Let us suppose that the simple mass/spring system shown in Fig. 10.1 is displaced by a small amount x0and suddenly released. The equation of the resulting motion in the absence of damping forces is
mx¨+kx=0 (10.1)
where k is the spring stiffness. We see from Eq. (10.1) that the mass, m, oscillates with simple harmonic motion given by
x=x0sin(ωt+ε) (10.2)
Fig. 10.1Oscillation of a mass/spring system.
Fig. 10.2Oscillation of annmass/spring system.
in whichω2=k/m andε is a phase angle. The frequency of the oscillation isω/2π cycles per second and its amplitude x0. Further, the periodic time of the motion, that is the time taken by one complete oscillation, is 2π/ω. Both the frequency and periodic time are seen to depend upon the basic physical characteristics of the system, namely the spring stiffness and the magnitude of the mass. Therefore, although the amplitude of the oscillation may be changed by altering the size of the initial disturbance, its frequency is fixed. This frequency is the normal or natural frequency of the system and the vertical simple harmonic motion of the mass is its normal mode of vibration.
Consider now the system of n masses connected by (n−1) springs, as shown in Fig. 10.2. If we specify that motion may only take place in the direction of the spring axes then the system has n degrees of freedom. It is therefore possible to set the system oscillating with simple harmonic motion in n different ways. In each of these n modes of vibration the masses oscillate in phase so that they all attain maximum amplitude at the same time and pass through their zero displacement positions at the same time. The set of amplitudes and the corresponding frequency take up different values in each of the n modes. Again these modes are termed normal or natural modes of vibration and the corresponding frequencies are called normal or natural frequencies.
The determination of normal modes and frequencies for a general spring/mass sys- tem involves the solution of a set of n simultaneous second-order differential equations of a type similar to Eq. (10.1). Associated with each solution are two arbitrary con- stants which determine the phase and amplitude of each mode of vibration. We can therefore relate the vibration of a system to a given set of initial conditions by assigning appropriate values to these constants.
A useful property of the normal modes of a system is their orthogonality, which is demonstrated by the provable fact that the product of the inertia forces in one mode and the displacements in another results in zero work done. In other words displace- ments in one mode cannot be produced by inertia forces in another. It follows that the normal modes are independent of one another so that the response of each mode to an externally applied force may be found without reference to the other modes. Therefore by considering the response of each mode in turn and adding the resulting motions we can find the response of the complete system to the applied loading. Another useful characteristic of normal modes is their ‘stationary property’. It can be shown that if an elastic system is forced to vibrate in a mode that is slightly different from a true normal mode the frequency is only very slightly different to the corresponding natural frequency of the system. Reasonably accurate estimates of natural frequencies may therefore be made from ‘guessed’ modes of displacement.
We shall proceed to illustrate the general method of solution by determining nor- mal modes and frequencies of some simple beam/mass systems. Two approaches are
10.1 Oscillation of mass/spring systems 329 possible: a stiffness or displacement method in which spring or elastic forces are expressed in terms of stiffness parameters such as k in Eq. (10.1); and a flexibility or force method in which elastic forces are expressed in terms of the flexibilityδof the elastic system. In the latter approachδis defined as the deflection due to unit force; the equation of motion of the spring/mass system of Fig. 10.1 then becomes
mx¨+x
δ =0 (10.3)
Again the solution takes the form x=x0sin(ωt+ε) but in this caseω2=1/mδ. Clearly by our definitions of k andδthe product kδ=1. In problems involving rotational oscil- lations m becomes the moment of inertia of the mass andδthe rotation or displacement produced by unit moment.
Let us consider a spring/mass system having a finite number, n, degrees of freedom. The term spring is used here in a general sense in that the n masses m1, m2,. . ., mi,. . ., mnmay be connected by any form of elastic weightless member.
Thus, if miis the mass at a point i where the displacement is xiandδijis the displacement at the point i due to a unit load at a point j (note from the reciprocal theoremδij=δji), the n equations of motion for the system are
m1xă1δ11+m2xă2δ12+ ã ã ã +mixăiδ1i+ ã ã ã +mnxănδ1n+x1=0 m1xă1δ21+m2xă2δ22+ ã ã ã +mixăiδ2i+ ã ã ã +mnxănδ2n+x2=0 . . . . m1xă1δi1+m2xă2δi2+ ã ã ã +mixăiδii+ ã ã ã +mnxănδin+xi =0 . . . . m1xă1δn1+m2xă2δn2+ ã ã ã +mixăiδni+ ã ã ã +mnxănδnn+xn=0
⎫⎪
⎪⎪
⎪⎪
⎬
⎪⎪
⎪⎪
⎪⎭
(10.4)
or
n j=1
mjx¨jδij+xi =0 (i=1, 2,. . ., n) (10.5) Since each normal mode of the system oscillates with simple harmonic motion, then the solution for the ith mode takes the form x=xi0sin(ωt+ε) so that
¨
xi= −ω2xi0sin(ωt+ε)= −ω2xi. Equation (10.5) may therefore be written as
−ω2 n j=1
mjδijxj+xi =0 (i=1, 2,. . ., n) (10.6) For a non-trivial solution, that is xi=0, the determinant of Eq. (10.6) must be zero.
Hence
(ω2m1δ11−1) ω2m2δ12 . . . ω2miδ1i . . . ω2mnδ1n ω2m1δ21 (ω2m2δ22−1) . . . ω2miδ2i . . . ω2mnδ2n . . . .
ω2m1δi1 ω2m2δi2 . . . (ω2miδii−1) . . . ω2mnδin
. . . . ω2m1δn1 ω2m2δn2 . . . ω2miδni . . . (ω2mnδnn−1)
=0
(10.7)
The solution of Eq. (10.7) gives the normal frequencies of vibration of the system. The corresponding modes may then be deduced as we shall see in the following examples.
Example 10.1
Determine the normal modes and frequencies of vibration of a weightless cantilever supporting masses m/3 and m at points 1 and 2 as shown in Fig. 10.3. The flexural rigidity of the cantilever is EI.
The equations of motion of the system are
(m/3)v¨1δ11+mv¨2δ12+v1=0 (ii) (m/3)v¨1δ21+mv¨2δ22+v2 =0 (iii) where v1 andv2 are the vertical displacements of the masses at any instant of time.
In this example, displacements are assumed to be caused by bending strains only; the flexibility coefficientsδ11,δ22andδ12(=δ21) may therefore be found by the unit load method described in Section 5.8. Then
δij=
L
MiMj
EI dz (iii)
where Miis the bending moment at any section z due to a unit load at the point i and Mj is the bending moment at any section z produced by a unit load at the point j. Therefore, from Fig. 10.3
M1=1(l−z) 0≤z≤l M2=1(l/2−z) 0≤z≤l/2 M2=0 l/2≤z≤l Hence
δ11= 1 EI
l
0
M12dz= 1 EI
l
0
(l−z)2dz (iv)
δ22= 1 EI
l
0
M22dz= 1 EI
l/2
0
l 2−z
2
dz (v)
δ12=δ21= 1 EI
l
0
M1M2dz= 1 EI
l/2
0
(l−z) l
2 −z dz (vi)
Fig. 10.3Mass/beam system for Example 10.1.
10.1 Oscillation of mass/spring systems 331
Integrating Eqs (iv), (v) and (vi) and substituting limits, we obtain δ11= l3
3EI δ22= l3
24EI δ12 =δ21= 5l3 48EI
Each mass describes simple harmonic motion in the normal modes of oscillation so thatv1=v01sin (ωt+ε) andv2=v02sin (ωt+ε). Hencev¨1= −ω2v1andv¨2= −ω2v2. Substituting for v¨1, v¨2, δ11, δ22 and δ12(=δ21) in Eqs (i) and (ii) and writing λ=ml3/(3×48EI), we obtain
(1−16λω2)v1−15λω2v2=0 (vii) 5λω2v1−(1−6λω2)v2=0 (viii) For a non-trivial solution
(1−16λω2) −15λω2 5λω2 −(1−6λω2)
=0
Expanding this determinant we have
−(1−16λω2)(1−6λω2)+75(λω2)2=0 or
21(λω2)2−22λω2+1=0 (ix)
Inspection of Eq. (ix) shows that
λω2=1/21 or 1 Hence
ω2= 3×48EI
21ml3 or 3×48EI ml3 The normal or natural frequencies of vibration are therefore
f1= ω1
2π = 2 π
3EI 7 ml3 f2= ω2
2π = 6 π
EI ml3
The system is therefore capable of vibrating at two distinct frequencies. To determine the normal mode corresponding to each frequency we first take the lower frequency f1 and substitute it in either Eq. (vii) or Eq. (viii). From Eq. (vii)
v1
v2 = 15λω2
1−16λω2 = 15×(1/21) 1−16×(1/21)
which is a positive quantity. Therefore, at the lowest natural frequency the cantilever oscillates in such a way that the displacement of both masses has the same sign at
Fig. 10.4The first natural mode of the mass/beam system of Fig. 10.3.
Fig. 10.5The second natural mode of the mass/beam system of Fig. 10.3.
the same instant of time. Such an oscillation would take the form shown in Fig. 10.4.
Substituting the second natural frequency in Eq. (vii) we have v1
v2 = 15λω2
1−16λω2 = 15 1−16
which is negative so that the masses have displacements of opposite sign at any instant of time as shown in Fig. 10.5.
Example 10.2
Find the lowest natural frequency of the weightless beam/mass system shown in Fig. 10.6. For the beam GJ=(2/3)EI.
The equations of motion are
mv¨1δ11+4mv¨2δ12+v1=0 (i) mv¨1δ21+4mv¨2δ22+v2=0 (ii) In this problem displacements are caused by bending and torsion so that
δij=
L
MiMj EI ds+
L
TiTj
GJ ds (iii)
10.1 Oscillation of mass/spring systems 333
Fig. 10.6Mass/beam system for Example 10.2.
From Fig. 10.6 we see that
M1=1x 0≤x≤l M1=1(2l−z) 0≤z≤2l M2=1(l−z) 0≤z≤l
M2=0 l ≤z≤2l 0≤x≤l T1 =1l 0≤z≤2l
T1 =0 0≤x≤l
T2 =0 0≤z≤2l 0≤x≤l Hence
δ11= l
0
x2 EI dx+
2l
0
(2l−z)2 EI dz+
2l
0
l2
GJ dz (iv)
δ22= l
0
(l−z)2
EI dz (v)
δ12=δ21= l
0
(2l−z)(l−z)
EI dz (vi)
from which we obtain
δ11= 6l3
EI δ22 = l3
3EI δ12=δ21= 5l3 6EI
Writingλ=ml3/6EI and solving Eqs (i) and (ii) in an identical manner to the solution of Eqs (i) and (ii) in Example 10.1 results in a quadratic inλω2, namely
188(λω2)2−44λω2+1=0 (vii)
Solving Eq. (vii) we obtain
λω2= 44±√
442−4×188×1 376
which gives
λω2=0.21 or 0.027
The lowest natural frequency therefore corresponds toλω2=0.027 and is 1
2π
0.162EI ml3 Example 10.3
Determine the natural frequencies of the system shown in Fig. 10.7 and sketch the normal modes. The flexural rigidity EI of the weightless beam is 1.44×106N m2, l=0.76 m, the radius of gyration r of the mass m is 0.152 m and its weight is 1435 N.
In this problem the mass possesses an inertia about its own centre of gravity (its radius of gyration is not zero) which means that in addition to translational displacements it will experience rotation. The equations of motion are therefore
mvδ¨ 11+mr2θδ¨ 12+v=0 (i) mvδ¨ 21+mr2θδ¨ 22+θ=0 (ii) where vis the vertical displacement of the mass at any instant of time andθ is the rotation of the mass from its stationary position. Although the beam supports just one mass it is subjected to two moment systems; M1at any section z due to the weight of the mass and a constant moment M2caused by the inertia couple of the mass as it rotates.
Then
M1=1z 0≤z≤l M1=1l 0≤y≤l M2=1 0≤z≤l M2=1 0≤y≤l Hence
δ11= l
0
z2 EI dz+
l
0
l2
EIdy (iii)
Fig. 10.7Mass/beam system for Example 10.3.
10.1 Oscillation of mass/spring systems 335 δ22=
l
0
dz EI +
l
0
dy
EI (iv)
δ12=δ21= l
0
z dz EI +
l
0
l
EI dy (v)
from which
δ11= 4l3
3EI δ22= 2l
EI δ12 =δ21= 3l2 2EI Each mode will oscillate with simple harmonic motion so that
v=v0sin (ωt+ε) θ=θ0sin (ωt+ε) and
¨
v= −ω2v θ¨ = −ω2θ Substituting in Eqs (i) and (ii) gives
1−ω2m4l3
3EI v−ω2mr23l2
2EIθ=0 (vi)
−ω2m3l2 2EIv+
1−ω2mr22l
EI θ=0 (vii)
Inserting the values of m, r, l and EI we have
1− 1435×4×0.763
9.81×3×1.44×106ω2 v−1435×0.1522×3×0.762
9.81×2×1.44×106 ω2θ=0 (viii)
− 1435×3×0.762
9.81×2×1.44×106ω2v+
1− 1435×0.1522×2×0.76
9.81×1.44×106 ω2 θ=0 (ix) or
(1−6×10−5ω2)v−0.203×10−5ω2θ=0 (x)
−8.8×10−5ω2v+(1−0.36×10−5ω2)θ=0 (xi) Solving Eqs (x) and (xi) as before gives
ω=122 or 1300 from which the natural frequencies are
f1= 61
π f2 = 650 π From Eq. (x)
v
θ = 0.203×10−5ω2 1−6×10−5ω2
Fig. 10.8The first two natural modes of vibration of the beam/mass system of Fig. 10.7.
which is positive at the lowest natural frequency, corresponding toω=122, and negative forω=1300. The modes of vibration are therefore as shown in Fig. 10.8.