gear tooth load the of gear tooth between the gear tooth should be less than the wear strength 3)Scoring-the excessive heat is generated when there is a excessive surface pressure, high speed or supply of lubricant fails. it is stick-slip phenomenon in which alternate shearing and welding takes place rapidly at high spots. Remedies- by proper designing of the parameters such as speed, pressure and proper flow of lubricant, so that the temperature at the rubbing faces is within the permissible limits. 4)Abrasive wear- the foreign particles of lubricants such as dirt, dust or burr enter between the tooth and damage the form of tooth Remedies- by providing filters for the lubricating oil or using high viscosity lubricant oil which unable the formation of thicker oil film and hence permits easy passage of such particles with ought damage of gear tooth surface. 5)Corrosive wear- due to presence of corrosive elements such as additives present in the lubricating oils. Remedies- proper anti corrosive additives should be used. ----------------------------------------------------------------------------------------------------- |
| Que.No | Marks | |
|---|---|---|
| Q 1 j ) |
2 |
Question:State four types of keys.Answer: (i) Sunk key: Rectangular sunk key, Square sunk key and Parallel sunk key (ii) Gib-head key (iii) Feather key Any four (iv) Woodruff key types of keys (v) Saddle key: Flat saddle key, Hollow saddle key (vi) Tangent key (vii) Round key ----------------------------------------------------------------------------------------------------- |
| Q 1 m ) |
2 |
Question:What are the requirement of a good coupling?Answer: A good coupling should have the following requirements: (i) It should be easy to connect and disconnect. (ii) It should transmit the full power from one shaft to another shaft without losses. (iii) It should hold the shafts in perfect alignment. (iv) It should reduce the transmission of shock loads from one shaft to another shaft. (v) It should have no projecting parts ----------------------------------------------------------------------------------------------------- |
| Q 3 b ) |
8 |
Question:In a rigid flaninged complended to transmit 20K.W at 700.r.p.mAnswer:
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| Q 3 c ) |
4 |
Question:What are the considerations in design of dimensions of formed and parallel key having rectangular cross section?Answer:
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| Q 6 a ) |
8 |
Question:Effect of Keyway on strength of shaftAnswer:
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| Q 6b)(ii) |
8 |
Question:Design consideration while designing the spur GearAnswer: 1) The power to be transmitted 2) The velocity ration or speed of gear drive. 3) The central distance between the two shafts 4) Input speed of the driving gear. 5) Wear characteristics of the gear tooth for a long satisfactory life. 6) The use of space & material should be economical. 7) Efficiency & speed ratio 8) Cost ----------------------------------------------------------------------------------------------------- |
| Q 6c)(ii) |
8 |
Question:State of the Classification of shaft coupling :Answer:
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| Que.No | Marks | |
|---|---|---|
| Q 1b)(ii) |
6 |
Question:The shaft running at 125 r.p.m. transmits 440 kW. Find the diameter of shaft (d) if allowable shear stress in shaft material is 55 N/mm2 and the angle of twist must not be more than 1 on a length of 16(d). The modulus of rigidity G = 0.80 105 N/mmAnswer:
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| Q 1 iii ) |
4 |
Question:State any four factors to be considered while selecting the coupling.Answer: Following factors should be consider while selecting coupling 1) Cyclic operation 2) Duration or life 3) Misalignment of shafts 4) Required torque and desired speed 5) Direction of rotation 6) Protection against overload 7) Operating conditions ----------------------------------------------------------------------------------------------------- |
| Q 2 c ) |
8 |
Question:A belt pulley is fastened to a 90 mm diameter shaft running at 300 r.p.m. by means of a key 20 mm wide and 140 mm long. Allowable stress for the shaft and key material are 40 N/mm2 in shear and 100 N/mm2 in crushing. Find the power transmitted and the depth of the key required.Answer:
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| Q 3 c ) |
4 |
Question:Prove that, for a square key, the permissible crushing stress is twice the permissible shear stress.Answer:
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| Q 3 d ) |
4 |
Question:Why a coupling should be placed as close to a bearing as possibleAnswer: Coupling should be placed as close to a bearing because of following reasons i. It gives minimum vibrations ii. Bending load on the shaft can be minimized iii. It increases power transmission stability. iv. To avoid deflections of shaft. ----------------------------------------------------------------------------------------------------- |
| Q 4a)(ii) |
4 |
Question:State the effect of key-way on the strength of shaft with suitable diagramAnswer:
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| Q 4b)(ii) |
6 |
Question:State any six design considerations while designing the spur gearAnswer: 1) The power to be transmitted 2) The velocity ration or speed of gear drive. 3) The central distance between the two shafts 4) Input speed of the driving gear. 5) Wear characteristics of the gear tooth for a long satisfactory life. 6) The use of space & material should be economical. 7) Efficiency & speed ratio 8) Cost ----------------------------------------------------------------------------------------------------- |
+| Que.No | Marks | |
|---|---|---|
| Q 1a)(iii) |
4 |
Question:Draw a neat sketch of flexible flange coupling and label its main components.Answer:
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| Q 1b)(ii) |
6 |
Question:A hollow shaft is required to transmit 50 kW power at 600 rpm. Calculate its inside and outside diameters if its ratio is 0.8. Consider yield strength of material as 380N/mm2 and factor of safety as 4.Answer: Given : P= 50 KW = 50000W Speed = 600rpm k=Di/do = 0.8 σyt= 380 N/mm2 Factor of safety= 4 Design stress σt=σyt/fos =380/4 =95 Shear stress = τ =σt/2 = 95/2 =47.5N/mm2 Torque transmitted by hollow shaft T = P x 60/2πN T = 50000 x 60/2π x600 T = 795.67 N-m T= 795670 Nmm T= π/16 Xτ X do3 (1-k 4 ) 795670 =π/16 X 47.5 X do3 ( 1-0.84 ) Do3=144529.313 Do = 53 mm say 55 mm Di = 0.8X 55 = 44mm ----------------------------------------------------------------------------------------------------- |
| Q 3 c ) |
4 |
Question:State the ‘Lewis equation’ for spur gear design. State SI unit of each term in the equation.Answer: Lewis equation: WT = σw.b.π.m.y, WT= Tangential load acting at the tooth in N σw= bending stress in N/mm2 b= width of the gear face in mm m= module in mm y= lewis form factor. ----------------------------------------------------------------------------------------------------- |
| Q 3 e ) |
4 |
Question:Prove that for a square key sc = 2t where sc = crushing stress t = shear stress.Answer:
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| Q 4a)(iii) |
4 |
Question:State four important modes of gear failure.Answer: Modes of Gear Failure: ANY 4 modes 1. Bending failure. Every gear tooth acts as a cantilever. If the total repetitive dynamic load acting on the gear tooth is greater than the beam strength of the gear tooth, then the gear tooth will fail in bending, 2. Pitting. It is the surface fatigue failure which occurs due to many repetition of Hertz contactstresses. 3. Scoring. The excessive heat is generated when there is an excessive surface pressure, high speed or supply of lubricant fails. 4. Abrasive wear. The foreign particles in the lubricants such as dirt, dust or burr enter betweenthe tooth and damage the form of tooth. 5. Corrosive wear. The corrosion of the tooth surfaces is mainly caused due to the presence ofcorrosive elements such as additives present in the lubricating oils ----------------------------------------------------------------------------------------------------- |
| Q 4b)(i) |
6 |
Question:Explain the design procedure of shaft on the basis of torsional rigidity. State the equation with SI units. State two applications of this approach.Answer:
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| Que.No | Marks | |
|---|---|---|
| Q 1a)(iii) |
4 |
Question:Write Lewis equation for strength of gear tooth. State the meaning of each termAnswer:
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| Q 1b)(ii) |
6 |
Question:A shaft 1000 mm long is supported between two bearings. A pulley of 250 mm diameter is keyed at 400 mm distance away from left hand bearing. The power transmitted by shaft is 10 kW at 800 r.p.m. The pulley gives power to another pulley vertically below it having an angle of contact between pulley and belt as 180°. The weight of pulley is 300 N. The coefficient of friction between belt and pulley is 0.15. Take shear stress for shaft material as 60 MPa. Find the diameter of the shaft.Answer:
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| Q 2 a ) |
8 |
Question:Explain the design procedure of handlever with neat sketch.Answer:
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| Q 2 b ) |
8 |
Question:Explain the design procedure of bush pin type flexible coupling with neat sketchAnswer: It consist of two shafts, two key, flanges, key, pin, rubber bush, brass bush, & pin ,following are the designations of various coupling dimensions which are use for the design procedure.
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| Q 2c)(ii) |
8 |
Question:Design ‘‘C’’ clamp frame for a total clamping force of 20 kN. The cross-section of the frame is rectangular and width to thickness ratio is 2. The distance between the load line and natural axis of rectangular cross section is 120 mm and the gap between two faces is 180 mm. The frame is made of cast steel for which maximum permissible tensile stress is 100 N/mmAnswer:
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| Q 3 c ) |
4 |
Question:Prove that for square key equally strong in shear and crushing, the permissible crushing stress is twice the permissible shear stressAnswer:
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| Q 3 e ) |
4 |
Question:How keys are classified? Give detailed classification of keys with neat sketches; also state their applications.Answer: Keys are classified on the basis of shape and application of keys.………………are 1) Sunk Key : a) Rectangular sunk key b) square sunk key c) Gib head key d) feather key e) Woodruff key 2) Saddle key a) Flat saddle key b) hollow saddle key 3) Round key 4) Splines …………………: 1) Sunk Key : used for heavy duty application a) Rectangular sunk key : for preventing rotation of gears and pulleys on shaft b) Gib headed key: used where key to be removed frequently. c) feather key: Machine tool 2) Saddle key: for light duty or low power transmission 3) Round key : Used for low Power drive 4) Splines: Where providing axial movement between shaft and mounted member
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| Q 4b)(i) |
6 |
Question:State the different modes of failure of gear teeth and their possible remedies to avoid the failure.Answer: 1. Bending failure. 2. Pitting. 3. Scoring. 4. Abrasive wear. 5. Corrosive wear Remedies to avoid failure: 1. Bending failure. In order to avoid such failure, the module and face width of the gear is adjusted so that the beam strength is greater than the dynamic load. 2. Pitting. In order to avoid the pitting, the dynamic load between the gear tooth should be less than the wear strength of the gear tooth. 3. Scoring. This type of failure can be avoided by properly designing the parameters such as speed, pressure and proper flow of the lubricant, so that the temperature at the rubbing faces is within the permissible limits. 4. Abrasive wear. This type of failure can be avoided by providing filters for the lubricating oil or by using high viscosity lubricant oil which enables the formation of thicker oil film and hence permits easy passage of such particles without damaging the gear surface. 5. Corrosive wear.. In order to avoid this type of wear, proper anti-corrosive additives should be used. ----------------------------------------------------------------------------------------------------- |
