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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/mm

State any four factors to be considered while selecting the coupling.

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

Explain the design procedure of shaft on the basis of torsional rigidity. State the equation with SI units. State two applications of this approach.

State four important modes of gear failure.

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.

Prove that for a square key sc = 2t where sc = crushing stress t = shear stress.

State the ‘Lewis equation’ for spur gear design. State SI unit of each term in the equation.

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.

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.

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

Draw a neat sketch of flexible flange coupling and label its main components.

(i) Explain different causes of gear tooth failure and suggest possible remedies to avoid such failures

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.

Explain effect of keyways on strength of shaft. Name one type of key which does not affect strength of shaft

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

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