×
Hit this exam harder than you have ever done before.
--Your friends at LectureNotes
Close

Note for Kinematics Machines - KM by Anurag Mairal

  • Kinematics Machines - KM
  • Note
  • 228 Views
  • Uploaded 2 months ago
0 User(s)

Share it with your friends

Leave your Comments

Text from page-1

Subject: Kinematics of Machines Unit V A. Pivots and Collars, Thrust Bearing V_1. A pivot bearing of a shaft consists of a frustum of a cone. The diameters of the frustum are 20 cm and 40 cm and its semi-cone angle is 60°. The shaft carries a load of 40 kN and rotates at 240 rpm. Determine (i) the magnitude of the pressure intensity on the bearing surface, and (ii) the horse-power lost in friction. Given µ =0.025. Make suitable assumption if required giving reasons for the same. Prove the formula used. V_2. A thrust bearing supports a load of 240 kN on collars 40 cm external diameter and 30 cm internal diameter. If µ = 0.07, calculate the power lost in friction at the bearing if the shaft rotates at 90 rpm. If the permissible bearing pressure intensity is not to exceed 30 N/cm2, calculate the number of collars required. V_3. The thrust of the propeller shaft of a marine engine is taken up by 8 collars whose external and internal diameters are 660 mm and 420 mm respectively. The thrust pressure intensity is 0.4 MN/m2 and may be assumed uniform. The coefficient of friction between the shaft and the collars is 0.4. If the shaft rotates at 90 r.p.m. find (i) total thrust on collars and (ii) power absorbed by friction at bearing. [651.4 kN, 68 kW] B. Braking of Vehicle, Brakes, Dynamometer V_4. The following data refers to a car in which brakes are applied to the front wheels: Wheel Base = 2.8 m; Centre of Mass from Rear Axle = 1.3 m; Centre of Mass above Ground Level = 0.96 m; Coefficient of Friction between Road and Tyres = 0.4 If the speed of the car be 40 km per hour, find the distance travelled by the before coming to rest when the car (i) moves up an incline 1 in 16 (ii) moves down an incline 1 in 16 (iii) moves on a level track. [22.5m, 40.89m, 29.03m] V_5. State the principle and working of any two types of dynamometers with neat sketches. V_6. A band and block brake having 14 blocks, each of which subtends an angle of 15° at the centre, is applied to a drum of 1 meter effective diameter. The drum and flywheel mounted on the same shaft weigh 20 kN and combined radius of gyration of 50 cms. The two ends of the band are attached to pins on opposite sides of the brake-lever at distances of 3 cm and 12 cm from the fulcrum. If a force of 200 N is applied at a distance of 75 cm from the fulcrum, find (i) maximum braking torque, (ii) angular retardation of drum, and (iii) time taken by the system to come to rest from the rated speed of 360 rpm. Take µ = 0.25. [2547.2 N-cm, 4.9976 rad/sec2, 7.54 sec] V_7. An internal expanding shoe brake has a diameter of 320 mm and width of 30 mm. The cam forces are equal. Maximum pressure is not to exceed 80 kN/m2. If the angle subtended by the shoe are φ1 = 15°, φ2 = 145°, actuating force applied at a distance of 220 mm from pivot, distance of pivot

Text from page-2

from the centre of drum is 125 mm and µ = 0.32, determine the actuating force and braking torque. Assume the rotation of drum is clock-wise.[175.7 N, 48 N-m] V_8. For the shoe brake shown in Fig.V8, the diameter of the brake drum is 400 mm and the angle of contact 96°. The applied force is 3 kN on each arm and the coefficient of friction between the drum and the lining is 0.35. Taking the dimensions a = 500 mm, b = 200 mm and c = 60 mm, determine the maximum torque transmitted by the brake. [1314 N-m] Fig. V8 Unit IV A. Gears IV_1. State and prove law of gearing for constant velocity ratio and show how the involute teeth profile satisfies the condition. Derive an expression for the velocity of sliding between a pair of involute teeth. IV_2. Derive the formula for the length of path of contact for two meshing spur gears having involute profile. IV_3. The following data refer to two mating involute gears of 20° pressure angle Number of Teeth on pinion : 20 Gear Ratio :2 Speed of Pinion (rpm) : 250 Module (mm) : 12 If the addendum on each wheel is such that the path of approach and path of recess on each side are of half the maximum possible length, find (i) Addendum for pinion and gear. (ii) The length of arc of contact. (iii) Maximum velocity of sliding during approach and recess. [aG = 7.77 mm, aP = 19.47 mm, 65.51 mm, Vs = 805.86 mm/sec, Vs’ = 1611.72 mm/sec]

Text from page-3

IV_4. Two 20° involute spur gears have a module of 10 mm. The addendum is one module. The larger gear has 50 teeth and the pinion 13 teeth. Does the interference occur? If it occurs, to what value should the pressure angle be changed to eliminate interference? IV_5. Two gear wheels mesh externally and are to give a velocity ratio of 3 to 1. The teeth are of involute type; module = 6 mm; addendum = one module, pressure angle = 20°. The pinion rotates at 90 rpm. Find (i) Minimum number of teeth on pinion to avoid interference on it and corresponding number on wheel. (ii) Lengths of path and arc of contact. (iii) Number of pairs of teeth in contact. (iv) The maximum velocity of sliding B. Gear Trains I_1. An epicyclic gear consists of a pinion, a wheel of 40 teeth and an annulus with 84 internal teeth concentric with the wheel. The pinion gears with the wheel and the annulus. The arm that carries the axis of the pinion rotates at 100 rpm. If the annulus is fixed, find the speed of the wheel; if wheel is fixed, find the speed of the annulus. Draw the gear train arrangement. Solution: Let TP = Number of Teeth on Pinion = ? TS = Number of Teeth on Wheel = 40 TA = Number of Teeth on Annulus = 84 The gear train arrangement is drawn below Assuming the gears have same module

Text from page-4

𝑇𝐴 𝑇𝑆 = + 𝑇𝑃 2 2 Hence 𝑇𝑃 = 𝑇𝐴 𝑇𝑆 84 40 − = − = 22 2 2 2 2 Revolutions Table Action Arm, a Arm Fixed, S +1 rev 0 Arm Fixed, S +x rev 0 Arm, +y rev 𝑦 Revolutions of Arm and Gears S P 𝑇𝑆 1 − 𝑇𝑃 40 𝑥 − 𝑥 22 20 𝑥+𝑦 𝑦− 𝑥 11 A 𝑇𝑆 𝑇𝑃 × 𝑇𝑃 𝑇𝐴 40 − 𝑥 84 10 𝑦− 𝑥 21 − From the given conditions (a) Arm rotates at 100 rpm, Annulus is fixed NA = 0; y = 100 𝑁𝐴 = 𝑦 − 10 10 𝑥 = 0 𝑜𝑟 100 − 𝑥 = 0 𝑜𝑟 𝑥 = 210 21 21 𝑁𝑆 = 𝑥 + 𝑦 = 210 + 100 = 𝟑𝟏𝟎 𝒓𝒑𝒎 (Answer 1) (b) Arm rotates at 100 rpm, Wheel is fixed NS = 0; y = 100; 𝑁𝑆 = 𝑥 + 𝑦 = 0 𝑜𝑟 𝑥 + 100 = 0 𝑜𝑟 𝑥 = −100 𝑁𝐴 = 𝑦 − 10 21 𝑥 = 100 − 10 21 (−100) = 𝟏𝟒𝟕. 𝟔 𝒓𝒑𝒎 (Answer 2) I_2. The pinion S as shown in Fig.1 has 15 teeth, and is rigidly fixed to a motor shaft. The wheel P has 20 teeth and gears with S and also with a fixed annulus wheel A. The pinion C has 15 teeth and is fixed to the wheel P. C gears with the annular wheel D, which is keyed to a machine shaft. P and C can rotate together on a pin carried by an arm a which rotates about the shaft on which S is fixed. Find the speed of the machine shaft if the motor rotates at 1000 rpm.

Lecture Notes