Mechanical Vibrations Single Degree-Of-Freedom Damped Free Vibration (Chapter TREE)

CHAPTER THREE Free Vibrations Single Degree-of-Freedom Damped free vibration

Damping Force is Linear and Proportional to Velocity

Viscous damping force is expressed by the equation:

𝐹𝑑 = 𝑐𝑥̇ where 𝑐 is the viscous damping coefficient (N.sec/m or lbf.sec/ft)

At rest, 𝑥 = 0 (static equilibrium)

𝑚𝑔 = 𝑘𝛿

Lecture notes on Mechanical Vibrations website : www.abdulrahmanbahaddin.epu.edu.krd

3-1

Apply Newton’s 2nd Law

∑ 𝐹 = 𝑚𝑥̈ ∑ 𝐹𝑥↓+ = 𝑚𝑥̈ 𝑚𝑔 − (𝑘𝑥 + 𝑘𝛿 ) − 𝑐𝑥̇ = 𝑚𝑥̈ 𝑚𝑔 − 𝑘𝑥 − 𝑘𝛿 − 𝑐𝑥̇ = 𝑚𝑥̈

Equation of motion:    

(𝑚𝑔 = 𝑘𝛿) (static equilibrium)

𝑚𝑥̈ + 𝑐𝑥̇ + 𝑘𝑥 = 0

2nd order Differential equation Homogeneous Linear Constant coefficients

Form of Solution:

𝑥 (𝑡) = 𝑋𝑠𝑖𝑛(𝜔𝑡 + ∅)


𝑜𝑟

𝑥 (𝑡) = 𝐶𝑒 𝑠𝑡

3-2

EOM of viscously damped free vibration system:

𝑚𝑥̈ + 𝑐𝑥̇ + 𝑘𝑥 = 0 Assume 𝑥 (𝑡) = 𝐶𝑒 𝑠𝑡 𝑥̇ (𝑡) = 𝑠𝐶𝑒 𝑠𝑡 𝑥̈ (𝑡) = 𝑠 2 𝐶𝑒 𝑠𝑡

𝑚𝑠2 𝐶𝑒𝑠𝑡 + 𝑐𝑠𝐶𝑒𝑠𝑡 + 𝑘𝐶𝑒𝑠𝑡 = 0 (𝑚𝑠2 + 𝑐𝑠 + 𝑘)𝐶𝑒𝑠𝑡 = 0 For non –trivial solution

𝑚𝑠2 + 𝑐𝑠 + 𝑘 = 0

𝑜𝑟

𝐶𝑒𝑠𝑡 = 0

But 𝐶𝑒 𝑠𝑡 ≠ 0

∴

𝑚𝑠2 + 𝑐𝑠 + 𝑘 = 0 𝑠1,2 =

Roots of the equation:

−𝑐±√𝑐 2 −4𝑚𝑘 2𝑚

𝑐 2 − 4𝑚𝑘 > 0

→

𝑠1 , 𝑠2 (𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠)

𝑐 2 − 4𝑚𝑘 = 0

→

𝑟 = −𝑐/2𝑚

𝑐 2 − 4𝑚𝑘 < 0

→

𝑠1 , 𝑠2 (𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑛𝑢𝑚𝑏𝑒𝑟𝑠)


3-3

𝑠1,2

−𝑐 √𝑐 2 − 4𝑚𝑘 = ± 2𝑚 2𝑚

𝑥 (𝑡) = 𝐶1 𝑒 𝑠1 𝑡 + 𝐶2 𝑒 𝑠2 𝑡 𝐼𝑓 𝑠1 , 𝑠2 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑒𝑞𝑢𝑎𝑙 𝐶1 𝑎𝑛𝑑 𝐶2 𝑎𝑟𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠  𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑎 𝑐𝑎𝑠𝑒 𝑤ℎ𝑒𝑛 ,

𝑐2 − 4𝑚𝑘 = 0

Solving for c:

𝑐 = 2√𝑘𝑚 = 𝐶𝑐𝑟 𝐶𝑐𝑟 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑑𝑎𝑚𝑝𝑖𝑛𝑔

𝐶𝑐𝑟 = 2√𝑘𝑚 𝐶𝑐𝑟 = 2𝑚𝜔𝑛

𝜔𝑛 = √

𝑘 𝑚 𝜔𝑛 : 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦

𝑐

𝜁=𝐶

𝑐𝑟

𝜁 : 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 𝑟𝑎𝑡𝑖𝑜


3-4

𝑠1,2

𝜔𝑛 = √

−𝑐 √𝑐 2 − 4𝑚𝑘 = ± 2𝑚 2𝑚

𝑘 𝑚

𝜔𝑛 . √𝑚 = √𝑘

𝐶𝑐𝑟 = 2√𝑘𝑚 = 2√𝑘. √𝑚 = 2𝜔𝑛 . √𝑚. √𝑚 𝐶𝑐𝑟 = 2𝜔𝑛 . 𝑚 2𝑚 =

𝜁=

𝑠1,2

𝑠1,2

𝐶𝑐𝑟 𝜔𝑛

𝑐 𝑐 𝑐 = = 𝐶𝑐𝑟 2𝑚𝜔𝑛 2√𝑘𝑚

√𝑐 2 − 4𝑚𝑘 = ± 𝐶𝑐𝑟⁄ 𝐶𝑐𝑟⁄ 𝜔𝑛 𝜔𝑛 −𝑐

√𝑐 2 − 4𝑚𝑘 −𝑐 = 𝜔 ± 𝜔𝑛 𝐶𝑐𝑟 𝑛 𝐶𝑐𝑟 −𝑐

𝑐2

4𝑚𝑘

𝑐𝑟

𝑐𝑟

𝑐𝑟

𝑠1,2 = 𝐶 𝜔𝑛 ± √𝐶


2 − 𝐶

2

𝜔𝑛

(𝐶𝑐𝑟 = 2√𝑘𝑚)

3-5

𝑠1,2

−𝑐 𝑐2 4𝑚𝑘 = 𝜔𝑛 ± √ 2 − 𝜔 𝐶𝑐𝑟 4𝑚𝑘 𝑛 𝐶𝑐𝑟

𝑐2

−𝑐

𝑠1,2 = 𝐶 𝜔𝑛 ± √𝐶 𝑐𝑟

𝑐𝑟

2

− 1 𝜔𝑛

(𝜁 =

𝑐 𝐶𝑐𝑟

)

𝑠1,2 = −𝜁𝜔𝑛 ± √𝜁 2 − 1 𝜔𝑛

𝜁=1

𝜁<1

𝜔𝑑 = √𝜁 2 − 1 𝜔𝑛

𝜁>1

𝜔𝑑 : 𝑑𝑎𝑚𝑝𝑒𝑑 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑐𝑛𝑦

With these definitions , EOM becomes

𝑚𝑥̈ + 𝑐𝑥̇ + 𝑘𝑥 = 0

𝑥̈ + 2𝜁𝜔𝑛 𝑥̇ + 𝜔𝑛 2 𝑥 = 0

⇛

𝜁2 − 1 < 0

⇛

0<𝜁<1

⇛

𝑈𝑛𝑑𝑒𝑟𝑑𝑎𝑚𝑝𝑒𝑑 𝑚𝑜𝑡𝑖𝑜𝑛

𝜁2 − 1 > 0

⇛

𝜁>1

⇛

𝑂𝑣𝑒𝑟𝑑𝑎𝑚𝑝𝑒𝑑 𝑚𝑜𝑡𝑖𝑜𝑛

𝜁2 − 1 = 0

⇛

𝜁= 1

⇛

𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑎𝑚𝑝𝑒𝑑 𝑚𝑜𝑡𝑖𝑜𝑛


3-6

Case (1) : Underdamped motion (complex conjugate roots) 𝜁2 − 1 < 0

⇛

0<𝜁<1

𝑠1,2 = −𝜁𝜔𝑛 ± √𝜁 2 − 1 𝜔𝑛 𝑠1,2 = −𝜁𝜔𝑛 ± √−1 √1 − 𝜁 2 𝜔𝑛

(𝑗 = √−1 )

𝑠1,2 = −𝜁𝜔𝑛 ± 𝑗 √1 − 𝜁 2 𝜔𝑛

𝜔𝑑 = √𝜁 2 − 1 𝜔𝑛

𝜔𝑑 : 𝑑𝑎𝑚𝑝𝑒𝑑 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑐𝑛𝑦

𝑠1,2 = −𝜁𝜔𝑛 ± 𝑗 𝜔𝑑 Solution of diff. equation:

𝑥 (𝑡) = 𝐶1 𝑒 𝑠1 𝑡 + 𝐶2 𝑒 𝑠2 𝑡 𝑥(𝑡) = 𝐶1 𝑒 (−𝜁𝜔𝑛 +𝑗 𝜔𝑑 )𝑡 + 𝐶2 𝑒 (−𝜁𝜔𝑛 −𝑗 𝜔𝑑)𝑡

This simplifies to :

𝑥 (𝑡) = 𝑒 −𝜁𝜔𝑛 𝑡 (𝐴 cos 𝜔𝑑 𝑡 + 𝐵 sin 𝜔𝑑 𝑡) Where A & B are arbitrary constants to be found from initial conditions Lecture notes on Mechanical Vibrations website : www.abdulrahmanbahaddin.epu.edu.krd

3-7

 𝑥(𝑡) = 𝑒 −𝜁𝜔𝑛𝑡 (𝐴 cos 𝜔𝑑 𝑡 + 𝐵 sin 𝜔𝑑 𝑡) 𝐴 = 𝑋𝑠𝑖𝑛𝜙 , 𝐵 = 𝑋𝑐𝑜𝑠𝜙 𝑥(𝑡) = 𝑒 −𝜁𝜔𝑛𝑡 (𝑋𝑠𝑖𝑛𝜙 cos 𝜔𝑑 𝑡 + 𝑋𝑐𝑜𝑠𝜙 sin 𝜔𝑑 𝑡) 𝑥(𝑡) = 𝑒 −𝜁𝜔𝑛𝑡 (𝑋𝑠𝑖𝑛(𝜔𝑑 𝑡 + 𝜙))

𝒙(𝒕) = 𝑿𝒆−𝜻𝝎𝒏 𝒕 𝒔𝒊𝒏(𝝎𝒅 𝒕 + 𝝓)


3-8

Case (2) : Critically damped motion 𝜁2 − 1 = 0

⇛

(Real equal roots)

𝜁=1

𝑠1,2 = −𝜁𝜔𝑛 ± √𝜁 2 − 1 𝜔𝑛 𝑠1,2 = −(1)𝜔𝑛 ± √1 − (1)2 𝜔𝑛

𝑠1,2 = −𝜔𝑛

𝑠1 = −𝜔𝑛 𝑠2 = −𝜔𝑛 Solution of diff. equation:

𝑥 (𝑡) = 𝐶1 𝑒 𝑠1𝑡 + 𝐶2 𝑡𝑒 𝑠2𝑡 𝑥 (𝑡) = (𝐶1 + 𝐶2 𝑡)𝑒 −𝜔𝑛 𝑡 𝐶1 𝑎𝑛𝑑 𝐶2 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 𝑡𝑜 𝑏𝑒 𝑓𝑜𝑢𝑛𝑑 𝑓𝑟𝑜𝑚 𝑖𝑛𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑜𝑛𝑠


3-9


3-10

Case (3) : Overdamped motion 𝜁2 − 1 > 0

⇛

(Real unequal roots)

𝜁>1

𝑠1,2 = −𝜁𝜔𝑛 ± 𝜔𝑛 √(𝜁 2 − 1) 𝑠1 = −𝜁𝜔𝑛 + 𝜔𝑛 √(𝜁 2 − 1) 𝑠2 = −𝜁𝜔𝑛 − 𝜔𝑛 √(𝜁 2 − 1) Solution of diff. equation:

𝑥 (𝑡) = 𝐶1 𝑒 𝑠1 𝑡 + 𝐶2 𝑒 𝑠2 𝑡 𝑥 (𝑡) = 𝐶1 𝑒

[−𝜁𝜔𝑛 +𝜔𝑛 √(𝜁2 −1) ]𝑡

+ 𝐶2 𝑒

[−𝜁𝜔𝑛 −𝜔𝑛 √(𝜁2 −1) ]𝑡

𝑥 (𝑡) = 𝐶1 𝑒 −𝜁𝜔𝑛𝑡 𝑒 𝜔𝑛√(𝜁 −1) 𝑡 + 𝐶2 𝑒 −𝜁𝜔𝑛𝑡 𝑒 −𝜔𝑛√(𝜁 −1) 𝑡 2

2

𝑥 (𝑡) = 𝑒 −𝜁𝜔𝑛𝑡 (𝐶1 𝑒 𝜔𝑛√(𝜁 −1) 𝑡 + 𝐶2 𝑒 −𝜔𝑛√(𝜁 −1) 𝑡 ) 2

2

𝐶1 and 𝐶2 are obtained from initial condition 𝑡 = 0 → 𝑥(0), 𝑥̇ (0)


3-11


3-12

Characteristic roots on complex plane: Roots of auxiliary equation 𝑠1,2 = −𝜁𝜔𝑛 ± 𝑗 √1 − 𝜁 2 𝜔𝑛 𝑠1,2 = −𝜁 ± 𝑗 √1 − 𝜁 2 𝜔𝑛


0<𝜁<1

3-13

LOGARITHMIC DECREMENT Rate of decay of free vibration is a measure of damping present in a system. Greater is the decay, larger will be the damping. A method to experimentally measure damping ratio in under-damped systems.

𝑥(𝑡) = 𝑒 −𝜁𝜔𝑛 𝑡 (𝐴 cos 𝜔𝑑 𝑡 + 𝐵 sin 𝜔𝑑 𝑡) or

𝒙(𝒕) = 𝑿𝒆−𝜻𝝎𝒏 𝒕 𝒔𝒊𝒏(𝝎𝒅 𝒕 + 𝝓)

𝑡𝑛+1 = 𝑡1 + 𝑛𝜏𝐷 𝜏𝐷 =

2𝜋

𝜔𝑑

=

2𝜋 𝜔𝑛 √1 − 𝜁 2 𝜏𝐷 : 𝑑𝑎𝑚𝑝𝑒𝑑 𝑝𝑒𝑟𝑖𝑜𝑑


3-14

Log Decrement 𝑥1 𝑋𝑒 −𝜁𝜔𝑛 𝑡1 𝑠𝑖𝑛(𝜔𝑑 𝑡1 + 𝜙) = 𝑥𝑛+1 𝑋𝑒 −𝜁𝜔𝑛 𝑡𝑛+1 𝑠𝑖𝑛(𝜔𝑑 𝑡𝑛+1 + 𝜙)

𝑥1 𝑒 −𝜁𝜔𝑛 𝑡1 = 𝑥𝑛+1 𝑒 −𝜁𝜔𝑛 𝑡𝑛+1 𝑥1 𝑒 −𝜁𝜔𝑛 𝑡1 = 𝑥𝑛+1 𝑒 −𝜁𝜔𝑛 (𝑡1 +𝑛𝜏𝐷 ) 𝑥1 𝑒 −𝜁𝜔𝑛 𝑡1 = 𝑥𝑛+1 𝑒 −𝜁𝜔𝑛 𝑡1 . 𝑒 −𝜁𝜔𝑛 𝑛𝜏𝐷 ) 𝑥1 1 = −𝜁𝜔 𝑛𝜏 𝑛 𝐷 𝑥𝑛+1 𝑒 𝑥1 = 𝑒 𝜁𝜔𝑛 𝑛𝜏𝐷 𝑥𝑛+1 2𝜋

𝜁𝜔𝑛 𝑛( 𝑥1 𝜔𝑛 √1−𝜁2 =𝑒 𝑥𝑛+1


)

3-15

𝜁𝜔 𝑛 . 2𝜋

( 𝑛 𝑥1 2 = 𝑒 𝜔𝑛 √1−𝜁 𝑥𝑛+1 𝑛 2𝜋𝜁

( 𝑥1 2 = 𝑒 √1−𝜁 𝑥𝑛+1

)

)

Take ln (𝑙𝑜𝑔𝑒 ) both sides 𝑛 2𝜋𝜁

( ) 𝑥1 2 √ 1−𝜁 ln ( ) = ln [𝑒 ] 𝑥𝑛+1

𝑥1 𝑛 2𝜋𝜁 ln ( )= =𝛿 2 𝑥𝑛+1 √1 − 𝜁 𝛿

≡ 𝐿𝑜𝑔 𝐷𝑒𝑐𝑟𝑒𝑚𝑒𝑛𝑡

1 𝑥1 2𝜋𝜁 ln ( )= 𝑛 𝑥𝑛+1 √1 − 𝜁2 𝑥1 𝑛 2𝜋𝜁 )= 𝑥𝑛+1 √1 − 𝜁2

𝛿 = ln (

m and k can be determined easily by static tests. c?

Determined by the concept of Logarithmic decrement.


3-16

𝑊ℎ𝑒𝑛 𝜁 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙 , √1 − 𝜁 2 ≅ 1 , 𝑎𝑛𝑑 𝑎𝑛 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛

𝑥1 ) 𝛿 ≅ 2𝜋𝜁ln ( 𝑥𝑛+1 is obtained. Figure below shows a plot of the exact and approximate values of 𝛿 as a function of

𝜁.


3-17

EXAMPLE: 𝑋1 = 0.68 𝑋6 = 0.12

𝜁 =? SOLUTION: 𝑡+1=6 ∴𝑡=5

𝑥1 𝑛 2𝜋𝜁 )= 𝑥𝑛+1 √1 − 𝜁2

𝛿 = ln (

0.68 5 × 2𝜋𝜁 ln ( )= ≈ 5 × 2𝜋𝜁 2 0.12 √1 − 𝜁 𝜁 = 0.055

5.5%


3-18

EXAMPLE: A vibrating system consisting of a weight of 𝑊 = 10 𝑙𝑏 and a spring with stiffness 𝑘 = 20 𝐼𝑏/𝑖𝑛 is viscously damped so that the ratio of two consecutive amplitudes is 1.00 𝑡𝑜 0.85. Determine: I. The natural frequency of the undamped system. 2. The logarithmic decrement. 3. The damping ratio. 4. The damping coefficient. 5. The damped natural frequency.

SOLUTION: I. The undamped natural frequency of the system in radians per second is

𝜔𝑛 = √

𝑘 386𝑖𝑛/𝑠𝑒𝑐 2 √ = 20 𝑙𝑏/𝑖𝑛 = 27.78 𝑟𝑎𝑑/𝑠𝑒𝑐 𝑚 10 𝑙𝑏

or in cycles per second 𝑓=

𝜔 = 4.42 𝑐𝑝𝑠 2𝜋

2. The logarithmic decrement is 𝛿 = ln (

𝑥1 1 ) = ln = 0.163 𝑥2 0.85

3. The damping ratio is approximately equal to:

𝜉=

𝛿 0.163 = = 0.026 2𝜋 2𝜋

4. The damping coefficient is obtained

𝑐 = 𝜉. 𝐶𝑐𝑟 = 2 × 0.026√(10 × 20)/386 = 0.037 𝑙𝑏. 𝑠𝑒𝑐/𝑖𝑛


3-19

5. The natural frequency of the damped system is 2

𝜔𝐷 = 𝜔𝑛 √1 − 𝜉 2

𝜔𝐷 = 27.78√1 − (0.026) = 27.77 𝑟𝑎𝑑/𝑠𝑒𝑐


3-20

EXAMPLE: The following data are given for a vibrating system with viscous damping: 𝑤 = 10 𝐼𝑏, 𝑘 = 30 𝐼𝑏/𝑖𝑛., and 𝑐 = 0.12 𝐼𝑏/𝑖𝑛./𝑠. Determine the logarithmic decrement and the ratio of any two successive amplitudes.

SOLUTION: The undamped natural frequency of the system in radians per second is

𝜔𝑛 = √

𝑘 30 × 386 =√ = 34.0 𝑟𝑎𝑑/𝑠𝑒𝑐 𝑚 10

The critical damping coefficient Ce and damping factor 𝜁 are

𝐶𝑐𝑟 = 2 ×

𝜁=

10 × 34 = 1.76 𝑙𝑏/𝑖𝑛./𝑠𝑒𝑐 386 𝑐 0.12 = = 0.0681 𝐶𝑐𝑟 1.76

The logarithmic decrement,

𝛿=

𝑛 2𝜋𝜁

√1 − 𝜁

2

=

𝑛 2𝜋(0.0681)

√1 −

(0.0681)2

= 0.429

𝑥1 = 𝑒 𝛿 = 𝑒 0.429 = 1.54 𝑥2


3-21

EXAMPLE: Show that the logarithmic decrement is also given by the equation

𝑥0 ln ( ) 𝑛 𝑥1 amplitude after 𝑛 𝛿=

1

where 𝑥𝑛 represents the cycles have elapsed. Plot a curve giving the number of cycles elapsed against 𝜁 for the amplitude to diminish by 50 percent.

SOLUTION: The amplitude ratio for any two consecutive amplitudes is

𝑥0 𝑥1 𝑥2 𝑥𝑛−1 = = =⋯= = 𝑒𝛿 𝑥1 𝑥2 𝑥3 𝑥𝑛 The ratio

𝑥0 𝑥𝑛

can be written as

𝑥0 𝑥0 𝑥1 𝑥2 𝑥𝑛−1 = ( )( )( ) = ⋯ = ( ) = (𝑒 𝛿 )𝑛 = 𝑒 𝛿𝑛 𝑥𝑛 𝑥1 𝑥2 𝑥3 𝑥𝑛 from which the required equation is obtained as

𝛿=

𝑥0 ln ( ) 𝑛 𝑥𝑛 1

To determine the number of cycles elapsed for a 50-percent reduction in amplitude, we obtain the following relation from the preceding equation:

𝛿 ≅ 2𝜋𝜁 =

1 𝑛

ln(2) =


0.693 𝑛

3-22

𝑛𝜁 =

0.693 = 0.110 2𝜋

The last equation is that of a rectangular hyperbola and is plotted in Figure below.


3-23

EXAMPLE: A platform of weight 𝑊 = 4000 𝐼𝑏 is being supported by four equal columns that are clamped to the foundation as well as to the platform. Experimentally it has been determined that a static force of 𝐹 = 1000 𝐼𝑏 appliled horizontally to the platform produces a displacement of ∆ = 0.10 𝑖𝑛. It is estimated that damping in the structures is of the order of 5% of the critical damping. Determine for this structure the following: 1. Undamped natural frequency. 2. Absolute damping coefficient 3. Logarithmic decrement. 4. The number of cycles and the time required for the amplitude of motion to be reduced from an initial value of 0.1 𝑡𝑜 0.01 𝑖𝑛.

SOLUTION: 1. The stiffness coefficient (force per unit displacement) is computed as 𝑘=

𝐹 1000 = 10000 𝑙𝑏/𝑖𝑛 ∆ 0.1

and the undamped natural frequency 𝑘 𝑘. 𝑔 10000 × 386 𝜔=√ =√ =√ = 31.06 𝑟𝑎𝑑/𝑠𝑒𝑐 𝑚 𝑊 4000

2. The critical damping is

𝐶𝑐𝑟 = 2√𝑘. 𝑚 = 2√10000 × 4000/386 = 643.8 𝑙𝑏. 𝑠𝑒𝑐/𝑖𝑛 and the absolute damping

𝑐 = 𝜉 𝐶𝑐𝑟 = 0.05 × 643.8 = 32.19 𝑙𝑏. 𝑠𝑒𝑐/𝑖𝑛 3. Approximately, the logarithmic decrement is

𝑥0 ) = 2𝜋𝜉 = 2𝜋(0.05) = 0.314 𝑥1

𝛿 = ln (


3-24

and the ratio of two consecutive amplitudes

𝑥0 = 1.37 𝑥1 4. The ratio between the first amplitude 𝑥0 and the amplitude 𝑥𝑘 after 𝑘 cycles may be expressed as

𝑥0 𝑥0 𝑥1 𝑥𝑘−1 = ( ).( )⋯ 𝑥𝑘 𝑥1 𝑥2 𝑥𝑘 Then taking the natural logarithm, we obtain

ln

𝑥0 = 𝛿 + 𝛿 + ⋯ + 𝛿 = 𝑘𝛿 𝑥𝑘 ln

𝑘=

0.1 = 0.314𝑘 0.01

ln 10 0.314

= 7.73 → 8 𝑐𝑦𝑐𝑙𝑒𝑠

The damped frequency 𝜔𝐷 is given by 𝜔𝐷 = 𝜔𝑛 √1 − 𝜉 2 = 31.06√1 − (0.05)2 = 31.02 𝑟𝑎𝑑/𝑠𝑒𝑐 and the damped period 𝑇𝐷 by 𝑇𝐷 =

2𝜋 2𝜋 = = 0.2025 𝑠𝑒𝑐 𝜔𝐷 31.02

Then the time for eight cycles is 𝑡 (8 𝑐𝑦𝑐𝑙𝑒𝑠) =8 Lecture notes on Mechanical Vibrations website : www.abdulrahmanbahaddin.epu.edu.krd

𝑇𝐷 = 1.62 𝑠𝑒𝑐 3-25

www.abdulrahmanbahaddin.epu.edu.krd


3-26

Mechanical Vibrations Single Degree-Of-Freedom Damped Free Vibration (Chapter TREE)

Recommend Documents