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Logarithmic decrement

ln(x/x)ln[ent/en(td)]lnend 12 nd

Since

2/(12)1/2

dn

then

[2 /(12)1/2 ]2 /(12)1/2

nn

If δ is obtained from measurement, ξ can be determined as

2 21/2 /[(2) ]

For ξ<<1

2 / 2

Note that δ=(1/(n-1))ln(x1/xn)

c=ξc =ξ2mω Cn

26,

| 30

Example 8.4: damping determination

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From given recorded vibration response, it is required to determine the

damping ratio.

26, | 31

Example 8.4

Solution From the plot,

x1 = 0.85, x2 = 0.45, x3 = 0.25, τd = 15.5-3.5 = 12 (s)

δ = ln(0.85/0.45) = 0.64

or δ = (1/2) ln(0.85/0.25) = 0.63

ξ = δ/[(2π)2 + δ2]1/2 = 0.64 /[(2π)2 + 0.642] 1/2= 0.101

Or ξ = δ/(2π) = 0.64/6.28 = 0.102

(x1 and x2)

(x1 and x3)

26,

| 32

Overdamped-Nonoscillatory motion

• If ξ>1 (c>cC), it is overdamped, and the roots are s1,2 = [- ξ ± (ξ2 -

1)1/2]ωn

• Solution is given as

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