Shaft.pdf
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- Question 1 In a reciprocating engine which has four cylinders in line, the reciprocating masses are each 0.5 kg per cylinder. The stroke is 120 mm, and the connecting rods are 450 mm long. The firing order of the engine is 1-4-2-3 and the cranks are set at 90° to each other. The cylinders are spaced 140 mm apart and the engine speed is 2000 rev/min. Find: 1.1. The maximum out-of-balance primary and secondary forces. 1.2. The maximum out-of-balance primary couples. Hint: (Take the centre of the engine as your reference)3. A motor is connected by a belt to pulley B. T, is the tight-side tensile force equal to 1800 N. The shaft spins at a constant speed, 650 rpm. The power delivered by the motor is 12.6 kW. On pulley A, connected to the load, the slack side of the belt has a tension equal to th of the tight side. b T T, B d shaft Dimensions are as follows: da 200 тm 300 тт | 250 тm |180 тm | 250 тm 26 тm dB dahaft a We analyzed a problem similar to this on Test 01 using static failure theories. Is this strictly correct? Because the shaft is spinning, would not a material particle see a time-varying reversal of the the bending stress as in the R.R. rotating beam test? Would a particle see a reversal of the torsional shear stress? Assume Sut = 370 MPa and Sy = 300 MPa. Ignore Marin factors. Ignore stress concentrations. Ignore direct shear stress (that due to V). Find the safety factors with respect to fatigue and yielding. Be guided by section 6-14 on combined loading.Q2 / A shaft transmitting 15kW at 200r.p.m. are supported by two bearings placed 1 m apart. A 600 mm diameter pulley is mounted at a distance of 300 mm to the right of left hand bearing and this drives a pulley directly below it with the help of belt having maximum tension of 2.25kN. Another pulley 400 mm diameter is placed 200 mm to the left of right hand bearing and is driven with the help of electric motor and belt having maximum tension of 3.376kN, which is placed horizontally to the right. The angle of contact for both the pulleys is180° and =0.24. Determine the suitable diameter for a solid shaft if the bending moment acting on the shaft is reversed, allowing working stress of 42MPA in shear for the material of shaft and have yield stress of 400MPA and ultimate stress of 500MP.. And design the keys which support each pulley if the material of the keys is the same for the material of shaft.
- der engine has been drawn 0 2 moment diagram for a multicylin ~ 2.4° horizontallv. e = 4500 N-m vertically and 1 mm 'mxemapwd areas between output torque curve and the mean resistance line taken in order from one end are 342. 23, 245. 303. 115. 232. 227. 164 mm2. when the engine is running at 150 r.p.-m- If the mass of the flywheel is 1000 kg and the tota! fluctuation of speed does not exceed 3% of the mean speed. find the minimum value of the radius of gyration 16.4/ The turningA horizontal shaft AD supported in bearings at A and B and carrying pulleys at C and D is totransmit 75 kW at 500 r.p.m. from drive pulley D to off-take pulley C, Calculate the diameter of shaft. The data given is : P1 = 2 P2 (both horizontal), Q1 = 2 Q2 (bothvertical), radius of pulley C = 220 mm, radius of pulley D = 160 mm, allowable shear stress = 45 MPa.[Ans. 100 mm] Given CD=350mm CA=100mm AB=150mm BD=100mmA solid constant-diameter circular shaft is subjected to the torques of TA = 480 lb-ft, TB = 1060 lb-ft, Tc= 860 lb-ft, and Tp = 280 lb-ft, acting in the directions shown. The bearings shown allow the shaft to turn freely. (a) Plot a torque diagram showing the internal torque in segments (1), (2), and (3) of the shaft. Use the sign convention presented in %D Section 6.6. (b) If the allowable shear stress in the shaft is 5800 psi, what is the minimum acceptable diameter for the shaft? TB TA Tc (1) Tp (3) В C D Answer: (a) T1 = Ib-ft T2 = Ib-ft T3 = Ib-ft (b) d = in.
- Consider a rotary motion axis driven by an electric motor. The rotary load is directly connected to the motor shaft with gear speed reducer of gear ratio = 5 and gear efficiency = 0.75. The rotary load is a solid cylindrical shape made of Aluminium material, d=50 mm, l=35 mm, p=2710 kg/m³. The desired motion of the load is a periodic motion as shown in figure. Motor rotor Desired Motion T. Gear ra Load T tame The total distance to be travelled is % of a revolution. The period of motion is tcyc=300 msec., and dwell portion of it is taw=100 msec., and the remaining part of the cycle time is equally divided between acceleration, constant speed and deceleration periods, t,-t,-ta=50 msec. Determine the required motor size for this application.12. A horizontal shaft AD supported in bearings at A and B and carrying pulleys at C and D is to transmit 75 kW at 500 r.p.m. from drive pulley D to off-take pulley C, as shown in Fig. 14.18. P₁ A IPE -P₂ -600 600- 300- 9₂ All dimensions in mm. Fig. 14.18 Calculate the diameter of shaft. The data given is: P,-2 P, (both horizontal), Q, 2 Q₂ (both vertical), radius of pulley C-220 mm, radius of pulley D 160 mm, allowable shear stress - 45 MPa. [Ans. 100 mm]For the free body diagram of the shaft shown below, the rolling contact bearings are located at A and C. For the bearing at A, the desired design load for the bearing is Tangential Direction for Gear Y-X axes: Radial Direction for Gear Z-X axes: 1000 Ibf 500 Ibf A A В 500 Ibf 500 Ibf 250 Ibf 250 Ibf
- Assuming there is no axial thrust, find the reactions @ A & C. Y Note: Force actlig on gear B is parallel to z-axis. Force acting on year D is parallel to y-ax's. 50 70015 3010 25 in 86916 201 10015. A pulley is keyed to a shaft midway between two anti-friction bearings. The bending moment at the pulley varies fiom – 170N-m to 510 N-m and the torsional moment in the shaft varies from 55 N-m to 165 N-m. The frequency of the variation of the loads is the same as the shaft speed. The shaft is made of cold drawn steel having an ultimate strength of 540 MPa and a yield strength of 400 MPa. Determine the required diameter for an indefinite life. The stress concentration factor for the keyway in bending and torsion may be taken as 1.6 and 1.3 respectively. The factor of safety is 1.5. Take size [Ans. 36.5 mm] factor = 0.85 and suface finish factor = 0.88. [Hint. Assume 0, = 0.5 0,;1, = 0.5 0,; t, = 0.55 0,]A flange coupling is to connect two 57mm shafts. The hubs of the coupling are each 111m indiameter and 92mm hub length. Six 16- mm bolts in a 165-mm diameter bolt circle connects theflanges. The key is 14mm high, 14mm wide and 86mm long. Coupling is to transmit 45kW at 160 rpm.For all parts, yield point value in shear is one-half the yield point value in tension or compression thatis 448MPa. Determinea. Shearing stress of the key and its factor of safetyb. Bearing stress of the key and its factor of safetyc. Shearing stress in bolts and the factor of safety based on yield point