Mechinical Vibration

-∫ π

m K Kδ

m

FORCED VIBRATIONS INTRODUCTION x When a mechanical system undergoes free vibrations, an initial force (causing some displacement) is impressed upon the system, and the system is allowed to vibrate under the mg influence of inherent elastic properties. The system however, comes to rest, depending upon the amount of damping in the system. In engineering situations, there are instances where in an external energy source causes vibrations continuously acting on the system. Then the system is said to undergo forced vibrations, as it vibrates due to the influence of external energy source. The external energy source may be an externally impressed force or displacement excitation impressed upon the system. The excitation may be periodic, impulsive or random in nature. Periodic excitations may be harmonic or non harmonic but periodic. The amplitude of vibrations remains almost constant. Machine tools, internal combustion engines, air compressors, etc are few examples that undergo forced vibration. 3.2 FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION Consider a spring mass damper system as shown in Figure 3.1 excited by a sinusoidal forcing function F=Fo Sinϖt

k

Cx∙

kx

C

F

x

m F = F0 Sin ωt F Fo

Figure 3.1

Let the force acts vertically upwards as shown in FBD. Then the Governing Differential Equation (GDE) can be written as m88 = - Kx8 - C x8 + F x

m 88 + C x8 + Kx = F --------------- (3.1) x is a linear non homogeneous II order differential equation whose solution is in two parts. 1. Complementary Function or Transient Response Consider the homogenous differential equation, m +x88C + Kx x8 = 0 which is incidentally the GDE of a single DOF spring mass dampersystem. It has been shown in earlier discussions that for different conditions of damping, the response decays with time. Thus the response is transient in nature and therefore termed as transient response. For an under dampedω tsystem the complementary function or transient response. n

xc = X1 ēζ ω t

xc = X1 ēζn

Sin (ωdt + φ ) (A Sin ωdt + B Cos ωdt)------------ (3.2)

2. Particular Integral or Steady State Response This response neither builds up nor decays with time. It is steady state harmonic oscillation having frequency equal to that of excitation. It can be determined as follows. Consider non-homogenous differential equation m 88 + Cx8 + Kx8 = Fo Sin ωt x

----------- (3.3)

The particular integral or steady state response is a steady state oscillation of the same frequency ω as that of external excitation and the displacement vector lags the force vector by some angle. Let x = X Sin (ωt - φ ) be the trial solution X: Amplitude of oscillation φ : Phase of the displacement with respect to the exciting force (angle by which the displacement vector lags the force vector). ∴ Velocity = x8 x8 = ωX. Cos (ωt - φ ) = ωX Sin [90 + (ωt - φ )] Acceleration 2 x88 = - ω X. Sin (ωt - φ ), substitute these values in GDE, (equation 3.1) We get -m ω2 X Sin (ωt - φ ) + Cω Sin [90 + (ωt - φ )] + KX Sin (ωt - φ ) = Fo Sin ωt 2 m ω X Sin (ωt - φ ) - Cω Sin [90 + (ωt - φ )] - KX Sin (ωt - φ ) + Fo Sin ωt = 0 ---------- (3.4) The four terms in the above equation represent both in magnitude and direction, the four forces namely: inertia force, damping force, spring force and impressed force, taken in

order, acting on the system and their sum is equal to zero. Thus they satisfy the D’Alemberts principle. Σ F = 0. Now, if vector representation as shown in Figure 3.2, is employed to denote these forces the force polygon shown in Figure 3.3 should close. Represent the force vectors and draw the force polygon as given below.

Figure 3.2

) (ωt - φ ) Reference axis

Figure 3.3 Impressed force: Fo Sin ωt: acts at an angle ωt from the reference axis. Displacement vector X: Lags the force vector by an angle φ and hence shown at (ωt - φ ) from the reference axis.

Spring force: - KX Sin (ωt - φ ): which means that the vector – KX acting at (ωt - φ ) or KX acting in opposite direction to (ωt - φ ) = at [90 + (ωt - φ )] Damping force: - CωX [Sin (90 + (ωt - φ )] - CωX acting at 90 + (ωt - φ ) or CωX acting in opposite direction to [90 + (ωt - φ )] 2 Inertia force: mω X Sin (ωt - φ ) Vector mω2X acting at (ωt - φ ) From the force polygon, in figure 3.3 Consider the triangle OAB. OA2 = OB2 + BA2 = (CωX)2 + (KX-mω2X)2 Fo2 = X2 (Cω)2 + x2 (K-mω2)2 Fo2 = X2 [(K-mω2)2 + (Cω)2] F0 X=

--------- (3.5)

√(K-mω2)2 + (cω)2

and tan φ = OB BA Cω tan φ = K-mω2

=

Cωx KX-mω2x

=

Cω K-mω2

∴ φ = tan-1

Cω K-mω2

-------- (3.6)

If X and φ are expressed in non-dimensional form it enables a concise graphical presentation of results. Therefore, divide the numerator and the denominator by K. F0 /K ∴X = √(1-

2

mω2 ) +( K

cω)2 Kk

tan φ =

Cω /K mω)2 (1K

Further, the above equations can be expressed in terms of the following quantities Fo K = Xst

- Zero frequency deflection

∴ Deflection of spring mass system under the steady force Fo should not be mistaken as Δst = mg m =K1 ωn2 K

C K

2ζ = Kωn

Thus X √(1 = -

Xst 2 ω ) + (2 ζ. ω) ωn2 ωn 2

2

ω ωn = r = frequency ratio Xst X= --------- (3.7) √ (1 - r2)2 +( 2 ζ. r)2 X whereXst is called magnification factor, amplification factor, or amplitude ratio. X M=Xst

:

It is the term by which Xst is to be multiplied to get the amplitude. 2ζ ωn

tanφ =

ω

( 1 -ωω ) 2

2ζr 2 1-r=

2

n

2ζr ... tanφ =1-r2

-------- (3.8)

Thus the steady state response xp = X Sin (ωt - φ ), in which X and φ are as given above. Total solution x = xc + xp For under damped conditions: as t ∞, xc 0 i.e., the transient response dies out. Complete solution consists only steady state response only. ∴ x = X Sin (ωt - φ ) --------- (3.9) As mentioned above, the transient vibrations die out very soon and hence the system vibrates ζ =The 0 behaviourζ =of0the system can be best understood by with steady response amplitudes. plotting frequency response curves as given below, in figure 3.4 and 3.5. Frequency Response Curves: Magnification Factor vs Frequency Ratio for Different amounts of Damping

Magn ificati on Facto rM= X/Xst

ζ = 0.25 ζ = 0.375 ζ = 0. 5 ζ = 0.707 ζ=1 ζ=2 Frequency Ratio r = (ω/ωn)

Figure 3.4 Phase lag vs frequency ratio for different amounts of damping.

Pha se An gle, φ ,

ζ = 0.25 ζ = 0.5 ζ = 0.707

ζ=0

ζ = 1.0 ζ = 2.0

0.5

1.0

1.5

2.0

2.5

3.0

Figure 3.5

Frequency Ratio r = (ω/ωn) The following characteristics of the magnification factor (M) can be observed. 1) For damped system (ζ =0); M as r 1. 2) Any amount of damping (ζ >0) reduces the magnification factor (M) for all values of forcing frequency. 3) For any specified value of r, a higher value of damping reduces the value of M 4) When the force is constant (r =0), M =1. 5) The amplitude of the forced vibrations becomes smaller with increasing value of forced frequency. i.e M 0,as r ∞. 6) For 0< ζ < 1/ √2 (0 < ζ 1/√2, the graphs of M decreases with increasing values of r. The following characteristics of the phase angle φ can be observed from the graph

1) For undamped system the phase angle is 00 for 01

x

Fo mω2X

Figure 3.8 At very large values of ω >φ approaches 1800, the inertia force becomes very large, where as the spring force and damping force vectors becomes negligibly small. The improved force is wholly utilized in balancing the inertia force. φ 1800 i.e., Fo = m ω2x X = Fo / m ω2 NUMERICAL EXAMPLES: 3.1) A machine part of mass 2.5 Kgs vibrates in a viscous medium. A harmonic exciting force of 30 N acts on the part and causes resonant amplitude of 14mm with a period of 0.22sec. Find the damping coefficient. If the frequency of the exciting force is changed to 4Hz, determine the increase in the amplitude of forced vibration upon removal of the damper. Data: m = 2.5Kg, F0 = 30N, X = 14mm, τ = 0.225sec Part 1: At Resonance ωn = forcing frequency = 2π / τ = 28.56 rad/sec At resonance: ω = ωn = 28.56 rad/sec ωn = √(K/m) = 28.56 rad/sec K = 2039 N/m Amplitude at resonance Fo/K X= √ [1 - r2] 2 + [2ζr] 2 As ω/ ωn = 1,

X = (F0/K)/2ζ = 0.014 ... ζ = 0.526

Damping coefficient = C = Cc ζ = 2m ωn ζ = 2*2.5*28.56*0.526 = 75.04 N/m/s C = 0.07504 Ns/m

Part (2): When f = 4 Hz Forcing ω = 2π *fn = 25.13 rad/sec Frequency ωn = 28.56 rad/sec, unchanged Amplitude of vibration with damper Xa = Fo/K √ [1 - r2] 2 + [2ζr] 2 = 0.01544m Amplitude of vibration without damper Xb = (30/2039)/(0.2258) = 0.0652m Increase in Amplitude = 0.0652 – 0.0155 = 0.0497m Amplitude = 49.7mm 3.2) A body having a mass of 15 kgs, is suspended from a spring which deflects 12mm due to the weight of the mass. Determine the frequency of free vibrations. What viscous damping force is needed to make the motion a periodic at a speed of 1mm/sec. If when, damped to this extent, a disturbing force having a maximum value of 100N and vibrating at 6Hz is made to act on the body. Determine the amplitude of ultimate motion. Solution: Data: m = 15Kg; F0 = 100 N; f = 6Hz; Δst = 12mm; (a) fn = (1/2π )√(g/ Δst) = 4.55Hz (b) The motion becomes aperiodic, when the damped frequency is zero or when it is critically damped (ζ = 1). ω = ωn = √(g/ Δ) = 28.59 rad/sec C = Cc = 2m ωn = 2*15*28.59 = 857 N/m/s = 0.857 N/mm/s Thus a force of 0.857 N is required at a rate of 1mm/s to make the motion a periodic. (c)

X=

F0

√(K-mω2)2 + (cω)2 ω = 2π f = 2π *6 = 37.7 rad/sec, f0 = 100 N fn = (1/2π )(√(K/m) ... K = 12,260 N/m X = 0.00298m = 2.98mm.

Condition for peak amplitude of vibration (Expression for peak amplitude) The frequency at which the maximum amplitude occurs can be obtained as follows. X Xst Xst 2 2 2 M= ∴ X = √ [1 - r ] + [2ζr] i.e., for a system acted upon by a known harmonic force, the amplitude depends only on (ω/ ωn). Hence for X to be maximum √ [1 – r2] 2 + [2ζr] 2 should be minimum. ∴

dx dx d(ω/ ωn) ∴ d(r)

([1 - r2] 2 + [2ζr] 2) = 0 2(1 - r2) 2 (-2r) + 4ζ2r = 0 2(1 - r2) + 4ζ2r = 0 42ζ2r = 0 = 2(1 - r2) 2ζ2 = 1-r2 r2 = 1 -2ζ2 r = √1 -2 ζ2 (ω/ ωn) peak = √1 -2 ζ2 ω ωn peak = √1 -2 ζ2

( (

)

ωp ωn

)

peak

= √1 -2 ζ2

---------- (3.10)

ωp = frequency at which peak amplitude occurs. Where ωp refers to the forcing frequency corresponding to the peak amplitude. No maximum or peak will1occur when the expression within the radical sign becomes negative i.e., for ζ > or √2 for ζ > 0.707.

(

ωp ωn

) = √1 -2 ζ

2

and peak amplitude is given by (X/Xst)max = 1/[2 ζ(√1- ζ)]

--------- (3.11)

3.3) A machine of mass 25 kgs, is placed on an elastic foundation. A sinusoidal force of magnitude 25N is applied to the machine. A frequency sweep reveals that the maximum

steady state amplitude of 1.3mm occurs when the period of response is 0.22sec. Determine the equivalent stiffness and damping ratio of the foundation. Solution: Data: F0 = 25N; m = 25 Kgs; Xmax = 1.3mm; τ = 0.22sec For a linear system, the frequency of response is same as frequency of excitation. ... Excitation frequency = ω = 2π f = 2π / τ = 28.6 rad/sec thus Xmax = occurs, when ω = 28.6 rad/s Condition for maximum amplitude to occur: r = √1 -2 ζ2 = ω/ωn ... ωn = ω /(√1 -2 ζ2 ) = 28.6/(√1 -2 ζ2 ) --------------(1) also we have, X/Xst =

for Xmax = r =√1 -2 ζ2

1 √ [1 - r2] 2 + [2ζr] 2

Xmax/Xst =

1 √ [1 – (1 -2 ζ2 )] 2 + [4ζ2(1 -2 ζ2 )]

=

1 2 ζ √ (1 -ζ2 )

Xmax/(F0/K) =

1 2 ζ √ (1 -ζ2 )

Xmaxmωn2/F0 =

1 2 ζ √ (1 -ζ2 )

25*0.013* ωn2/25 =

1 2 ζ √ (1 -ζ2 )

Now substitute for ωn2 from eq.(1);

0.013*28.6/(√1 -2 ζ2) =

1 2 ζ √ (1 -ζ2)

1.0633/(√1 -2 ζ2) =

1 2 ζ √ (1 -ζ2)

Squaring and rearranging, ζ4 - ζ2 +0.117 = 0 Z2 – Z + 0.117 = 0 where ζ2 = Z. Solving the quadratic equation ζ = 0.368, 0.93 The larger value of ζ is to be discarded because the amplitude would be maximum only for ζ < 0.707 ... take ζ = 0.368 ... natural frequency ωn =

ω √ (1 – 2(0.368)2 ) ωn = 33.5 rad/sec stiffness of the foundation, K = mωn2 = 25(33.5)2 = 28.05*103 N/m 3.4) A weight attached to a spring of stiffness 525 N/m has a viscous damping device. When the weight is displaced and released, without damper the period of vibration is found to be 1.8secs, and the ratio of consecutive amplitudes is 4.2 to 1.0. Determine the amplitude and phase when the force F=2Cos3t acts on the system. Solution: Data: K = 525 N/m; τ = 1.8secs: x1 = 4.2; x2 = 1.0; F = F0sinωt = 2cos3t ... F0 = 2N, ω = 3 rad/sec X= Fo/K √ [1 - r2] 2 + [2ζr] 2 ωn = 2π / τ = 3.49rad/sec δ = ln(4.2/1.0) = 1.435 ζ= δ = 0.22 2 2 √ (4π + δ ) r = ω/ωn = 2/3.49 = 0.573 r2 = 0.328 X=

2/525 √ [1 – 0.328] 2 + [4*0.484*0.328]

X = 5.3mm

φ = tan-1(2ζr) (1-r2) φ = tan-1(2*0.22*0.573) (1-0.328) -1 φ = tan (0.375) φ = 20.560 3.5) The damped natural frequency of a system as obtained from a free vibration test is 9.8 cps. During a forced vibration test with a harmonic excitation on the same system, the frequency of vibration corresponding to peak amplitude was found to be 9.6 cps. Determine the damping factor for the system and natural frequency. ωd = 9.8 cps, ωp = 9.6 cps.

(ωp / ωn) = √1 -2ζ2 ωn = ωd/√1 -2ζ2

∴ ωp√1 -2ζ2 /ωd = √1 -2ζ2 Solving for ζ: ζ = 0.196 ωn = ωd/ √1 -2ζ2 = 10 cps. 3.6) A reciprocating pump of mass 300 Kgs is mounted at the middle of a steel plate of thickness 12 mm and width 500 mm and length 2.5 m damped along two edges as shown. During the operation of the pump, the plate is subjected to a harmonic excitation of F(t) = 50 cos 60 t N. Determine the amplitude of vibration of the plate.

. 12 m = 300 Kgs F0 = 50 N ω = 60 K = 192EI/l3 = 176.94*103 N/m ζ=0

2.5 m

500

X = F0 /(K-m ω2)2 X = 6.13*10-8mm

Vibrations Due to Reciprocating and Rotating Masses Unbalance in rotating machine is one of the common causes of vibration. The centrifugal force generated (meω 2) due to the rotation of the body is proportional to the square of the frequency of rotation. This CF varies with speed of rotation and is different from the

harmonic excitation discussed in previous articles in which the maximum force is independent of frequency.

ω ω

Fig. 3.9 Model of Reciprocating Machine

Fig. 3.10 Model of Rotating Machine

Let:

m: mass of unbalanced mass. M: Total mass including unbalanced mass. e = eccentricity of unbalanced mass. = crank radius of reciprocating machine = stroke / 2. The force due to the unbalanced mass is as shown in the FBD. The GDE ∴ mx8 + Cx8 + Kx = meω 2 Sin ω t = F0 Sin ω t where F0 = meω 2 Let the steady state response be x = X. Sin (ω t - φ ) from the previous discussion we have X Xst

1 √[1-mr ] + [2 ζ r]2

=

2 2

where Xst = F0 / K, r = ω /ω here F0 = meω 2.

n

meω 2 /K ∴ X = √[1-mr2]2 + [2 ζ r]2 X=

meω M

MX me

2

M K √[1-mr2]2 + [2 ζ r]2

ω 2 M/K √[1-mr2]2 + [2 ζ r]2

ω2*1 ω (K/M) ω

2 2

n

= MX me

=

√[1-mr2]2 + [2 ζ r]2

r2 √[1-mr2]2 + [2 ζ r]2

--------- (3.12)

and 2ζr φ = tan-1 1-r2

-------- (3.13)

The variation of MX/me with (r = ω /ω n) for different values of ζ is shown in figure 3.11 However, the variation of φ and r remains as earlier.

ζ=0

M X/ me

ζ = 0.1 ζ = 0.15 ζ = 0.25 ζ = 0.5 ζ = 1.0 r=ω/ω

n

Figure 3.11

The following observations can be made. Case (i) when ω ω n, r>>1, r ≈ ∞. MX/me = 1 (independent of ζ, effect of damping is negligible) At low speed meω 2 is zero and hence the curve starts from zero. It increases with increase in (ω /ω n) until the condition of resonance is achieved. At resonance MX/me = 1/2 ζ and thus

the amplitude X is limited, by the damping present in the system. When ω /ω n is very large MX/me approaches unity. Numerical Examples Unbalanced Rotating and Reciprocating Masses and Force Transmissibility. 1) A reciprocating machine of mass 75 Kgs is mounted on springs of stiffness 11.76*10 5 N/m and a damper of damping factor 0.2. The slider of mass 2 Kgs within the machine has a reciprocating motion with a stroke of 0.08 m. The speed is 3000 rpm. Assuming the motion of the piston to be harmonic, determine 1. Amplitude of vibration of the machine. Solution: M = 75 Kgs; m = 2 Kgs, K = 11.76*105 N/m. For vibrations due to rotating unbalance Amplitude of vibration MX r2 = me √[1- r2]2 + [2 ζ r]2 e = stroke/2 = 0.08/2 = 0.04 m ω = 2π (3000) / 60 = 314 rad/sec. ω n = √K/m = √11.76*10.5 /75 = 125 rad/sec. ω /ω n = r = 314 /125 = 2.51 75 (X)/2(0.04) = (2.51)2 / √(1-2.52)2 + (2*0.2*2.51)2 X = 0.00125 m = 1.25 mm

Vibration Isolation and Transmissibility In machines vibrations are caused due to unbalanced masses. These vibrations are transmitted to the foundation upon which the machines are installed. If the transmission of vibrations to the foundations is not avoided the adjoining machines also set to vibrate. To minimize the forces transmitted to the foundation machines are usually mounted on springs or dampers or some other vibration isolation material. Vibration isolation is measured in terms of the motion or force transmitted to the foundation. The lesser the force or motion transmitted the greater the vibration isolation. Force Transmissibility or Transmissibility Ratio In the case of forced vibrations, it is defined as the rating force transmitted to that impressed upon the system. it is a measure of the effectiveness of a isolating material. For a spring mass damper system under harmonic excitation X=

Xst √[1- r2]2 + [2 ζ r]2

φ = tan-1

( 21-rζ r ) 2

--- (a) and --- (b)

The forces are transmitted to the foundation or structure through the springs and dampers provided in the system. Thus the force transmitted to the foundation are the spring force KX and the damping force cω x. Hence the total force transmitted to the foundation is the vector sum of KX and Cω X as shown in the Figure.

Impressed force

m Spring force (KX) Ft = Force transmitted

Damping force (CωX) Foundation

= √(KX)2 + (Cω X)2 = √(KX)2 + (Cω X)2. (KX)2 / (KX)2 = (KX) √1 + (Cω /K)2

Substituting for X from (a) F0 Ft = K. K √1(Cω /K)2 √[1- r2]2 + [2 ζ r]2

=

F0 √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2

Ft √ 1+ (2 ζ r)2 = F0 √[1- r2]2 + [2 ζ r]2

= Transmissibility Ratio (TR)

A plot of transmissibility ratio ω /ω n is shown in figure given below.

ζ = 0.2 ζ = 0.5 ζ = 0.6 ζ = 0.2

0

Ft/F

ζ=0

ζ = 0.6

ζ=0 Frequency Ratio r = (ω/ωn) The following observations can be made. Case (i) when ω /ω n = 0, r = 0, TR = 1, (independent of ζ ) (ii) ω = ω n, r = 1, resonance. √ 1+ 4 ζ 2 TR = , dependent on ζ 2ζ If ζ = 0, TR = ∞ Case (iii) when ω /ω n = √2, Ft/F0 = 1, independent of ζ Case (iv) when ω /ω n >>> , r

∞

Ft / F0 = TR = 0 Discussions: When ω /ω n = 0, i.e., the force is steadily applied, TR = 1, irrespective of the amount of damping produced in the systems. When ω /ω n = 1, it is condition of resonance. The force transmitted is infinity. If damping is used the magnitude of transmitted force is

reduced. When ω /ω n < √2 the transmitted is always greater then the impressed force. When ω /ω n = √2, for all the values of damping the force transmitted is equal to impressed force. When ω /ω n > √2, the transmitted force is always less than the impressed force. it also implies that TR decreases with decreasing values of ζ. Thus, an undamped spring is superior to a damped spring in reducing force transmissibility. But certain amount of damping is necessary for ω to pass through the resonance condition. As seen from the above order to isolate vibrations due to external force, ω /ω n should be very large, i.e., >√2. For a given value of ω /ω n should be very small. The static deflection of the spring should be as high as possible. These conditions will be satisfied by materials like steel springs, rubber, cork, felt etc., which are generally used as vibration isolators. Numerical Examples on Unbalanced Rotating and Reciprocating Masses and Force Transmissibility. 1. A reciprocating machine of mass 75 Kgs is mounted on springs of stiffness 11.76*105 N/m and a damper of damping factor 0.2. The slider of mass 2 Kgs within the machine has a reciprocating motion with a stroke of 0.08 m. The speed is 3000 rpm. Assuming the motion of the piston to be harmonic. 2. Amplitude of vibration of the machine. 3. Transmissibility ratio. 4. Force transmitted to the foundation. 5. Is vibration isolation achieved? If so how. Solution: M = 75 Kgs: m = 2 Kgs, K = 11.76*105 N/m. For vibrations due to rotating unbalance Amplitude of vibration MX r2 = 2 2 me √[1- r ] + [2 ζ r]2 e = stroke/2 = 0.08/2 = 0.04 m ω = 2π (3000) / 60 = 314 rad/sec. ω n = √K/m = √11.76*10.5 /75 = 125 rad/sec. ω /ω n = r = 314 /125 = 2.51 75 (X)/2(0.04) = (2.51)2 / √(1-2.52)2 + (2*0.2*2.51)2 X = 0.00125 m = 1.25 mm Transmissibility Ratio: (TR)

TR = TR = =

Ft F0

=

√ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2

√1 + (2*0.2*2.51)2 √(1-2.52)2 + (2*0.2*2.51)2

√ 1+ 1.0080 √28.09 + 2.008 =

√ 2.008 √29.09

TR = 0.1861 Force transmitted to the foundation Ft Also TR = F 0 Ft = (TR) F0

= (0.1861) * meω 2 = (0.1861) * 2 * 0.04 (314)2

Ft = 1467.9 N Vibration Isolation: Vibration isolation is achieved as only 18.6 % of the maximum shaking force (F0) is transmitted to the foundation. This is because the operating range of frequency ratio ω /ω n = r < √2 (2.51> 1.41). Ft As r >>>>> √2, 0 F0 2. A mass of 100 Kg, is mounted on a spring support having a spring stiffness of 20000 N/m and a damping coefficient of 100 NS/m. The mass is acted upon by a harmonic force of 39 N at the undamped natural frequency of the set up. Find 1. Amplitude of vibration of the mass. 2. Phase difference between the force and displacement. 3. Force transmissibility ratio. 3. A refrigerator of mass 35 Kgs operating at 480 rpm is supported on 3 springs. If only 10% of the shaking force is to be transmitted to the foundation what should be the value of K. √ 1+ (2 ζ r)2 Ft = √[1- r2]2 + [2 ζ r]2 F0 assuming that no damped used ζ = 0 TR = 1/√(1- r2) = 1/ ± (1- r2) Ft ω = 2π* 480 / 60 = 16 π rad/sec, TR = 0.1 = F 0 1 0.1 = ± [1- (16 π/ω n)2] ∴ ± 0.1 – 0.1 (16 π/ω n)2 = 1

When positive sign is considered - 0.1 (16 π/ω n)2 ± 1 – 0.1 (16 π/ω n)2 = 0.9/-0.1 = - 9 ∴ (16 π/ω n) = √-9

which is not possible

Taking the negative sign - 0.1 + 0.1 (16 π/ω n)2 = 1 16 π/ω n = (1+0.1/0.1) = √11 ∴ ω n = 15.15 rad/sec. ω n K = √K/m ∴ Keq = 8.037 N/m ∴ K = 8.037/3 = 2.679 N/m. 4. A machine supported symmetrically on four springs has a mass of 80 Kgs. The mass of the reciprocating mass is 2.2 Kgs which move through a vertical stroke of 100 mm with SHM. Neglecting damping, determine the combined stiffness of the springs so that the force transmitted to the foundation is 1/20th of the impressed force. the machine crank shaft rotates at 800 rpm. If, under actual working conditions, the damping reduces the amplitudes of successive vibrations by 30%, find (a) The force transmitted to the foundation at 800 rpm (b) The force transmitted to the foundation at resonance. (c) The amplitude of vibrations at resonance. M = 80 Kgs, m 2.2 Kg TR = 1/20 = 0.05 N = 800 rpm e = 100/2 = 50 mm. ω = 2 πN/60 = 83.78 rad/sec. 1. In the absence of damping TR = 1/ (r2 – 1) 1 0.05 = 83.782 ωn

(

)

- 1 ∴ ω n = 18.28 rad/sec

ω n2 = K/m K = Mω n2 = 26.739 N/m 2. When damping is present δ = In (X1/x2) = In (1/1-0.3) = 2 π ζ / √1 – ζ2

∴ ζ = 0.0567 Ft = Force transmitted to foundation at 800 rpm. Ft √ 1+ (2 ζ r)2 = F0 √[1- r2]2 + [2 ζ r]2

= 0.0563 = TR

Ft = F0 * TR = TR * meω 2 = 43.47 N 3. At resonance ω /ω n = 1 TR = Ft = TR F0

√ 1+ (2 ζ r)2 2ζ

= 8.875

∴ (Ft) Res = F0 * TR = meω n2 * TR = 2.2 * 0.05 * (18.78)2 * 8.875 (Ft)Res = 326.25 N 4. Amplitude of vibration at resonance. Force transmitted at Resonance = Stiffness = 32.6.25/26.739 = 12.2 mm Forced Vibrations due to Excitation of Base Some times the base or support of a spring-mass damper system undergoes harmonic excitation as shown in figure.

+x M

y = Y Sin ω t

+y Base Base excitation of the mass from Let y: denotes the displacement of the base and x: denotes the displacement static equilibrium position at a given instant t; such that Absolute Amplitude of the Mass (X) y = Y sin ω t

x

KX

. CX

x

y

Ky

. Cy

KX Ky

. CX . Cy

x

. .

K (x-y) C(x – y) Now from NSL mx88

= - K (x-y) – C(x8 - y8 ) = - Kx + Ky – Cx8 + y8

∴ m88 + Cx8 + Kx = Cy8 + Ky governing differential equation. x Let the steady state response x = X Sin (ω t - φ ) x8 = ω X cos (ω t - φ ) = ω X Sin [90 + (ω t - φ )] 88 = - ω 2 X Sin (ω t - φ ) x also

y = Y Sin ω t y8 = ω Y Cos ω t = ω Y Sin [90 + ω t] 2 8y8 = - ω Y Sin (ω t), substituting these values in GDE

- m ω 2X Sin (ω t - φ ) + Cω X Sin [90 + (ω t - φ )] + KX Sin (ω t - φ ) = Cω Y Sin [90 ω t] + K Y Sin ω t. m ω 2X Sin (ω t - φ ) - Cω X Sin [90 + (ω t - φ )] – KX Sin (ω t - φ ) + Cω Y Sin [90 ω t] + K Y Sin ω t = 0 Thus Σƒ = 0. The forces can be represented as shown, and the force polygon should close. From the triangle OAB.

KY CωY Y Mω2X A ωt CωY (ωt - φ ) φ X F0 KY (ωt - φ ) 0

KX-Cω2X

KX

B

KX CωX

mω2X

OA2

= AB2 + BO2

F02

= (KX - mω 2X)2 + (Cω X)2

X

Y

CωX

= [KX - mω 2X (KX/KX)]2 + [Cω X (KX/KX)]2 F02

= (KX)2 [(1- mω 2/K)2 + (Cω /K)2]

Also F02 = (KY)2 + (Cω Y)2 = (KY)2 [1 + Cω Y/ KY)2] = (KY)2 [1 + Cω / K)2] ∴ (KY)2 [1 + Cω / K)2] = (KX)2 [(1 - mω 2/ K)2 + (Cω / K)2] Taking square roots and rearranging the terms KY √[1 + Cω / K)2] = KX √[1- r2]2 + [2 ζ r]2 The ratio X/Y is called displacement transmissibility. This equation is similar to that of transmissibility ratio, and all the observation and discussions are same as that discussed under transmissibility ratio and the frequency response curve given below. Displacement transmissibility is defined are the ratio of displacement transmitted to the mass to the displacement impressed upon the base.

ζ=0

X/ Y

ζ = 0.2 ζ = 0.5 ζ = 0.6 ζ = 0.2

ζ = 0.5

ζ=0 r = (ω/ωn) Relative Amplitude: If Z represents the relative motion of the mass with respect to the support we have Z = x-y ∴ x = (z + y) Substituting this in the governing differential equation. m88+ Cx8 + Kx = Cy8 + Ky mx88 + CZ8 + KZ = - m8y8 Z and substituting for ‘y’ from y = Y Sin ω t m88 + CZ8 + KZ = m ω 2Y Sin ω t Z Comparing this with mx88+ Cx8 + Kx = meω 2 Sin ω t, of reciprocating and rotating unbalance. The steady state relative amplitude Z and the phase angle lag φ between the excitation and relative displacement. The relative motion frequency response which is similar to reciprocating and rotating unbalance as given below, can be used in designing vibration measuring instruments.

Z/Y

Z/ Y = r2 / √[1- r2]2 + [2 ζ r]2 φ = tan-1 [2 ζ r/ 1- r2]

ζ=0

ζ = 0.1 ζ = 0.15 ζ = 0.25 ζ = 0.5 ζ = 1.0 r=ω/ω

n

Force Transmitted: Force is transmitted to the base through the spring and dampers. If ‘Z’ represents the relative displacement then Force transmitted = Ft = √(KZ)2 + (Cω X)2 Ft = Z√K2 (Cω )2 The force transmitted to the base is also determined by Ft = m ω 2X Numerical Examples on Base Excitation 1. Figure (given in problem No.3) shows a simple model of motor vehicle that can vibrate in vertical direction while traveling over a rough road. The vehicle has a mass of 1200 Kg. The suspension system has a spring constant of 400 KN/m and a damping ratio of ζ = 0.5. If the vehicle speed is 100 Km/hr. Determine the displacement amplitude of the vehicle. The road surface varies sinusoidally with an amplitude Y = 0.05 m and a wave length of 6 m. Given m = 1200 Kg Speed 100 Km/hr K = 400 * 103N/m Y = 0.05 m ζ = 0.5 wave length = 6 m = period Model: Single degree freedom damped system base excitation. Frequency of base excitation: ƒ = Speed/ Length of one cycle (100*1000)/3600 m/sec = 6m ƒ = 4.63 Hz. ∴ Frequency of base excitation (100*1000) ω = 2πƒ = 2π 3600 * 6 m

= 29.0087 rad/sec.

Natural frequency = ω n = √ K/m = √400* 103 / 1200 = 18.2574 rad/sec. ∴ Frequency ratio = r = ω /ω n = 1.5903 √ 1+ (2 ζ r)2 ∴ X = √[1- r2]2 + [2 ζ r]2 = 0.8493 Y ∴ Displacement amplitude of the vehicle X/Y = 0.8493 ∴ X = 0.8493 * (0.05) X = 0.0425 m 2. A precession grinding machine is supported on an isolator that has a stiffness of 1 MN/m and a viscous damping constant of 1KN-S/m. The floor on which the machine is mounted is subjected to a harmonic disturbance due to the operation of an unbalanced engine in the vicinity of grinding machine. Find the maximum acceptable displacement of the floor if resulting amplitude of vibration of grinding wheel is to be restricted to 10 -6m. Assume that the grinding wheel and machine are rigid bodies of total weight 5000 N.

Grinding Wheel Grinding Machine

.

x = X sin(ωt - φ )

X= 10-6 m

y = Y sin10π t

Isolators

Y =? Solution: W = 5000 N K = 1*106 N/m X = 10-6m √ 1+ (2 ζ r)2 X Y = √[1- r2]2 + [2 ζ r]2 ω n = √K/m = √106/509.6

∴ m = w/g = 509.6 Kg W = mg;m = 5000/9.81 Kg C = 103 N-S/m Y=? y = Y Sin 10πt ∴ ω = 10 π ω = 31.4 rad/sec

ω n = 44.29 rad/sec

≈ 423 rpm

C = Cc ∴ ζ = C/Cc = 1032.mω n = 1032*509.6*44.29 Ζ = 0.0222 ω = 10 π = 31.4 rad/sec r = 31.4 / 44.29 = 0.7093 X √ 1+ [2(0.0222*0.7093)]2 ∴ Y = √(1- 0.70932)2 + (2 ζ r)2 = √1+(0.0314)2 = 1/ √(0.246 + 1.0) = 1.1166 X Y

= 1.1166 given that X = 10-6m

∴ Y = X/1.1166 = 10-6/1.1166 = 8.955*10-7m Y = 8.95*10-7 mm 3. A trailer has 1000 Kg mass when fully loaded and 250 Kg when empty. The suspension has a stiffness of 350 kN/m. The damping factor is 0.5. The speed of the trailer is 100 Km/hr. The road varies sinusoidally with a wave length of 5 m. Determine the amplitude ratio of the trailer: 1.When fully loaded. 2.When empty. Data: Mass of empty trailer = 250 Kg, ζ = 0.50 Mass of loaded trailer = 1000 Kg, k = 350 kN/m Speed of trailer = 100 Km/hr = 100*1000/3600 = 27.77 m/sec

x m One cycle

y = Y Sin ω t

C,K

y

100 km/hr

Y Road profile

Time period = τ = wave length /velocity = 5/27.77 sec. = 0.18 sec Forcing frequency, w = 2π/ τ = 2π / 0.18 = 34.896 rad/sec. 1. Empty trailer: Natural frequency of empty trailer wn = √(k/m) = √(350*103/250) = 37.416 rad/sec Frequency ratio, r = w/wn = 34.896/37.416 = 0.933 The ratio of amplitude of vibration of empty trailer to that of road surface is given as √ 1+ (2 ζ r)2 X = √[1- r2]2 + [2 ζ r]2 Y = 1.3676/0.9419 = 1.4518

2. Loaded trailer: When the trailer is fully loaded the natural frequency is given by wn = √(k/m) = √(350*103/1000) = 18.708 rad/sec Frequency ratio r = w/wn = 34.896/18.708 = 1.8653 √ 1+ (2 ζ r)2 X = √[1- r2]2 + [2 ζ r]2 Y = 2.116 / 3.1026 = 0.6819 [X/Y]empty = 1.4518 [X/Y]loaded = 0.6819 The amplitude of vibration reduces as the mass of the system (loaded trailer) is increased. 4. An aircraft radio of mass 20 Kgs is to be isolated from engine vibrations, which is vibrating with amplitude of 0.05 mm at 500 cpm. The radio is mounted on four isolators, each having a spring scale of 31400 N/m, and damping factor of 392 NS/m. a. What is the amplitude of vibration of the radio? b. What is the amplitude of vibration of the radio relative to the engine. c. What is the dynamic load on each isolator due to vibration. Part A m = 20 Kgs K = 4*31400 = 125600 N/m C = 4*392 = 1568 NS/m y = Y Sin ω t ∴ Y = 0.05 mm, ω = 2πƒ = 2π* 500/ 60 rad/sec ω = 52.5 rad/sec. ω n = √K/m = √125600/20 = 79.2 rad/sec. ω /ω n = 52.5 / 79.5 = 0.662 ζ = C/2 √Km = 0.496 X -5 Y = X = 0.069 mm = 6.9*10 m Part B X √ 1+ (2 ζ r)2 = Y √[1- r2]2 + [2 ζ r]2

Z = 0.025 mm = 2.5*105m Part C Ft = Z √K2+ (Cω )2 = 2.5*10-5 √(1568*52.4)2 + 1256002 Ft = 3.8 N ∴ Total force transmitted. ∴ Ft on each isolator = 3.8/4 = 0.95 N The dynamic load can also be computed using Ft = mω 2X = (20) (52.4)2*6.9*10-5 Ft = 3.8 N

Chapter 5 Vibration Measuring Instruments 5.1 Introduction In practice the measurement of vibrations becomes necessary due to following reasons. 1. To determine natural frequencies, modal shapes and damping ratios. The measurement of frequencies of vibration and forces developed is necessary to design active vibration isolation systems. 2. The theoretically computed vibration characteristics of a machine or structure may be different from the actual values due to the assumptions made in the analysis. (To verify the analytical models). 3. Periodic measurement of vibration characteristics of machines and structures becomes essential to ensure adequate safety margins. (Preventive maintenance). 4. Measurement of input and resulting output vibration characteristics of a system helps in identifying the system in terms of its mass, stiffness and damping. 5.2 Vibration Measurement Scheme Figure 5.1 shows the basic features of a vibration measurement scheme. 1. Vibrating machine or structure. 2. Vibration transducer or pick up. 3. Signal conversion instrument. 4. Display / recording. 5. Data analysis. 2

3

1

4

5

Figure 5.1 Vibration Measurement Scheme The motion of a vibrating body is converted in to an electrical signal by the vibration transducer or pick up. The transducer transforms changes in mechanical quantities such as displacement velocity, acceleration in to changes in electrical quantities such as voltage or current. (Electrodynamic pick up, electromagnetic pick up, piezo electric pick up, inductive displacement pick up, LVDT pick up, capacitive pick up). Since the output signal of a transducer is too small to be recorded directly, a signal conversion instrument is used to amplify the signal to the required value (Amplifier). The output from the signal conversion

instrument can be displayed on a display unit or stored in a computer for later use (Oscilloscope, A to D converters, milli voltmeters, computers etc., the data can then be analyzed to determine the desired vibration characteristics of the machine. Depending upon the quantity measured the vibration measuring instrument is called a vibrometer, a velocity meter, an accelerometer, a phase meter or a frequency meter. To summarise, following are the guidelines. 1. Displacement measurements may be useful for studying low frequency vibrations, where corresponding velocity and acceleration measurements are too small for practical purposes. 2. Velocity measurements may be useful at intermediate frequencies where displacement measurements are likely to be small to measure conveniently. 3. Acceleration measurements may be useful at high frequencies. Instead of the above, vibration analyzers can also be used. Several commercial vibration analyzers are available today. They consist of a vibration pick up and an FFT (Fast Fourier Transformation) analyser, a balancing kit for phase measurement and an inbuilt computer. The pick up essentially a piezo electric type with a natural frequency of 25 kcps. (KHz). Built in double integration is also available for displacement plots. FFT converts time domain signal to a signal in frequency domain to identify the frequencies of concern. 5.3 Vibration pick ups: Seismic Instruments The commonly used vibration pick ups are called seismic instruments. The basic element of many vibration measuring instrument is a seismic unit which is basically a spring massdamper system mounted on a vibrating body on which measurements are to be made as shown in Figure 5.2.

Casing

x C y

Figure 5.2Seismic Unit Depending on the frequency range utilized displacement, velocity or acceleration is indicated, by the relative motion of the suspended mass with respect to the case.

Behaviour of Seismic unit Consider the equation of motion of spring-mass-damper system, subjected to base excitation, as shown in Figure 5.3.

x = X Sin (ω t-φ )

M

y = Y Sin ω t

Base Figure 5.3 ..mx = - C .(x-y) . – K (x-y) if displacement the equation of motion becomes .. Z = x-y; . relative . mZ + CZ + KZ = mω 2Y Sin ω t from this r2 Z 2 2 = √[1- r ] + [2 ζ r]2 Y φ = tan-1 [2 ζ r/1-r2] The parameters that influence Z/Y and φ are: (1) frequency ratio r = ω /ω n. (2) Damping factor ζ, as shown in the Figure 5.4. Range for Vibrometer

Range for Accelerometer

ζ=0

Figure 5.4. Frequency response curves

ζ = 0.15 ζ = 0.25 ζ = 0.5

n

ζ = 1.0

r = ω / ω

Z/ Y

ζ = 0.1

Figure 5.4 also shows the range of frequencies corresponding to which a seismic instrument act as a vibrometer or an accelerometer. Type of instrument is determined by the useful range of frequencies with respect to the natural frequency (ω n) of the instrument. The relative displacement Z, may represent the displacement or acceleration depending upon ω n of the seismic unit and frequency of vibrating body, ω .

5.4 Vibrometer or Seismometer It is an instrument with low natural frequency. Therefore, ω >>>>> ω

n

r >>>> 1, r is very large. Z/Y ≈ 1, in particular when r > 3 Z/Y ≈ 1, (independent of ζ ) ∴ Z=Y Relative displacement of the seismic mass = displacement of base. ∴ Z = X-Y, X = 0, ∴ Z=Y Hence the seismic mass remains stationary. It remains undisturbed in space. The supporting casing moves the vibrating body. Thus the relative displacement between the casing and the mass is the true displacement of the casing. Like wise, the relative velocity between the casing and the mass is the true velocity of casing. Usually, the relative motion Z is converted into electric voltage. The seismic mass is a magnet moving relative to the coils fixed to the case, as shown in Figure 5.6.

x

0 0 0

0 0 0

Seismic mass

y

Figure 5.6 The voltage generated is proportional to the rate of cutting of magnetic field. Therefore the output of the instrument is proportional to the velocity of the vibrating body. Such instruements are called velometers. A typical instrument of this kind may have a natural frequency of 1 Hz to 5 Hz and a useful range of 10 Hz to 2000 Hz. The sensitivity of such instruments may be in the range of 20 mV/cm to 350 mV/cm. Both the displacement and acceleration are available from the velocity type transducer by means of the integrator or the differeniator provided in most signal conditioner units.

Limitation of Vibrometers In order to have r >>>1, ω n should be very small. This means that, the mass must be very large and the spring must have a very low stiffness. Therefore, a vibrometer is a spring-massdamper system with a very large mass and a flexible spring. This results in bulky instrument, which is not desirable in many applications. In practice, a vibrometer may not have a large value of r, and hence the value of Z, may not be exactly equal to Y. In such cases the true value of Y, can be computed from: r2 Z = 2 2 √[1- r ] + [2 ζ r]2 Y 5.5 Accelerometer It is an instrument with high natural frequency. When the natural frequency of the instrument is high compared to that of the vibrations to be measured, the instrument indicates acceleration. Then ω >> 1 900 < φ < 1800

G O

φ C

(c) 900 < φ < 1800 Disc rotates with heavy side inside. d. When φ = 1800

Irrespective of amount of damping, the point G approaches O. The system tends to be more stable and it is the desirable conditions.

G

φ C

O

(d) φ = 1800 Figure 5.19 shows the phase at different rotational speeds. Gφ C

O

G φ C O

(a) φ < 90

(b) φ = 90

O

G

φ C

(c) 900 < φ < 1800

O

φ G C

(d) φ = 1800

Figure 5.19 Numerical Example 5. A disc of mass 5 kg is mounted midway between two bearings which are 480 mm apart, on a horizontal steel shaft 9 mm in diameter. The CG of the disc is displaced by 3 mm from its geometric centre. Equivalent viscous damping at the centre of the disc is 48 Ns/m. If the shaft rotates at 675 rpm determine (a) the maximum stress in the shaft. (b) What is the power required to drive the shaft at this speed. (c) Also compare the maximum bending stress with the dead load stress in the shaft. Data: m = 5 kgs, e = 3 mm,

l = 480 mm, d = 9 mm, E = 200 GPa (assumed) C = 48 Ns/m, N = 675 rpm,

Part (a) ω n = √(g/δ st); δ

st

= Wl3/48EI = 1.79 mm = 1.79 * 10-3m

ω n = √[9.81/1.79*10-3] = 74.03 rad/sec ω = 2π N/60 = 70.686 rad/sec

r = ω /ω n = 0.955 K = mω n2 = 27402.2 N/m = 27.4 * 103 N/m C = 48 Ns/m ζ = C/2mω n = 0.0648 ∴ X/e = r2/√ [(1 –r2)2 + (2ζ r)2] X = 15.8 mm = 15.8 * 10-3 m Dynamic load Fd = √[(KX)2 + (Cω X)2] = 497.56 N Total bending load = FB = (5 * 9.81) + 497.56 = 546.6 N Maximum bending stress σ b = (32 M)/π d3, M = Wl/4 (bending moment, simply supported) W = Fb = 546.6 N σ b = (32*546.6*480)/π (4) (9)3 = 916.4 N/mm2 Part (b) T = Torque: Damping force * X = Damping torque = (Cω X) X = Cω X2 = (48*70.686*15.8*10-3) * 15.8*10-3 T = 1.102 Nm Power = 2π NT/60 N = 675 rpm P = 77.9 Watts Part (c) Bending stress due to dead load. σ b = (32 M)/π d3, M = Wl/4 (bending moment, simply supported) W = mg = 5 * 9.81 = 49.05 N

σ σ

= 82.24 N/mm2 bmax/σ b, dead load = 916.4/82.04 = 11.14 b

Session-XI (17.5.05) BKS

6.Vibrations of Two Degree of Freedom Systems

6.1 Introduction The modeling method discussed in previous chapters employed only one coordinate to describe the motion of the system completely. But general mechanical systems require several degrees of freedom for a meaningful model. Systems modeled with two independent co-ordinates to describe their motion are called two Degree of Freedom systems. There are two equations of motion for a two DOF system, one for each mass. They are generally in the form of coupled differential equations- i.e., each equation involves all the coordinates. If a harmonic solution is assumed for each co-ordinate, the equations of motion lead to a frequency equation that gives two natural frequencies for the system. If a suitable initial excitation is given the system vibrates at one of these natural frequencies. During free vibrations at one of the natural frequencies, the amplitude of two degrees of freedom (coordinates) are related in a specific manner and the configuration is called principal mode, or normal mode or natural mode of vibration. Thus a two DOF system has two normal modes of vibration corresponding two natural frequencies. 6.2 Free vibrations of two DOF system: Consider a two DOF system as shown in Figure 6.1, executing free vibrations. Let an initial displacement X1 be given to mass m1 and X2 to mass m2. Figure 6.2 shows the corresponding free body diagram.

K1 m1 K2

X1

m2 K3 Figure 6.1

X2

Let X2 > X1 K1 X1

K1 X1 m1

m1 X1

K2 X2

X1

K2 X1

K2 (X2 – X1)

K2 X2

K2 X1

K2 (X2 – X1)

m2

m2 K3 X2

X2 K3 X2

Figure 6.2 Based on Newton’s second law of motion ∑ƒ = mX For mass m1 ..

..

m1x1 = - K1x1 + K2 (x2-x1)

..

m1x1 + K1x1 – K2 x2 + K2x1 = 0

..

m1x1 + x1 (K1 + K2) = K2x2

----- (1)

for mass (2)

..

m2x2 = - K3x2 – K2 (x2 – x1)

.. ..

m2x2 + K3 x2 + K2 x2 – K2 x1 m2 x2 + x2 (K2 + K3) = K2x1

----- (2)

X2

Let us assume that under steady state conditions the solutions for x1 and x2 be harmonic therefore, assume x1 = X1 sin ω t, x2 = X2 sin ω t

..

x1= - ω 2X1 sin ω t,

..

x2 = - ω 2 X2 sin ω t

Substitute these in (1) and (2) - m1ω 2X1 sin ω t + (K1 + K2) X1 sin ω t = K2 X2 sin ω t - m2 ω 2X2 sin ω t + (K2 + K3) X2 sin ω t = K2 X1 sin ω t. Removing sin ω t through out and re arranging the terms. X1/X2 = K2/(K1 + K2 – m1ω 2) = [(K2 + K3) – m2ω 2]/K2 Cross multiplying K22 = (K1 + K2 – m1ω 2) (K2 + K3 – m2ω 2) On simplification we get m1m2 ω 4 – [m1 (K2 + K3) + m2 (K1 + K2)] ω + [K1K2 + K1K3 + K2K3] = 0

2

The above equation is quadratic in ω 2 and gives two values of ω 2 and therefore the two positive values of ω correspond to the two natural frequencies ω n1 and ω n2 of the system. The above equation is called frequency equation since the roots of the above equation give the natural frequencies of the system. Discussions: Let K1 = K3 = K m1 = m2 = m Then the frequency equation becomes m2ω 4 – 2 m (K + K2) ω 2 + (K2 + 2KK2) = 0 Let: ω 2 = Ω ∴Ω 2 = ω 4, m2 Ω 2 – 2 m (K + K2) Ω + (K2 + 2 KK2) = 0 ∴ m2 Ω 2 – 2 m (K + K2) Ω + (K2 + 2KK2) = 0 The roots of the above equation are as follows: Let a = m2, b = -2 m (K + K2); c = (K2 + 2KK2) ∴Ω 1,2 = [- b ± √(b2 – 4ac)]/2a = [- (-2m) (K + K2) ± √[-2m (K+K2)]2 – 4 (m2) (K2 + 2KK2)]/2m2

= [+ 2m (K +K2)]/2m2 ± [√4m2[(K2 + k22 + 2 KK2) – (K2 + 2KK2)]/4m4 = (K+ K2) /m ± √(K22/m2) = (K +K2) /m ± K2/m ∴Ω 2 = (K + 2K2) /m ω

2

n2

= (K + 2K2) /m ∴ ω

n2

= √[(K + 2K2) /m]

Ω 1 = (K + K2) /m – K2 / m = K/m ω

2

n1

∴ω

= K/m n1

= √(K/m)

ω n1 is called the first or fundamental frequency or I mode frequency, ω n2 is called the second or II mode frequency. Thus the number of natural frequencies of a system is equal to the number of degrees of freedom of system.

Session-XII (18.5.05) BKS Two DOF System (contd.) Modes Shapes: From X1/X2 = K2/(K+K2) -mω 2 = (K2 + K) - mω 2/K2 Substitute ω

n1

in any one of the equation.

(X1/X2)ω n1 = K2 / K+ K2 – m . K/m (X1/X2)ω n1 = 1 (X1/X2)ω n2 = K2 / K + K2 – m(K+ 2K2/m) = K2/-K2 = -1 (X1/X2)ω n2 = -1 The displacements X1 and X2 corresponding to the two natural frequency of the system can be plotted as shown in Figure 6.3, which describe the mode in which the masses vibrate. Such a diagram is called principal mode shape of the system. When the system vibrates in principal mode the masses oscillate in such a manner that they reach maximum displacements simultaneously and pass through their equilibrium points simultaneously or all moving parts of the system oscillate in phase with one frequency. Since the ratio X1/X2 is important rather than the amplitudes themselves, it is customary to assign a unit value of amplitude to either X1 or X2. When this is done, the principal mode is referred as normal mode of the system.

m2

m1

K1 X1

X2

m1 K2

I Mode m1

m2

.

X1 K3 Figure 6.3

Node

X2 m2

6.3 Discussion on Natural frequencies and mode shapes: Observation 1: It can be seen from the figure when the system vibrates in first mode, the amplitude of two masses remain same. The motion of both the masses are in phase i.e., the masses move up or down together. When the system vibrates in II mode the displacement of two masses have the same magnitude with opposite signs. Thus the motions of m1 and m2 are 1800 out of phase. Observation 2: When the system vibrates in first mode, the length of the middle spring remains constant, this spring (coupling spring) is neither stretched nor compressed. It moves bodily with both the masses and hence totally ineffective as shown in Figure 6.4. Even if the coupling spring is removed the two masses will vibrate as 2 SDOF system with ω n = √(K/m). Where as when it vibrates in II mode, the midpoint of the middle spring remains stationary for all the time. Such a point which experiences no vibratory motion is called a node, as shown in Figure 6.5. Observation 3: When the two masses are given equal initial displacements in the same direction and released, they will vibrate in I mode. When they are given equal initial displacements in opposite direction and released they will vibrate in II mode as shown in Figures 6.4 and 6.5

K1

K1 X1

m1

K1

m1 K2

X1

m1

K2 K2 X2

m2

m2

K3 K3

Figure 6.4

X2

m2 K3

K1

K1 X1

m1 K2

Nod e

m2

X2 K3

K1

m1

.

.

m1

K2

Nod e

m2

K2

m2 K3

K3

Figure 6.5 If unequal displacements are given in any direction, the motion will be superposition of two harmonic motions corresponding to the two natural frequencies.

Numerical Example 1. Obtain the frequency equation for the system shown in Figure. Also determine the natural frequencies and mode shapes when K1 = 2K, K2 = K, m1 = m, m2 = 2m. K1 X1

K1 X1

K1 m1 m1

X1 K2

m2

X1 K2 (X2 – X1)

K2 X2

K2 X1

K2 X2

X2

X1

m1

K2 (X2 – X1)

K2 X1 m2

X2

m2

X1

From NSL .. for mass (1) m1X1 = - K1X1 + K2 (X2 – X1) .. = - K1X1 + K2 X2 – K2X1 m1X1 + X1 (K1 + K2) = K2 X2 ----- (1) For .. mass (2) m2X2 = - K2 (X2 – X1) .. = - K2 X2 + K2 X1 m2 X2 + K2X2 = K2 X1

----- (2)

= A Sin ω t X1 = - ω 2 A Sin ω t,

X2 = B Sin ω t X2 = - ω 2 B Sin ω t

..Let X

1

Substitute these in (1) and (2) -m1ω 2 A Sin ω t + (K1 + K2) A Sin ω t = K2 B Sin ω t A (K1 + K2 – m1ω 2) = K2B A/B = K2 / [(K1 + K2 – m1ω 2)] ----- (3) - m2 ω 2B Sinω t + K2B Sin ω t = K2 A Sin ω t (K2 – m2ω 2) B = K2 A A/B = [K2 – m2ω 2] / K2 ----- (4) Equating (3) and (4) K2 / (K1 + K2 – m1ω 2) = [K2 – m2ω 2] /K2 K22 = (K1 + K2 – m1ω 2) (K2 – m2ω 2) K22 = (K1 + K2) K2 – m1ω 2 K2 – m2ω 2 (K1 + K2) + m1 m2ω m1 m2 ω 4 - ω 2 [m1 K2 + m2 (K1 + K2)] + K1 K2 = 0

4

Put ω 2 = Ω m1 m2 Ω2 – Ω [m1 K2 + m2 (K1 + K2)] + K1K2 = 0 Or Ω = [[m1 K2 + m2 (K1 + K2)] ± √ [{m1K2 + m2(K1+K2)}2]- 4 m1 m2K1K2]] / 2m1m2 Frequency equation of the system To determine the natural frequencies Given K1 = 2 K, K2 = K m1 = m, m2 = 2m

Ω = [mK + 2m (2K +K) ± √[mK + 6mK)2 – 4m 2mK2K]] / 2m . 2m = [7 mK ± √[(7mK)2 – 4 (4m2K2)]] / 4m2 = [7mK ± √(49m2K2 – 16m2K2] / 4m2 Ω = [7mK ± 5.744 mK] /4m2 Ω1 = ω ω

n1

n2

= [7 mK – 5.744 mK] /4m2 = 1.255 mK /4m2 = 0.3138 K/m

= 0.56 √(K/m)

Ω2 = ω ω

2 n1

2 n2

= [7mK + 5.744 mK] /4m2 = 3.186 K /m

= 1.784 √(K/m)

To determine the mode shapes: I mode shape: Substituting ω n12 = 0.3138 K/m A/B = [K2 – m2 ω 2] /K2 = [K2 –m2. ω

2

n1

]/K2

A/B = (K – 2m.0.3138 K/m)/K = 1-2(0.3138)] A/B = 0.3724 If A = 1, B = 2.6852 II mode: Substituting ω A/B = [K2 –m2ω

2

n2

= 3.186 K/m

2

n2

] /K2

= (K – 2m. 3.186 K/m) / K = (1 – 3.186 *2) = - 5.372 A/B = - 5.372, if A = 1, B = - 0.186

A =1

.

B = 2.6852

A =1

. . . B = -0.186

I Mode

2. Determine the natural frequency and the corresponding mode shapes for the system shown in figure X1

X2

K1

K2

K3

m1 •

m2 •

•

•

Given K1 = 3K, K2 = 2K, K3 = K m1= m, m2 = 2m Free body diagram

X1 K1X1

m1

X2 K2X1

K2X1 K2X2

m2

K3X2

K2X2

K2 (X2 –X1) K1X1 ω

n1

= √(K/m)

K2 (X2 –X1) ω

K3X2 n2

= √(5.5 K/m)

Session-XIII (20.5.05) BKS Two DOF systems (contd.) 3. Determine the Natural frequencies and ratio of amplitudes of the system shown in Figure. Solution similar to example No. 1 ω

n1

= 0.517 √(K/m)

(A/B)ω n1 = 0.731

ω

n2

= 1.931 √(5.5 K/m) K

(A/B)ω n2 = -0.2732 m

2K

2m

4. Same as above Given m1 = 1.5 kg

m2 = 0.8 kg K1

K1 = K2 = 40 N/m ω

n1

= 9.39rad/sec

(A/B)ω n1 = -0.765

ω

n2

= 3.88 rad/sec

m1

(A/B)ω n2 = 0.696 K2

m2

5. Determine the natural frequencies of the system shown in figure. Also determine the ratio of amplitudes and locate the nodes for each mode of vibration. Assume that the tension ‘T’ in the string remains unchanged, when the masses are displaced normal to the string.

l

l

l

m1

m2

m1 T T x1

T α

x1 - x2

β

m2 T x2 γ

Masses in displaced position

T cos α

x1

T

α T sin α

x1

m1

T cos β

β

T

T (x1 - x2) β T sin β T cos β

T sin β m2

x2 T cos γ

γ T sin γ

Free body diagram NSL. For mass (1)

T

..

mx1 = - T Sin α - T Sinβ

Sin α = x1/l, Sinβ = x2/l (x1 – x2/l

= - T x1 /l – T (x1-x2)/l = - Tx1/l – Tx1/l + Tx2/l

..

∴ mx1 + 2Tx1/l = Tx2/l

----- (a)

NSL. For mass (2)

..

mx2 = - T Sinγ + T Sinβ = - Tx2/l + T. (x1 –x2)/l

..

∴ mx2 + 2Tx2/l = Tx1/l Let

ω t, ..x1 = A Sin 2

x1 = - ω A Sin ω t

----- (b)

..

x2 = B Sin ω t x2 = -ω 2 B Sin ω t,

Substitute in (a) and (b) - m1ω 2 A Sin ω t + (2T/l) A Sin ω t = (T/l) B Sin ω t. Removing sin ω t throughout A [(2T/l) – m1ω 2)] = B. (T/l) ∴ A/B = (T/l)/ [(2T/l) – m1ω 2)]

----- (a1)

Similarly - m2 ω 2. B Sin ω t + (2TB Sin ω t)/l = (T. A Sin ω t)/l B [(2T/l) – m2ω 2)] = A. (T/l) ∴ A/B = [(2T/l) – m2ω 2)]/T/l

----- (a2)

Equating (a1) and (a2) and cross multiplying (T/l)/ [(2T-lm1ω 2)/l] = [(2T – lm2ω 2)/l]/(T/l) ∴ T2 = (2T – lm1 ω 2) (2T – lm2ω 2) T2 = 4T2 – 2Tlm1ω 2 – 2Tlm2ω 2 + l2m1m2ω

4

∴ l2m1m2 ω 4 – 2Tl (m1 + m2) ω 2 + 3T2 = 0 Let Ω = ω

2

∴ l2m1m2 Ω 2 – 2Tl (m1 + m2) Ω + 3T2 = 0

Frequency Equation

Ω 1, 2 = [2Tl (m1 + m2) ± √[{2T (m1 + m2)l}2 – 4 l2m1 m2 3T2)] / 2 m1 m2 l2 Let = m1 = m2 = m ∴Ω 1.2 = [2Tl (m + m) ± √[2T (2ml)2 – 4l2m2. 3T2)] / 2. m2. l2 = 4mTl ± √[(4mTl)2 – 12 m2 l2 T2] / 2m2l2 On further simplification Ω 1 = ω n12 = T/ml ∴ ω Ω 2 = ω n2 = 3T/ml ∴ ω

= √(T/ml) n2 = √(3T/ml) n1

Mode Shape: A/B = (T/l)/[(2T/l) – m1 ω 2] I mode: A/B = 1

if A = 1, B = +1

II mode: A/B = -1 A/B = -1

if A = 1, B = -1

(A/B)ω n1 = 1

m1

m2

X1

X2

I Mode m1

.

X1 Node

Semi Definite Systems or Degenerate System

II Mode

Eg: Coupled locomotive

(A/B)ω n1 = -1

m2

Systems for which one of the natural frequencies is equal to zero are called semi definite systems. X1

X2 K

m1 •

m2 •

•

•

FBD:

x2 > x1 X1

X2 KX2

KX2

m1

KX1

KX1

K (X2 –X1)

m1

K (X2 –X1)

Mass..(1) ∴ m1 x1 = K (x2 –x1)

..

m1 x1 + Kx1 = Kx2

----- (1)

..

m2 x2 = - K (x2 – x1)

..

m2 x2 + Kx2 = Kx1 Let

ω t, ..x1 = A Sin 2

x1 = - ω A Sin ω t

----- (2)

..

m2

x2 = B Sin ω t x2 = - ω 2 B Sin ω t,

m2

Substitute in (1) and (2) m1 (- ω 2 A Sin ω t) + K A Sin ω t = K B Sin ω t Further simplifications leads to A/B = (K)/ [K – m1 ω 2]

----- (3)

m2 (- ω 2 B Sin ω t) + K B Sin ω t = K A Sin ω t Further simplifications leads to A/B = [K – m2 ω 2] / (K)

----- (4)

Cross multiplying and simplifying further m1 m2 ω 4 – K (m1 + m2) ω 2 = 0

Frequency equation

ω 2 [m1 m2 ω 2 – K (m1 + m2)] = 0 Finding the roots we get the natural frequencies ω 1=ω

n1

=0

ω 2=ω

n2

= √[{K(m1 + m2)}/(m1 * m2)]

When one of the roots of the frequency equation is zero, one of the natural frequencies is zero. Such systems are referred as semi definite systems. The system will move as a rigid body without any distortion of spring. The amplitudes of two masses are equal. They are also referred as free-free system. Mode Shapes: I mode: (A/B)ω n1 = (K)/ [K – m1 ω 2] ω

n1

=0

m1 0

m2 0 A =1

B =1

I mode

(A/B)ω n1 = 1

II mode: (A/B)ω n2 = (K)/ [K – m1 ω 2]

0 m1 A=1 Node

. B=-1

II mode

0 m2

ω

n2

= √[{K(m1 + m2)}/(m1 * m2)]

if m1 = m2 =m Then (A/B)ω n2 = -1 6. Determine the natural frequency and mode shapes of the system shown in Figure. Given m1 = 10 kgs, m2 = 15 kgs, K = 320 N/m X1

X2 K

m1 •

m2 •

•

•

Solution: It is a free –free system Free body diagram

x2 > x1 X1 m1

X2 KX2

KX2 KX1

Frequencies ∴ω ω

n2

ω

n2

n1

m1

= √[K (m1 + m2)/(m1 * m2)] = √[{320(10 + 15)}/ (10*15)] = 7.30 rad/sec

Mode Shapes I mode

KX1

K (X2 –X1)

=0

m2

K (X2 –X1)

m2

(A1/A2)ω n1 = 1.0, if A = 1, B = 1 m1 0

m2 0 A =1

B =1

I mode II mode (A1/A2)ω n2 = (K)/[K – m1ω

2

n2

]

= 320 / [320 – 10 * (7.30)2] = - 1.49 if A = 1, B = -0.671 0 m1

.

A=1

Node

0 m2

B = - 0.671

II mode 7. An electric train made of two cars each of mass 2000 kgs is connected by couplings of stiffness equal to 40 * 106 N/m. Determine the natural frequency of the system.

m1

m2 K

Coupled Cars Solution: This is an example similar to problem No. 6 only the answer are given here. Given m1 =m2 = 2000 kgs. K = 40* 106 N/m

ω

n1

ω

n2

ω

n2

=0 = √(2K/m) = √(2*40*106) /2000 = 200 rad/sec

Analysis of Two DOF Torsional Systems

Figure above shows a two rotor system which can be represented as follows.

J1

Kt

Free body diagram is as given below.

J2

θ

θ

1

Kt θ

2

Kt θ

1

J1

J2 Kt θ

θ

Kt θ

2

1

θ

1

Kt (θ

Kt (θ

1

- θ 2)

- θ 2)

1

θ

θ

Kt (θ

1

θ

1

and θ

2

2

- θ 2)

1

Kt (θ

in CCW direction looking from left.

NSL for Rotor (1)

..

J.. 1 θ 1 = - Kt (θ 1 - θ 2) J1 θ 1 + Kt θ 1 = Kt θ For .. rotor (2) J.. 2 θ 2 = + Kt (θ 1 - θ 2) J2 θ 2 = + Kt θ 1 – Kt θ

..

J2 θ Let,

2

2

+ Kt θ

2

= Kt θ

2

----- (1)

2

1

----- (2)

1

- θ 2)

2

..θ

θ 2 = B sin ω t 1 = A sin ω t, .. θ 1 = - ω 2 A sin ω t, θ 2 = - Bω 2 sin ω t Substituting the above in 1 and 2 and simplifying we get the amplitude ratios and frequency equation as follows. A/B = Kt/[Kt – J1ω 2] ----- (a1) A/B = [Kt – J2ω 2] / Kt

----- (a2)

Frequency equation J1 J2 ω 4 – (J1 + J2) Kt ω 2 = 0 ω 2[J1J2ω 2 – (J1 + J2) Kt] = 0 ∴ω

2

= 0,

ω

n1

=0

and or J1 J2 ω 2 – (J1 + J2) Kt = 0 ω

n2

= [(J1 + J2) Kt] /J1 * J2

ω

n2

= √[{(J1 + J2) Kt}/ J1 *J2]

2

8. Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends. The masses of the discs are m 1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm. The length of the shaft is 3 m and its diameter = 10 cm. Modulus of rigidity for shaft material of the shaft G = 0.83 * 1011 N/m2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm. Solution: Part (1) For free body diagram and expression for frequencies refer previous discussion. m1 = 500 kg m2 = 1000 kg D1 = 1.25 m D2 = 1.9 m l = 3.00 m d = 0.10 m G = 0.83 * 1011 N/m2 Two rotor system is a semi definite system whose natural frequency is given by ω n1 = 0 ω n2 = √[{(J1 + J2) Kt}/ J1 *J2]

ω n = √[Kt(J1 +J2)/J1J2] J1 = ½ m1 R12 = 98 kg – m2,

J2 = ½ m2 R22 = 453 kgm2

Kt = GIp/l = [0.83 * 1011 / 3.00] *[π (d4)/ 32] = 2.725 * 105 N-m/ rad ω

n2

= 58.1 rad/sec

Part (2): Since the diameters are different along the length equivalent stiffness is to be determined as follows.

J1

Kt1

J2

Kt2

Kte J1

J2

Equivalent System Given, d1 = 10 cm, d2 = 20 cm, l1 = l2 = 1.5 m ∴ 1/Kte = 1/Kt1 + 1/Kt2 Kte = 5.13 * 105 N-m/rad ω

n2

= 79.597 rad/sec

∴ Hence there is 37% increase in the natural frequency of the system. 9. Determine the frequency equation, natural frequency and mode shapes for a double pendulum shown in figure.

Given m1 = m2

l1 = l2 = l

l1 m1 l2 m2

Free body diagram

θ

l1

1

T1

m1 T2 θ

x1

2

l2 T2

m2

m1g x2 m2g

T1 T1Cos θ θ

1

1

T2Sin θ T1Sin θ

θ

1

2

T2 θ 2 T2Sin θ

2

Applying NSL for mass (1)

..

but x1 =lθ

..

m1 lθ

1

..∴ x

1

1

1

..

+ T2 sin θ

At mass (1)

2

1

+ T2 sin θ

T1 cosθ

At mass (2) T2 cosθ θ

2

= lθ.. 1, .. x2..= lθ 1 + lθ x2 = lθ 1 + lθ 2

= - T1 sin θ

2

1

2

2

----- (a)

= mg + T2 cosθ

2

----- (1)

= mg

being very small cosθ = 1

T2 = mg

----- (2)

∴ T1 = mg + mg T1 = 2mg

..

2

2

m2g

Considering only the oscillation

m1 x1 = - T1 sin θ

T2Cos θ

T2

m1g T2Cos θ

2

----- (3)

∴ mlθ

..

lθ

1

1

= - 2 mg sin θ

+ 2gθ

1

- gθ

2

1

+ mg sin θ

=0

2

----- (b)

Similarly for mass (2)

..

= - T2 sin θ

mx2

= - mg sin θ

.. ml (θ

..

..

lθ

1

2

.. 1

T2 cos θ

2

= - mgθ

+ θ 2) + mg θ

+ lθ

2

+ gθ

2

2

2

= m2 g,

T2 = mg

2

=0

=0

----- (c)

Equations (b) and (c) represent GDE Let θ

..

θ

1

1

= A sin ω t

θ

2

= B sin ω t

θ

2

= - B ω 2 sin ω t

..

= - ω 2 A sin ω t,

Substitute in (b) and (c) - lω 2 A sin ω t + 2 g A sin ω t – g B sin ω t = 0 A (2g - lω 2) = Bg A/B = g/[2g-lω 2]

----- (b1)

-lω 2 A sin ω t - lω 2 B sin ω t = - g B sin ω l A/B = (lω 2 – g)/lω

2

A/B = [g - lω 2] / lω

2

----- (c1)

Equating b1 and c1 and cross multiplying we get frequency equation. l2ω 4 – 4glω 2 + 2g2 = 0 Let Ω =ω

2

∴ l2 Ω 2 – 4glΩ + 2g2 = 0 The roots are Ω 1 = 0.5857 g/l = ω

2

n1

frequency equation

ω

n1

= 0.7655 √(g/l)

Ω2=ω

2 n2

∴Ω 2 = ω

= 3.414 g/l n2

= 1.847 √(g/l)

Mode shapes I mode (A/B)ω n1 = g /[2g - lω

2 n1

] = 1/1.4143

A/B = 1/1.4143 ∴ A = 1, B = + 1.4143 II mode (A/B)ω n2 = g /[2g - lω

A = 1,

2 n2

] = 1/-1.4143

B = -1.414

Session- ()CSM –Ch-1

1.INTRODUCTION 1.1 The study of vibration A body is said to vibrate if it has periodic motion. Mechanical vibration is the study of oscillatory motions of bodies. Vibrations are harmful for engineering systems. Some times vibrations can be useful. For example, vibratory compactors are used for compacting concrete during construction work. Excessive vibration causes discomfort to human beings, damage to machines and buildings and wear of machine parts such as bearings and gears. The study of vibrations is important to aeronautical, mechanical and civil engineers. It is necessary for a design engineer to have a sound knowledge of vibrations. The object of the sixth semester course on mechanical vibrations is to discuss the basic concepts of vibration with their applications. The syllabus covers fundamentals of vibration, undamped and damped single degree of freedom systems, multidegrees of freedom systems and continuous systems. 1.2 Examples of vibration 1.Beating of heart 2. Lungs oscillate in the process of breathing 3. Walking- Oscillation of legs and hands 4. Shivering- Oscillation of body in extreme cold 5. Speaking - Ear receives Vibrations to transmit message to brain 6. Vibration of atoms 7. Mechanical Vibrations 1.3 Classification of vibrations One method of classifying mechanical vibrations is based on degrees of freedom. The number of degrees of freedom for a system is the number of kinematically independent variables necessary to completely descibe the motion of every particle in the system. Based on degrees of freedom, we can classify mechanical vibrations as follows: 1.Single Degree of freedom Systems 2.Two Degrees of freedom Systems 3.Multidegree of freedom Systems 4.Continuous Systems or systems with infinite degrees of freedom Another broad classification of vibrations is: 1. Free and forced vibrations 2. Damped and undamped vibrations. Sometime vibration problems are classified as: 1.Linear vibrations 2. Non-linear vibrations 3. Random vibrations

4.Transient vibrations A system is linear if its motion is governed by linear differential equations. A system is nonlinear if its motion is governed by nonlinear differential equations. If the excitation force is known at all times, the excitation is said to be deterministic. If the excitation force is unknown, but averages and standard derivations are known,the excitation is said to be random. In this case the resulting vibrations are also random. Some times systems are subjected to short duration nonperiodic forces. The resulting vibrations are called transient vibrations. One example of a nonperiodic short duration excitation is the ground motion in an earthquake The main causes of vibrations are: 1. Bad design 2. Unbalanced inertia forces 3. Poor quality of manufacture 4. Improper bearings (Due to wear & tear or bad quality) 5. Worn out gear teeth 6. External excitation applied on the system The effects of vibrations are as follows: 1. Unwanted noise 2. Early failure due to cyclical stress(fatigue failure) 3. Increased wear 4. Poor quality product 5. Difficult to sell a product 6. Vibrations in machine tools can lead to improper machining of parts 1.4 Basic terms associated with vibrations FREE VIBRATIONS Vibrations under free or natural conditions. No disturbing forces. Example: - Simple Pendulum

Fig 1.1 (a) Simple pendulum

FORCED VIBRATIONS Vibration due to impressed disturbing force Examples 1.Electric bell-clipper oscillation under electromagnetic force. 2.I.C Engines-vibrations due to unbalanced inertia forces

DEGREES OF FREEDOM

m1

m1 Single D.O.F Fig 1.1 (a)

m2

m1

m2

Two D.O.F

Fig 1.1(b)

m3 Three D.O.F Fig 1.1( c)

Cantilever Beam

Continuous system

Infinite Degrees of Freedom

Fig 1.1( d )

1.5 SIMPLE HARMONIC MOTION (S H M) The oscillations of the mass shown in fig 1.1 (a ) are described as simple harmonic motion. . Simple harmonic motion is represented graphically in fig 1.2

X-Displacement A-amplitude T-Periodic Time f-Frequency f=1/T ω =Frequency in radians per second t= time

X

A

x

t

ω A

ω ²A

t

t

X X= A sin ω t Fig 1.2 SHM Simple harmonic motion is characterized by periodic oscillation about the equilibrium position. Each oscillation is one cycle. For S.H.M the time taken to execute one cycle, the period, is constant. The frequency of motion is the number of cycles executed in a fixed period of time, usually 1 second. The amplitude, the maximum displacement from equilibrium position, is also constant in S.H.M.

. X=Velocity = A ω cos ω t =A ω sin ( ω t + π /2 )

.. X=Acceleration = - ω ² A sin ω t = ω ² A sin (ω t + π ) = -ω ² x

PROPERTIES OF OSCILLATORY MOTION Peak value- Indicates space requirement. An indication of maximum stress in the vibrating part Average Value - Average value for complete sine wave is zero For half sine wave X = 2A / π A-Amplitude Mean square value - For sine wave X² =1/2 A² RMS Value = A /√ 2 Problem 1 The frequency of Vibrations of a machine is 150 Hz. Determine a) Its frequency in rad/sec. b)Time Period of oscillations. If the amplitude of vibrations is 0.8 mm, determine the acceleration a) In m/s² b) In terms of g Solution: Given f = 150 Hz , A= 0.8 mm w= ? T=? a = ? (in m/s²) a = ? (in terms of g) w=2 π f = 2 π (150) = 942 Rad/sec T= 1/f=1/150= 0.0066 sec = 6.66 milli seconds x = A sin (ω t +Φ ) = 0.8 sin ( 942 t + Φ )

. x = 0.8 (942) cos (942 t +Φ ) ..

x = - 0.8 (942)2 Sin (942 t + Φ) a = (x ) max = 0.8 (942)2 mm/s2 = 710.61 m/s2 = 710.61/9.81= 72.43 g Problem 2.

A body suspended from a spring vibrates vertically up and down between two positions 3 and 5 cms above the ground. During each second it reaches the top position (5 cms above ground) 15 times. Find the time period, frequency, circular frequency and amplitude of motion. Solution:

Amplitude = (5-3)/2 =1 cm. f = Frequency =15 cps T = Period = 1/15 Sec

5 ω

3

3

ω = Circular Frequency = 2 π f = 2 π (15) =30 π rad/sec

1.6 Addition of harmonic motions of same frequency x1= A1 Sin ω t x2 = A2 Sin (ω t + Φ) X = x1+ x2 = A1 Sin ω t + A2 Sin (ω t + Φ) X = Sin ω t (A1+ A2 Cos Φ ) + Cos ω t (A2 Sin Φ) Let A1+ A2 Cos Φ = A Cos θ -------- 1 A2 Sin Φ = A Sin θ -------- 2 X = Sin ω t (A Cos θ ) + Cos ω t (A Sin θ ) X= A Sin (ω t + θ ) A2 (Sin2 θ + Cos2 θ ) = (A1 + A2 Cos Φ)2 + (A2 Sin Φ)2 A = √ A12 + A22 + 2 A1 A2 Cos Φ) From equations 1 and 2 we also get Tan θ = A2 Sin Φ/ (A1 + A2 Cos Φ)

Graphical Method for addition of two harmonic motions

A A2

φ

θ

A1

ω t

SUM OF HARMONIC MOTIONS

Same frequency but different phase angles

Sum of two harmonic motions of slightly different frequencies and same amplitude

Beats Is also a harmonic motion of the same frequency

Continuous build up and decrease in amplitude

1.7 BEATS The phenomenon of beats occurs when two harmonic motions of slightly different frequencies and same amplitude are added. When the two harmonic motions are in the same phase, the resultant amplitude will be maximum. On the other hand, when the two motions are out of phase, they will provide minimum amplitude vibration. X1 = A Sin ω 1t X2 = A Sin ω 2t X = X1+ X2 =A Sin ω 1t + A Sin ω 2t Let

= 2 A Sin (ω 1+ ω 2)t Cos (ω 1- ω 2)t X = Sin [(ω 1+ ω 2)t ]/2 When B= 2 A Cos [(ω 1- ω 2)t]/2 The Frequency of beats is (ω 1- ω 2)/2 π

Hz

Graphical representation of Beats

1.8 Fourier series analysis Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 1.3. Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series. The problem becomes a multifrequency excitation problem. The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term. Example :- Excitation force is periodic

Force 5000 N

0.2

0.5

0.7

1

1.2

Time in sec Force Developed during punching operation With the help of Fourier series vibration analysis of such problems can be done

Fourier Series ∞

X(t)= ao/2 +∑ (an cos nω t) + bn sin nω t) n=1 ω = 2 π / T = Fundamental frequency ao, a1,a2,……b1,b2….. are coefficients of infinite series (a1cos ω t+ b1sin ω t) is First Harmonic 2π /ω 2π /ω ao= ω / π x(t) dt , an= ω / π x (t) cos (nω t)dt o o

∫

bn= ω / π

∫

x(t) sin(nω t)dt

Problem 1.

Develop the Fourier Series for the curve shown in figure

The function is defined as y=x (t)

-π