Vibration lab manual 1
- 1. AKSUM UNIVERSITY COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING VIBRATION LAB MANUAL Compiled by A.Syed Bava Bakrudeen, Associate Professor, Mechanical Department, Aksum University, Ethiopia LIST OF EXPERIMENTS 1. FREE VIBRATION OF SPRING MASS SYSTEM 2. FORCED VIBRATION OF SPRING MASS SYSTEM 3. TORSIONAL VIBRATION 4. WHIRLING SPEED (SINGLE ROTOR SYSTEM) 5. WHIRLING SPEED (TWO ROTOR SYTEM) 1. FREE VIBRATION OF SPRING MASS SYSTEM 1
- 2. THEORY: The vibrations of the system executes under no damping condition is known as undamped vibration. When no external force is acts on a body after giving an initial displacement then the body is said to be under free or natural vibrations. The frequency of the free vibrations is called free or natural frequency and denoted by “fn”. Simple pendulum is an example of undamped free vibrations. AIM: To calculate the undamped natural frequency of a spring mass system subjected to free vibration. APPARATUS REQUIRED: Weight, Spring setup and Stop watch. FORMULA: 1. Stiffness k= (Load/Deflection) 2. fn(exp)=(1/T)Hz 3. fn(th)=(1/2π)(√(g/δ)) Hz PROCEDURE: 1. Fix the system at required position. 2. Add the required weight to the system and note down the deflection of the spring. 3. Make the spring to oscillate for 10 times. 4. Note the corresponding time taken for 10 oscillations and calculate the time period. 5. From the time period calculate the experimental natural frequency. GRAPH: 1. Load Vs Deflection 2.Load Vs fn(th) 3.Load Vs fn(exp) GRAPH OBSERVATION: When the load is increases deflection is also increases and fn(exp), fn(th) will be decreases. RESULT: The undamped natural frequency of a spring mass system is calculated and graph is drawn. TABULATION: 2
- 3. SI NO WEIGHT ADDED (Kg) DEFLECT ION δ (m) STIFF NESS K (kg/m) TIME FOR 10 OSCILL ATION (T) (Sec) EXPERI MENTAL TIME PERIOD (T) (Sec) EXPERI MENTAL NATURAL FREQUENCY (Hz) THEORE TICAL NATURAL FREQUEN CY (Hz) 1 4 100x10-3 40 20 2 0.5 1.576 MODEL CALCULATION: 1. Stiffness k= (Load/Deflection) = (4/100x10-3 )= 40 kg/m 2. fn(exp)=(1/T) = (1/3) = 0.5 Hz 3. fn(th) = (1/2π)(√(g/δ)) = (1/2π)(√(9.81/100x10-3 )) = 1.576 Hz 2. FROCED VIBRATION OF SPRING MASS SYSTEM THEORY: 3
- 4. The vibration of the system executes under damping system is known as damped vibrations. In general all the physical systems are associated with one or other type of damping. In certain cases amount of damping may be small in other case large. In damped vibrations there is a reduction in amplitude of over every cycle of vibration. This is due to fact that a certain amount of energy possessed by the vibrating system always dissipated in overcoming frictional resistances to the motion. The rate at which the amplitude of vibration decays depends upon the type and amount of damping in the system. Damped vibrations can be free vibrations or forced vibrations. Shock observer is an example of damped vibration. AIM: To calculate the undamped and damped frequency of a spring mass system subjected to forced vibration. APPARATUS REQUIRED: Spring system with damper, stopwatch and recording chart. PROCEDURE: 1. Connect the exciter to D.C motor. 2. Start the motor and allow the system to vibrate without damping. 3. Wait for 3 to 5 minutes for the amplitude to build for particular forcing frequency. 4. Adjust the position of strip chart recorder. Take the record of amplitude Vs time on the strip chart. 5. Take the record by changing the forcing frequency. 6. Repeat the graph for different damping. Damping can be changed adjusting the position of the exciter. 7. Plot the graph of amplitude Vs frequency for each damping condition. RESULT: 1. From the graph it can be observed that the amplitude of vibration decreases with time. 2. Amplitude of vibration is less with damped system as compared to undamped system. TABULATION-1: System with undamped vibration 4
- 5. SI No NUMBER OF OSCILLATI ONS (n) TIME FOR n OSCILL ATION (T) (Sec) EXPERI MENTAL TIME PERIOD (T) (Sec) EXPERI MENTAL NATURAL FREQUENCY (Hz) AMPLIT UDE (mm) 1 10 30 3 0.33 7 TABULATION-1: System with damped vibration SI No NUMBER OF OSCILLATI ONS (n) TIME FOR n OSCILL ATION (T) (Sec) EXPERI MENTAL TIME PERIOD (T) (Sec) EXPERI MENTAL NATURAL FREQUENCY (Hz) AMPLIT UDE (mm) 1 10 28 2.8 0.36 4.5 3. TORSIONAL VIBRATION THEORY: 5
- 6. When the particles of the shaft or disc move in a circle about the axis of the shaft, then the vibrations are known as torsional vibrations. The shaft twisted and untwisted alternatively and the torsional shear stresses are induced in the shaft, since there is no damping in the system these are undamped vibrations. There is no external force is acting on the body after giving an initial angular displacement then the body is said to be under free or natural vibration. Hence the given system is an undamped free torsional vibration. AIM: To determine the natural frequency of undamped torsional vibration of single rotor shaft system. APPARATUS REQUIRED: Rotor system with shaft, weight and stop watch. FORMULA: A. Experimental 1. Torsional stiffness (or ) Torsional rigidity Kt(exp) = (WxR)/Φ in N-m/ radian Φ – The angle twist in radian (1 radian = Φ0 x π/1800 ) 2. Periodic time, T (expt) = t/n (sec) 3. Frequency f (expt) = 1/T (Hz) B. Theoretical 1. Polar moment of inertia of shaft J = π d4 /32 in m4 . 2. Moment of inertia of disc I = (W/g) (D2 /8) in kg-m2 . Where W total weight of disc with mass added W= W1 + W2 W1 Weight of the disc in Newton. (If they don’t give take as zero) W2 Weight added in the disc in Newton. 3. Torsional stiffness Kt(theo) = ( G x J)/L in N-m 4. Periodic time, T (theo) = 2π x (√(I/Kt)) in Seconds 5. Frequency, f (theo) = (1/2π)x (√(Kt/I) in Hz PROCEDURE: 1. Fix the weight at convenient position and fix the one end of the shaft at the bracket by chuck. Twist the rotor through some angle and release. 2. Note down the time required for “n” oscillations. 3. Repeat the procedure for different length of the shafts. GRAPH: 1. Load Vs Deflection 2. Load Vs fn(th) 3. Load Vs fn(exp) TABULATION: 6
- 7. SI No Wei ght in N (W) Angle (Φ) Torsional Stiffness Kt (exp) in Nm/rad Time for “n= 10” oscillatio n (T) (Sec) Experim ental Time Period (Texp) (Sec) Experi mental Natural Freque ncy (fexp) (Hz) Moment of Inertia of the disc in kg-m2 . (I) Torsional Stiffness Kt (Theo) in Nm Theoreti cal Time Period (Texp) (Sec) Theoreti cal Natural Frequen cy (fexp) (Hz) Deg ree Rad 1 5 3.5 0.061 9.0164 20 2 0.5 0.0031 9.92 0.11 9 OBSERVATIONS: 1. Diameter of disc (D) = 220 mm, Radius of disc (R) =110mm 2. Length of shaft (L) = 1000 mm 3. Diameter of shaft (d) = 6mm 4. Modulus of rigidity of shaft (G) = 78000 N/mm2 (or) 7.8 x1010 N/m2 . MODEL CALCULATION: 1. Kt(exp) = (WxR)/Φ = (5x110x10-3 )/0.061 = 9.0164 Nm/rad 2. T (expt) = t/n = 20/10 = 2 sec 3. f (expt) = 1/T =1/2 =0.5 Hz. 4. J = π d4 /32 = π x (6x10-3 )4 /32 = 1.272 x10 -10 m4 . 5. I = (W/g) (D2 /8) = (5/9.81) x (0.222 /8) = 0.0031 kg-m2 . 6. Kt(theo) = ( G x J)/L = (7.8x1010 x 1.272 x10 -10 )/(1x1000x10-3 ) =9.92 N-m 7. T (theo) = 2π x (√(I/Kt)) = 2π x (√(0.0031/9.92))= 0.11 sec 8. f (theo) = (1/2π)x (√(Kt/I) =(1/ 2π) x (√(9.92/0.0031))= 9 Hz RESULT: 1. The experimental torsional stiffness is decreased with increased weight but the theoretical it is standard one. 2. The torsional stiffness = __________ Nm/rad (Exp) , __________ Nm (Theo). 3. The theoretical maximum natural frequency at ____ N is _____Hz. 4. The experimental maximum natural frequency at ____ N is _____Hz. 4. WHIRLING SPEED (SINGLE ROTOR SYSTEM) 7
- 8. THEORY: A certain speed, a rotating shaft or rotor has been found exhibit excessive lateral vibrations (transverse vibrations). The angular velocity of the shaft at which this occurs is called a critical speed or whirling speed or whipping speed. At a critical speed, the shaft deflection becomes excessive and may cause permanent deformation or structural damage. Therefore it is important to note that the machine should never be operated for any length of time at a speed close to a critical speed. AIM: To determine whirling speed of single rotor shaft theoretically and experimentally. APPARATUS REQUIRED: Whirling speed setup, stop watch and tachometer. FORMULAE: 1. Rigidity of the shaft calculated at centre position of the disc (c) = 48 EI/ L3 in N/m E= Young’s Modulus =210000 in N/mm2 . I = Moment of inertia = πd4 /64 in mm4 . d= Diameter of the shaft = 6mm. L= Distance between the length of the bearing in mm 2. Angular frequency (ω0) = (√(c/m)) in (1/s) where m= Mass of the disc = 1.0kg. 3. Theoretical critical speed Ncr(theo) = (30/π)x ω0 in rpm 4. Experimental critical speed Ncr(expt) = (Nu + No)/2 Nu= Under critical speed in rpm No= Over critical speed in rpm. PROCEDURE: 1. Take the apparatus and fix the rotor at the centre. 2. Fix the support bearing at required position and find the theoretical critical speed. 3. Rotate the shaft at under and over critical speeds and note down the reading, from that calculate the experimental critical speed. 4. Tabulate the value and plot the graph. GRAPH: 1. N cr (theo) Vs N cr (expt) RESULT: 1. The critical speed will decrease with increasing in distance between the bearing 2. The maximum critical speed at a distance L= _____________ mm Theoretical critical speed = _____________ rpm Experimental critical speed = _____________ rpm TABULATION: S. Distance Rigidity Natural Theoreti Under Over Experime 8
- 9. No between the bearing (L) in mm (c) in N/m Frequen cy ω0 in (1/s) cal critical speed Ncr(theo) in rpm critical speed Nu in rpm critical Speed No in rpm ntal critical speed Ncr(expt) 1 300 23,747 154.1 1471.55 1320 1360 1340 OBSERVATION: d= Diameter of the shaft = 6mm. E= Young’s Modulus =210000 N/mm2 . m – Mass of the disc = 1.0kg. MODEL CALCULATION: 1. Moment of inertia (I) = πd4 /64 = π64 /64 = 63.61mm4 . 2. Rigidity of the shaft (c) = 48 EI/ L3 = 48x210000x63.61/ (3003 ) = 23.747N/mm =23747 N/m. 3. Angular frequency (ω0) = (√(c/m)) = (√(23747/1)) = 154.1 in (1/s) 4. Ncr(theo) = (30/π)x(√(c/m)) = (30/π)x(√(23747/1)) =1471.55 rpm. 5. Ncr(expt) = (Nu + No)/2 = (1320 +1360)/2 =1340 rpm. Critical speed of sh 5. WHIRLING SPEED (TWO ROTOR SYSTEM) THEORY: 9
- 10. A certain speed, a rotating shaft or rotor has been found exhibit excessive lateral vibrations (transverse vibrations). The angular velocity of the shaft at which this occurs is called a critical speed or whirling speed or whipping speed. At a critical speed, the shaft deflection becomes excessive and may cause permanent deformation or structural damage. Therefore it is important to note that the machine should never be operated for any length of time at a speed close to a critical speed. In single rotor system it will produce only single critical speed, but in two rotor system it will produce two critical speeds. AIM: To determine whirling speed of two rotor shaft theoretically and experimentally. APPARATUS REQUIRED: Whirling speed setup, stop watch and tachometer. FORMULAE: 1. γ = (S/L) S= Distance between the discs in mm, L = Distance between the bearings in mm. 2. Rigidity of the shaft calculated at centre position of the disc (c) = 48 EI/ L3 in N/m E= Young’s Modulus =210000 in N/mm2 . I = Moment of inertia = πd4 /64 in mm4 . d= Diameter of the shaft = 6mm. 3. Coefficient of influence of two discs (a1,a2) a) a1 = α1/c where α1= 1-2γ2 +γ4 . b) a2 = α2/c where α2= 1-4γ2 +4γ3 -γ4 . 4. Natural frequencies of disc 1 and 2 a) ω1 = ω0 x (√(1/ (α1+ α2)) in (1/s) b) ω2 = ω0 x (√(1/ (α1- α2)) in (1/s) Where ω0 = (√(c/m)) 5. Theoretical critical speeds of disc 1 and 2 a) Ncr(theo)1 = (30/π)x ω1. b) Ncr(theo)2 = (30/π)x ω2. PROCEDURE: 1. Take the apparatus and fix the seaport bearings and disc rotors at the required position. 2. Calculate the theoretical critical speed. 3. Rotate the shaft at correct critical speeds and note down the readings. GRAPH: N cr (theo) Vs N cr (expt). TABULATION: The distance between the length of bearing (L) = ____470 mm______. The rigidity of the shaft (c) = _________6175 N/m_________. 10
- 11. S. No Distance between the Disc (S) in mm γ Coefficients Natural Frequencies (1/s) Theoretical critical speed Ncr(theo) in rpm Experimental critical speed Ncr(expt) α1 α2 ω1 ω2 Disc1 Disc2 Disc1 Disc2 1 117.5 0.25 0.878 0.808 60.52 297 578 2836 538 --- OBSERVATION: L= Distance between the length of the bearing = 470 mm d= Diameter of the shaft = 6mm. E= Young’s Modulus =210000 N/mm2 . m – Mass of the disc = 1.0kg. MODEL CALCULATION: 1. γ = (S/L) = (117.5/470) = 0.25 2. Moment of inertia (I) = πd4 /64 = π64 /64 = 63.61mm4 . 3. Rigidity of the shaft (c) = 48 EI/ L3 = 48x210000x63.61/ (4703 ) = 6.175N/mm =6175 N/m. 4. a. α1= 1-2γ2 +γ4 = 1- (2x0.252 )x(0.254 ) = 0.878 b. α2= 1-4γ2 +4γ3 -γ4 = α2= 1- (4x0.252 )+(4x0.253 )- (0.254 )= 0.808 5. ω0 = (√(c/m)) = (√(6175/1)) = 78.58 rpm 6. a. ω1 = ω0 x (√(1/ (α1+ α2)) = 78.58x (√(1/(0.878+ 0.808)) = 60.52 (1/s) b. ω2 = ω0 x (√(1/ (α1- α2)) = 78.58 x (√(1/(0.878- 0.808)) = 297 (1/s) 7. a. Ncr(theo)1 = (30/π)x ω1= (30/π)x 60.52 = 578 rpm b. Ncr(theo)2 = (30/π)x ω2= (30/π)x 297 = 2836 rpm 11
- 12. Critical Speed setup RESULT: 1. The maximum critical speed at a distance S= _____________ mm a. Theoretical critical speed of Disc1= _________ rpm, Disc2= _________ rpm. b. Experimental critical speed of Disc1 = _________ rpm, Disc2= _________ rpm. 12