STODOLA’METHOD

ωn2 = --- = [W1Y12 + W2Y22] [50 x 19212 + 75 x 6852] / (EI)2 Or ωn = 0.0811 √ (EI)

8.5.3. STODOLA’METHOD

Introduction:- This method is an iterative method and used for finding the fundamental (lowest) natural frequency of un-damped free vibrations of M-D-F systems

Procedure:- (i) Assume a reasonable deflection curve of the system. The static deflection curve itself can be assumed as in the case of Rayleigh’s method.

(ii) Using the above deflection curve, the inertia loading of the system is determined. This loading will be in terms of ωn2, where ωn is the fundamental natural frequency of the system.

(iii) Considering the system is loaded with the inertia loads, the new deflection curve is determined. This also will be in terms of ωn2.

If the assumed deflection curve of (i) above is similar to the calculated deflection curve of (iii), then the assumed shape of the deflection curve is correct and (iii) gives the value of ωn2. If the deflection curve of (i) and (iii) are not similar, then the calculated deflection curve of (iii) is used as the assumed deflection curve for the next iteration and the procedure is repeated till the assumed deflection curve and the calculated deflection curve are similar.

It can be shown that whatever deflection curve was assumed initially, we finally end up with the deflection curve corresponding to the fundamental mode.

Example 8.10 :- To determine the lowest natural frequency for the lateral vibrations of the beam shown in Fig. E8.10.

250 kg 150 kg 1.5 m 5.0 m 1.5 m

m

g = 250 x 9.81 N

Y

Y

Fig. E8.10: Schematic for example 8.10.

Fig. E8.10(a): Deflection curve when only m1g is acting on the beam.

a11 = Y11 / m1g = 3.96 / (EI) ; a21 = Y21 / m1g = 2.79 / (EI) Similarly when m2g alone acts at location 2 we have

a22 = Y22 / m2g = 0.188 / (EI) ; a12 = a21 by Maxwell’s theorem.

Trial 1 :- Assume Y1 = 1.0 and Y2 = 1.0

Inertia force at location 1 = F1 = m1ω2Y1 = 250 ω2.

Similarly F2 = 150 ω2. Therefore Y1’ = a11F1 + a12 F2

Or Y1’ = [3.96 x 250 + 2.79 x 150 ] ω2 / (EI) = 1408.5 ω2 / (EI) Similarly Y2’ = a21F1 + a22F2 = [2.79 x 250 + 0.188 x 150] ω2 / (EI) = 725.7 ω2 / (EI)

Therefore Y1’ : Y2’ = 1 : 0.515. This is different from the assumed deflection. Hence one more trial is required.

Trial 2 :- Assume Y1= 1.0 and Y2 = 0.515.

Then F1 = 250ω2 and F2 = 150 x 0.515 x ω2 = 77.25ω2

Therefore Y1’’ =[3.96 x 250 + 2.79 x 77.25] ω2 / (EI) = 1205.5 ω2 / (EI).

Similarly Y2’’ = [2.79 x 250 + 0.188 x 77.25] ω2 / (EI) = 712 ω2/ (EI) Therefore Y1’’: Y2’’ = 1 : 0.590.

Trail 3 :- Assume Y1 = 1.0 and Y2 = 0.590. Then the calculations as shown in trial 2 will give Y1’’’ = 1.0 and Y2’’’ = 0.577.

Trial 4 :- Assume Y1 = 1.0 and Y2 = 0.577. Then Y1”” = 1231.5 ω2/ (EI) and Y2”” = 713.8 ω2 / (EI).

Y1”” : Y2”” = 1.0 : 0.580 which is same as the assumed deflection within the acceptable accuracy. Hence iteration may be stopped.

Therefore 1231.5ω2 / (EI) = 1.0 or ω = 0.028 √ (EI).

Examples on Stodola’s Method

Example 8.11:- Determine the fundamental frequency for the system shown in Fig. E8.11 using Stodola’s method.

Fig. E8.11: Schematic for example 8.11.

^{4m 2m m}

3k k k

Solution : The displacement coefficients for the given system are determined as : a11 = 1/ 3k ; a2 = a31= a12= a13= 1/3k ; a22 = a32 = a23= 4/3k

Following the procedure as shown in trial 1 we get

X1” : X2” : X3” = 1 : 2.94 : 3.66 which is not the same as assumed profile.

Introduction :- In beam vibrations the natural frequencies of the second and higher modes are often considerably greater than that of the fundamental frequency. This fact will enable us to approximate the fundamental frequency with acceptable accuracy.

Dunkerly’sEquation:- In order to illustrate the Dunkerly’s method, let us consider the free vibrations of a three degree-of-freedom system. The governing equations in terms of flexibility influence coefficients can be written as

x1 = a11F1 + a12F2 + a13F3

x2 = a21F1 + a22F2 + a23F3 ………. 8.12 x3 = a31F1 + a32F2 + a33F3

For free vibrations the forces F1, F2 and F3 can be replaced with the .. .. ..

inertia forces − m1x1, - m2x2 and – m3x3.. If the system is vibrating

..

harmonically with frequency ω, then – mixi= miω2xi. Therefore equations 8.11 can be written as follows:

x1 = a11(m1ω2x1) + a12 (m2ω2x2) + a13(m3ω2x3)

x2 = a21(m1ω2x1) + a22(m2ω2x2) + a23(m3ω2x3) ……….8.12 x3 = a31(m1ω2x1) + a32(m2ω2x2) + a33(m3ω2x3)

Equations 8.12 can be written in matrix form as follows:

x1 a11m1 a12m2 a13m3 x1

Equations 8.12 are satisfied if the determinant of these equations vanishes:

(a11 – 1/ω2) (a12m2) (a13m3) (a21m1) (a22 – 1/ω2) (a23m3)

(a31m1) (a32m2) (a33 – 1/ω2)

Expanding the determinant we have the following frequency equation.

(1/ ω2)3 – [a11m1 + a22m2 + a33m3] (1/ω2)2

− [a12m2a21m1 + ……….](1/ω2) − [ ………] = 0 ………….8.14

We know from a theorem in algebra which states that if the coefficient of the highest term of the nth-degree equation is reduced to unity, the coefficient of the second highest term will be equal to the sum of the roots of the equation.

If the roots of Eq. 8.14 are 1/ω1, 1/ω2, and 1/ω3, the above equation can be factored into the following form:

Where the terms ω11, ω22, and ω33 are the natural frequencies of the system, with each mass acting separately in the absence of other masses.

Since ω2 and ω3 are natural frequencies corresponding to the higher modes and are larger than the fundamental frequency Eq. 8.16 can be approximated as

1 1 1 1

--- ≈ --- + --- + --- ...8.17 ω12 ω112 ω222 ω332

Eq.8.17 is called as the Dunkerly’s equation and has many useful applications as illustrated in the following examples.

Examples

Example 8.12:- Determine the fundamental frequency of a uniformly loaded cantilever beam with a concentrated mass M at the end equal to the mass of the uniform beam

Solution:-

Let ω11 be the natural frequency of the uniformly loaded beam by itself and ω22 be the natural frequency of the same beam when a mass M is acting at the end of the beam, neglecting the weight of the beam.

Example 8.12

For a uniformly distributed load of a cantilever beam we have ω112 = 3.515 2 [(EI) / (ML3)]

For the concentrated mass at the end of the mass less cantilever we have ω222 = 3.00 [(EI) / (ML3)]

The natural frequency of a given airplane wing -

In torsion is 1600 cpm. What will be the new natural frequency if a 500-kg bomb is hung at a position one-sixth of the semi span from the centre line of the airplane such that its moment of inertia about the torsional axis is 1800 N– cm – s2 ? The torsional stiffness of the wing at this point is 60 x 10 6 N – cm / rad.

L

M

M

Solution: Frequency of the bomb attached to the weightless wing is √ (60 x 106) f22 = (1 / 2π) √ (kt / J) = (1/2π)--- √ (1800)

Or f22 = 29.1 c.p.s. = 1745 c.p.m.

The new natural frequency with the bomb will be 1 1

1 / f12 = 1 / f112 + 1 / f222 = --- + 1600 2 1745 2 Or f1 = 1180 c.p.m.

Example 8.13:- The fundamental frequency of a uniform beam of mass M, simply supported is equal to π2 √ (EI / ML3). If a lumped mass m0 is attached to the beam at

x = L/ 3, determine the new fundamental frequency.

Solution:-By Dunkerly’s equation we have 1 / ω12 = 1 / ω112 + 1 / ω222.

The above equation can be written as follows:

1 / ω12 = 1 / ω112 + a22 m2 ...(1)

Where m2 is the mass of the concentrated weight or exciter and a22 is the influence coefficient of the structure at the point of attachment of the exciter. Multiplying Eq. (1) through by ω12 and rearranging we can write

1

(ω1 / ω11)2 = --- ………..(2) [ 1 + a22m2ω112]

a22 is the influence coefficient at x = L / 3 due to a unit load applied at the same point and from the knowledge of strength of materials

a22 = 8L3 / (6 x 81 EI)

m

x L