Mechanics of Machines Lab PDF

EXPERIMENT NO. 1 Object: To find out critical speed experimentally and to compare the whirling speed of a shaft. Appratus: Tachometer, shaft, End fixing arrangement etc. Theory: This apparatus is developed for the demonstration of a whirling phenomenon. The shaft can be tested for different end co...

EXPERIMENT NO. 1 Object: To find out critical speed experimentally and to compare the whirling speed of a shaft. Appratus: Tachometer, shaft, End fixing arrangement etc. Theory: This apparatus is developed for the demonstration of a whirling phenomenon. The shaft can be tested for different end conditions. The apparatus consists of a frame to support its driving motor, end fixing and sliding blocks etc. A special design is provided to clear out the testing of bearing of motor spindle from these testing shafts. The special design features of this equipment are as follow: a. Coupling A flexible shaft is used to drive the test shaft from motor. b. Ball bearing fixing ends. The end fixes the shaft while it rotates. This can be replaced within a short time with the help of this unit. The fixing ends provide change of end fixing condition of the rotating shaft as per the requirement. Shaft supplied with the equipment Polished steel shaft is supplied with the machine. The dimensions being as under: Shaft no. Diameter (approx) Length(approx) 1. 4.0 mm 900 mm 2. 4.7 mm 900 mm End fixing arrangement At motor end as well as tail end different end conditions can be developed by making use of different fixing blocks. 1. Supported end conditions - make use of end block with single self-aligning bearings. 2. Fixed end condition - make use of end block with double bearing. Guard’s d1 and d2 and d3: The guard’s d1, d2 and d3 can be fixed at any position on the supporting bar frame which fits on side supports. Rotating shafts are to be fitted in blocks in a and b stands. Speed control of driving motor: The driving motor is 230v, dc 1/6 hp, 3000 rpm, universal motor and speed control unit is a dimmer state of 240v, 2 amps, 50 c/s. Measurement of speed: To measure the speed of the rotating shaft a simple tachometer may be used on the opposite side of the shaft extension of the motor.  Whirling of Elastic Shaft: if L = length of the shaft in cm. E = young’s module kg/cm2 = 2.060 x 106 I = 2nd moment of inertia of the shaft cm4 w = weight of the shaft per unit length kg/cm. g = acceleration due to gravity of cm/sec2 = 981 Then the frequency of vibration for the various modes is given by the equation: f = k x √ (EIG/wL4) value of k End condition 1st mode 2nd mode fixed , supported fixed , fixed 1.47 2.56 1.57 2.46 Data: shaft dia i = cm4 w = kg/cm 4.0 mm x 10-4 0.15 x 10-2 4.7 mm x 10-4 0.19 x 10-2 Calculations: a) Both ends of shafts free (supported) 1st and 2nd mode of vibration can be observed of shafts with both rods. b) One end of shaft fixed and the other free; 1st and 2nd mode of vibration can be observed on shaft with 4.0 mm rod. c) Both ends of shaft fixed- 2nd mode of vibration cannot be observed on any of the shafts as the speeds are very high and hence beyond the range of the apparatus. Fixed – fixed Diameter of rod = 0.4 cm Weight = 150 grams =0.0015 kg/cm Young modulus e = 2.06 x 106 s.no. speed value i=(πd4)/64 weight fth = fact = rpm of k in kg/ √(EIG/wL4) rpm/time 1stmode cm 1 2 Diameter of rod = 0.47 cm Weight = 190 gram = kg/cm = 0.0019 Young modulus e = 2.06 x 106  S.no. Speed Value I=(πd4)/64 weight Fth = fact = rpm of k in kg/ √(EIG/wL4) rpm/time 2nd mode cm 1 2 Supported – fixed Diameter of rod = 0.4 cm Weight = 150 gram = kg/cm = 0.0015 Young modulus e = 2.06 x 106 S.no. Speed Value I=(πd4)/64 weight Fth = fact = rpm of k in kg/ √(EIG/wL4) rpm/time 1st mode cm 1 2 Diameter of brass rod = 0.47 cm Weight = 190 gram = kg/cm = 0.0019 Young modulus e = 2.06 x 106 S.no. Speed Value I=(πd4)/64 weight Fth = fact = rpm of k in kg/ √(EIG/wL4) rpm/time 2nd mode cm 1 2 Result : % Error Preacautions:  EXPERIMENT NO. 2 Object: (a) Determination of characteristic curve of a governor(spindle) speed against sleeve displacement.(For Porter Governor) (b) Plotting of governor characteristic curves of radius of rotation of the ball centre against controlling force. (For Porter Governor) Apparatus used: - Porter Governors. Introduction & theory:-The function of a governor is to regulate the mean speed of an engine, when there are variations in the load e.g. when the load on an engine increases, its speed decreases, therefore it becomes necessary to increase the supply of working fluid. When the load on the engine decreases, its speed increases and thus less working fluid is required. The governor automatically controls the supply of working fluid to the engine with the varying load conditions and keeps the mean speed within certain limits. The governors may, broadly, be classified as 1. Centrifugal governor 2. Inertia governor The centrifugal governors, may further be classified as follows: (1) Dead weight: Watt, Porter governor and Proell governor (2) Spring controlled governors: Hartnell governor, Hartung governor, Wilson-Hartnell governor and Pickering governor 1. Watt Governor:-The simplest form of a centrifugal governor is a Watt governor. It is basically a conical pendulum with links attached to a sleeve of negligible mass. The arms of the governor may be connected to the spindle in the following three ways : 1. The pivot P, may be on the spindle axis. 2. The pivot P, may be offset from the spindle axis and the arms when produced intersect at O. 3. The pivot P, may be offset, but the arms crosses the axis at O. 2. Porter Governor: - The porter governor is a modification of a Watt’s governor, with central load attached to the sleeve. The load moves up down the central spindle. This additional downward force increases the speed of revolution required to enable the balls to rise to any to any pre-determined level. 2. Proell Governor: - It is a modification of a Porter governor, in which balls are mounted on the extension of lower arms. Under normal conditions extension of lower arms remains vertical however it changes position with variation in speed.This governor is more sensitive than porter governor. Calculation :- Radius of rotation r can be calculated as follows: a) Find height h = (ho-x/2) b) Find α by using Cos α = h/L c) Then r = 50 + L Sin α 2 N = 895/h (For watt governor) Force can be calculated as follows: a) Find the angular velocity ω of the spindle By 2π N/ 60 Where N is the speed of spindle. b) Find the centrifugal force acting on the ball Force F = m ω2r For Hartnell Governor Radius of rotation r = ro + x.a/b Where a, b are length pof bell crank lever 2 N = m + M (1+q)/2 x 895 (For porter governor ), where, q = tan β/ tan α mh Observation:- For watt and porter governor Mass of the ball (m) = o.6 kg. Length of each link (L) = 125 mm Initial height of the governor (ho) = 94mm. Initial radius of rotation (ro )= 136 mm Weight of sleeve(M) = 0.6 kg For Watt Governor S. No. Sleeve Avg. Height(h) Cos α r = 50 Force F displacement(X) Speed(RPM) +L ,mm Sin α 1 10 2 20 3 30 4 40 5 50 For Porter Governor a) Weight of Sleeve= 0.5 kg S. No. Sleeve Avg. Height(h) Cos α r = 50 Force F displacement(X) Speed(RPM) +L ,mm Sin α 1 10 2 20 3 30 4 40 5 50 a) Weight of Sleeve= 1 kg S. No. Sleeve Avg. Height(h) Cos α r = 50 Force F displacement(X) Speed(RPM) +L ,mm Sin α 1 10 2 20 3 30 4 40 5 50 a) Weight of Sleeve= 1.5 kg S. No. Sleeve Avg. Height(h) Cos α r = 50 Force F displacement(X) Speed(RPM) +L ,mm Sin α 1 10 2 20 3 30 4 40 5 50 Performance Characteristic Curve: speed vs height of governor Force vs radius of rotation Precautions :-  DO NOT KEEP THE MAINS ON when trial is complete  increase the speed gradually.  take the sleeve displacement reading when the pointer remain steady.  see that at higher speed the load on sleeve does not hit the upper sleeve of the governor.  while closing the test bring the dimmer to zero position and then switch OFF VIVA – QUESTIONS: 1. What is the function of a governor ? 2.How does it differ from that of a flywheel ? 3.State the different types of governors. 4.What is the difference between centrifugal and inertia type governors ? 5.Explain the term height of the governor. 6.What are the limitations of a Watt governor ? 7.What is the stability of a governor ? 8. Define the Sensitiveness of governor. 9. Which of the governor is used to drive a gramophone ? 10.The power of a governor is equal to -------- --. 11.What is hunt?  EXPERIMENT NO. 3 Object :- To check experimentally the normal method of calculating the position of counter balancing weight in rotating mass systems. Apparatus Used:- Static & Dynamic balancing machines, a set of 6 blocks of different weights Theory:- Static Balancing: A system of rotating masses is said to be in static balance if the combined mass centre of the system lies on the axis of rotation. Whenever a certain mass is attached to a rotating shaft, it exerts some centrifugal force, whose effect is to bend the shaft and to produce vibrations in it. Dynamic Balancing: When several messes rotates in different planes, the centrifugal force, in addition to being out of balance, also forms couples. A system of rotating masses is in dynamic balance when there does not exit any resultant centrifugal force as well as resultant couple.In order to prevent the effect of centrifugal force, another mass is attached to the opposite side of the shaft. The process of providing the second mass in order to counteract the effect of the centrifugal force of the first mass, is called balancing of rotating masses. The following cases are important from the subject point of view : 1 Balancing of a rotating mass in the same plane.(Static Balancing) 2 Balancing of a rotating mass in different planes. (Dynamic Balancing) PROCEDURE:- Static Balancing: Remove the belt, the value of weight for each block is determined by clamping each block in turn on the shaft and with the cord and container system suspended over the protractor disc, the number of steel balls, which are of equal weight are placed into one of the containers to exactly balance the blocks on the shaft. When the block becomes horizontal, the number of balls N will give the value of wt. for the block. For finding out Wr during static balancing proceed as follow: 1 Remove the belt. 2 Screw the combined hook to the pulley with groove. This pulley is diff. than the belt pulley. 3 Attached the cord end of the pans to above combined hook. 4 Attached the block no.-1 to the shaft at any convenient position and in vertical downward direction. 5 Put steel balls in one of the pans till the blocks starts moving up. (upto horizontal position). 6 Number of balls gives the Wr value of block-1. Repeat this for 2-3 times and find the average no. of balls. 7 Repeat the procedure for other blocks. Dynamic balancing It is necessary to leave the machine before the experiment. Using the value of Wr, obtained as above, and if the angular positions and planes of rotation of three of four blocks are known, the student can calculate the position of the other block(s) for balancing of the complete system. From the calculations, the student finally clamps all the blocks on the shaft in their appropriate positions. Replace the motor belt , transfer the main frame to its hanging position and then by running the motor, one can verify that these calculations are correct and the balls are perfectly balanced. Observations:- For Static Balancing For static balancing, the Wr for unbalance are:- Block 1 2 3 4 5 6 No. M.r For Dynamic Balancing Consider the 4 blocks for finding the angular positions of these blocks for complete dynamic balancing, the table will be as follows:- S.No Block M.r Distance Couple Angular. No. (No. of Balls) (l)(cm) positions(θ) 1. 2. 3. 4. Calculation: - The balancing masses and angular positions may be determined graphically as given below:- 1. First of all, draw the couple polygon from the data which are calculated in table to some suitable scale. The vector distance represents the balanced couple. The angular position of the balancing mass is obtained by drawing, parallel to vector distance. By measurement we will find the angle.  4. Then draw the force polygon from the data, which are calculated in table to some suitable scale. The vector distance represents the balanced force. The angular position of the mass is obtained by drawing, parallel to vector distance. By measurement we will find the angle in the anticlockwise direction from x axes. Result:- Angular position of block are obtained from polygon and the magnitude of block U is also obtained Mr =……Adjust all angular and lateral position properly and find that the shaft rotates without vibrations. PRECAUTIONS:- 1 Do not run the motor for more time in unbalanced position. 2 Place the weight/balls gently in the pan. While placing the balls the pan should be hold gently and check that it should not jump its position. 3 Weight setting gauge should be check gently. 4. Couple should be represented by a vector drawn perpendicular to the plane of the couple. 5. Angular position measure carefully in clockwise direction. 6. Vector diagram should be represent with suitable scale.. VIVA QUESTIONS: 1. Why is balancing of rotating parts necessary for high speed engines ? 2. Explain the terms ‘static balancing’ and ‘dynamic balancing’. State the necessary conditions to achieve them. 3. Discuss how a single revolving mass is balanced by two masses revolving in different planes. 4. How the different masses rotating in different planes are balanced ? 5. Explain the method of balancing of different masses revolving in the same plane. 6.Why is balancing of rotating parts necessary for high speed engines ? 7. Explain the terms ‘static balancing’ and ‘dynamic balancing’. State the necessary conditions to achieve them. 8. Discuss how a single revolving mass is balanced by two masses revolving in different planes. 9. How the different masses rotating in different planes are balanced ? 10. Explain the method of balancing of different masses revolving in the same plane. \  Experiment No. 4 To study the undamped free vibration of equivalent spring mass system. To find the frequency of undamped free vibration of equivalent spring mass system. 1. DESCRIPTION: The equipment is designed to study free damped and undamped vibration. It consists of M.S. rectangular beam supported at one end by a trunion pivoted in ball bearing. The bearing housing is fixed to the side member of the frame. The other end of beam is supported by the lower end of helical spring; upper end of the spring is attached to screw, which engages with screwed hand wheel. The screw can be adjusted vertically in any convenient position and can be clamped with the help of lock nut. The exciter unit can be mounted at any position along the beam. Additional known weights may be added to the weight platform under side exciter. Figure 6.1: Undamped free vibration of equivalent spring mass system 2. EXPERIMENTAL PROCEDURE:  Support one end of beam in the slot of trunion and clamp it by means of screw.  Attached the other end of the beam to lower end of spring.  Adjust the screw to which the spring is attached with the help of hand wheel such that beam is horizontal in position.   Weight the exciter assembly along with discs, bearing and weights platform.  Clamp the assembly at any convenient position.  Measure the distance L1 of the assembly from pivot. Allow system to vibrate freely.  Measure the time for any 10 oscillations and periodic time and natural frequency of vibration.  Repeat the experiment by varying L1 and also putting different weights on platform. 3. Nomenclature: fact Actual frequency sec-1 ftheo Theoretical frequency sec-1 g Acceleration due to gravity m/Sec2 k Stiffness of the spring kg/m L Distance of spring from pivot. m L1 Distance of w from pivot. m Li Length of spring. m m Total mass of exciter assembly kg-sec2/m me Equivalent mass exciter assembly kg-sec2/m n No. of oscillations. t Time taken by ‘n’ oscillation sec Tact Actual time period sec Ttheo Theoretical time period sec w Weight attached on exciter assembly kg W Weight of exciter assembly along with wt. platform. kg 4. Observation & Calculation: Data: Li = K = 930 kg/m. W = 21 Kg. Observation Table: Sr. No W (kg) w (kg) Ll (m) L (m) n t (sec)  Theoretical Calculations: Ww m g  L2  me m  12   L   me Ttheo  2 K  1 ftheo  T theo Practical Calculations: t Tact  n 1 fact  Tact Result Table: Sr. No Tact. (sec) Ttheo. (sec) fact. (sec-1) ftheo. (sec-1) 5. Conclusion:  Experiment No.5 1. Aim: To study the Forced damped Vibration of Equivalent Spring Mass System. To find the frequency of forced damped equivalent spring mass system To find amplitude of vibration. 2. Description: It is similar to that described for exp. no. 6. The exciter unit is coupled to D.C. variable speed motor. RPM of motor can be varied with the speed control unit. Speed of rotation can known from the RPM indicator on control panel. It is necessary to connect the damper unit to the exciter. Amplitude of vibration can be recorded on strip chart recorder. Figure 7.1: - Forced damped Vibration of Equivalent Spring Mass System 3. Damping Arrangement:  Close the one hole of damper for light damping.  Close the two holes of damper for medium damping.  Close all the three holes of damper for heavy damping. 4. Experimental Procedure:  Arrange the set-up as described for exp. no. 6  Start the motor and allow the system to vibrate.  Wait for 1 to 2 minutes for amplitude to build the particular forcing frequency.   Adjust the position of strip chart recorder. Take the recorder of amplitude Vs. time on strip chart recorder by starting recorder motor. Press recorder platform on the pen gently. Pen should be wet with ink. Avoid excessive pressure to get good result.  Take record by changing forcing frequencies.  Repeat the experiment by adjusting the holes on the piston of damper can change different damping.  Plot the graph of amplitude vs. frequency for each damping conditions. 5. Nomenclature: fa Frequency. sec-1 K Stiffness of spring kg/m Li Length of spring. m N RPM of motor. RPM n Nos. of oscillations from graph. t Time for ‘n’ oscillations. sec 6. Observation & calculation: Data: Li = 0.126m K = 930 kg/m. Observation Table: Sr. No N (RPM) n t (sec) Calculations: t Tact  n 1 fact  Tact Result Table: Sr. No fa sec-1 Amplitude ( mm) 7. Conclusion: