3/7/2020
0.18
0.16
0.14
0.12
0.1
Acceleration
0.08
0.06
0.04
0.02
0
-0.02
-0.04
-0.06
-0.08
-0.1
0
Dr. Ghada Hekal
Associate professor
Civil Engineering Department
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40
Time
Menoufia University
1
2
Damping is any effect tends to reduce the
amplitude of vibration
In damping, the energy of the vibrating system is
dissipated by various mechanisms, for example:
•Friction of steel connections
•Opening and closing of microcracks in concrete
•Friction between structure itself and nonstructural
elements such as partition walls, door ways, window
frames…..etc.
3
4
It seems impossible to identify or describe
mathematically each of these energy dissipating
mechanisms in an actual building.
The actual damping can be idealized by a linear viscous
damper or dashpot. The damping coefficient C is
selected so that the energy it dissipates is equivalent to
the energy dissipated in all damping mechanisms,
combined, present in the actual structure. This
idealization is called viscous damping.
Unlike mass and stiffness, damping can't be
calculated from structural elements dimensions or
cross sections.
5
6
1
3/7/2020
mu(t ) +
=
P (t)
cu (t )
+
K u(t)
0
Viscous damper is a piston filled with a viscous
liquid. Damping force depends on velocity
* pull quickly→ larger resistance
* pull slowly→ smaller resistance
7
mu(t ) +
cu (t ) + K u(t) = 0
8
𝒂𝒚ሷ + 𝒃𝒚ሶ + 𝒄𝒚 = 𝟎
𝜆1,2 =
2
Put 𝑦ሷ = 𝜆 , 𝑦ሶ = 𝜆 𝑎𝑛𝑑 𝑦 = 1
𝒂𝜆2 + 𝒃𝜆 + 𝒄 = 𝟎
𝜆1,2 =
𝑏 2 − 4𝑎𝑐 = 0
𝑏 2 − 4𝑎𝑐 = 0
−𝑏 ± 𝑏 2 − 4𝑎𝑐
2𝑎
𝑏 2 − 4𝑎𝑐 > 0
𝑏 2 − 4𝑎𝑐 > 0
𝑏 2 − 4𝑎𝑐 < 0
9
9
𝑏 2 − 4𝑎𝑐 < 0
(𝝀𝟏 𝒂𝒏𝒅 𝝀𝟐 )
(𝝀)
Dr. Ghada Hekal
𝒚 = 𝑪𝟏 𝒆𝝀𝒙 + 𝑪𝟐 𝒙𝒆𝝀𝒙 𝒚 = 𝑪𝟏 𝒆𝝀𝟏 𝒙 + 𝑪𝟐 𝒆𝝀𝟐 𝒙
𝝀𝟏 = 𝜶 + 𝜷𝒊
𝝀𝟐 = 𝜶 − 𝜷𝒊
𝑦 = 𝑒 𝛼𝑥 𝐶1 𝑐𝑜𝑠β𝑥 + 𝐶2 𝑠𝑖𝑛β𝑥
10
Dr. Ghada Hekal
10
mu(t ) +
1, 2 =
cu (t ) + K u(t) = 0
To solve equation, Let:
 = λ , u = λ, u = 1
u
2
m2 + c + k = 0
c 2 − 4mk = 0
2
𝑐 = 4𝑚𝑘
𝑘
∵ 𝜔𝑛 2 = → 𝑘 = 𝑚𝜔𝑛 2
𝑚
𝑐 2 = 4𝑚2 𝜔𝑛 2
𝑐𝑐𝑟 = 2𝑚𝜔𝑛
1, 2
− c  c 2 − 4mk
=
2m
In this case ccr is called
coefficient of critical
damping.
Critically damped
system.
11
−𝑏 ± 𝑏 2 − 4𝑎𝑐
2𝑎
− c  c 2 − 4mk
2m
c 2 − 4mk  0
c > ccr
c 2 − 4mk  0
c < ccr
𝑐
>1
𝑐𝑐𝑟
c
1
ccr
𝜁>1
 1
ζ = damping
ratio
Overdamped
system
ζ = damping ratio
Underdamped
system
12
2
u(t)
3/7/2020
The system returns to equilibrium without oscillation
t, sec
Underdamped
Critically damped
Overdamped
The system returns to equilibrium as quickly as possible without oscillation
The system oscillates to the amplitude gradually decreasing to zero
13
In structural engineering, we need only to study the
underdamped systems because structures of interest
like buildings, bridges, dams, nuclear power plants,
offshore structures... etc. all fall into this category
because their damping ratio ζ is less than 0.10.
14
The solution of free undamped vibration
equation:
mu(t ) +
cu (t ) + K u(t) = 0
Is :
u(t)


 u (0) +  n u (0) 
u (t ) = e − t u (0) cos  D t + 
 sin  D t 



D


n
Where,
ωD : Damped cyclic frequency
16
𝑻𝑫 =
t, sec
 D = n 1 −  2
Undamped
UnderDamped
17
𝑻𝒏
𝟏 − 𝜻𝟐
Tn
TD
0.06
0.04
0.02
0
0
0.5
1
1.5
2
2.5
-0.02
-0.04
-0.06
18
19
3
3/7/2020
𝑢(𝑡) = 𝑒 −𝜁𝜔𝑛 𝑡 𝑢(0) cos 𝜔𝐷 𝑡 +
𝑢1
=
𝑢2
𝑢(0)
ሶ
+ 𝜁𝜔𝑛 𝑢(0)
sin 𝜔𝐷 𝑡
𝜔𝐷
𝑙𝑛
𝑢(0)
ሶ
+ 𝜁𝜔𝑛 𝑢(0)
sin 𝜔𝐷 𝑡
𝜔𝐷
𝑢(0)
ሶ
+ 𝜁𝜔𝑛 𝑢(0)
𝑢(0) cos 𝜔𝐷 (𝑡 + 𝑇𝐷) +
sin 𝜔𝐷 (𝑡 + 𝑇𝐷)
𝜔𝐷
𝑒 −𝜁𝜔𝑛 𝑡 𝑢(0) cos 𝜔𝐷 𝑡 +
𝑒 −𝜁𝜔𝑛 (𝑡+𝑇𝐷)
u1
𝑢1
= 𝑒 𝜁𝜔𝑛 𝑇𝐷 = exp( 𝜁𝜔𝑛 𝑇𝐷 )
𝑢2
𝑇𝑛
1 − 𝜁2
=
2𝜋
𝜔𝑛 1 − 𝜁 2
𝑙𝑛
𝑢1
2𝜋
= 𝜁𝜔𝑛
𝑢2
𝜔𝑛 1 − 𝜁 2
𝑙𝑛
𝑢1
=
𝑢2
u2
20
u1
u2
2𝜋𝜁
1 − 𝜁2
21
So, for any two successive amplitudes,
ln
∵ 𝑇𝐷 =
𝑢1
= 𝜁𝜔𝑛 𝑇𝐷
𝑢2
𝑙𝑛
𝑢1
= 𝜁𝜔𝑛 𝑇𝐷
𝑢2
𝑢𝑖
=𝛿=
𝑢𝑖+1
   2
2𝜋𝜁
If we relate two amplitudes several cycles apart
1 − 𝜁2
𝐹𝑜𝑟 𝑒𝑥𝑎𝑚𝑝𝑙𝑒: 𝑙𝑛
δ is called logarithmic decrement.
= 𝑙𝑛
Note that: δ is independent on time which means that the
ratio between each successive peaks is constant.
𝐹𝑜𝑟 𝜁 < 0.2
1 − 𝜁2 ≈ 1
   2
22
𝑢1
𝑢2
+ 𝑙𝑛
𝑢2
𝑢3
𝑢1
𝑢4
+ 𝑙𝑛
= 𝑙𝑛
𝑢1 𝑢2 𝑢3
. .
𝑢2 𝑢3 𝑢4
𝑢3
𝑢4
= 𝛿 + 𝛿 + 𝛿 = 3𝛿
𝛿=
1 𝑢1
ln
≈ 2𝜋𝜁
3 𝑢4
23
   2
If we relate two amplitudes several cycles apart
𝛿=
𝑢𝑖
𝑙𝑛
= 𝑗𝛿
𝑢𝑖+𝑗
1
𝑢𝑖
ln
≈ 2𝜋𝜁
𝑗 𝑢𝑖+𝑗
Because it is impossible to calculate damping analytically,
𝛿=
1
𝑢𝑖
ln
≈ 2𝜋𝜁
𝑗 𝑢𝑖+𝑗
we can calculate it experimentally by applying a free
vibration test on the system, measuring two amplitudes of
its response and calculating damping from the above
equation.
24
25
4
3/7/2020
A free vibration test is conducted to a water tank. A cable attached to
𝛿=
the tank applies a lateral force of 15t and pulls the tank horizontally by 5
1
𝑢𝑖
ln
≈ 2𝜋𝜁
𝑗 𝑢𝑖+𝑗
cm. the cable is suddenly cut, and the resulting free vibration is
recorded. At the end of three complete cycles, the time is 1 sec. and the
amplitude is 2 cm. from these data compute:
 =
u
1
ln i
2j ui + j
 =
u
1
ln i
2j ui + j
1- Damping ratio
2- Natural period
3- Effective stiffness
4- Effective weight
5- Damping coefficient
6- Number of cycles required for the amplitude to be reduced to 0.5 cm.
26
27
5cm
0.06
u1
0.04
u(t), m
P = 15 t
u2
u3
0.02
u4
0
0
0.5
1
1.5
2
2.5
-0.02
-0.04
-0.06
0.06
u1
u(t), m
0.04
u1= 5 cm
u2
u3
0
0
0.5
 =
u
1
ln i
2j ui + j
𝜁=
1
5
ln
2𝜋 × 3 2
u4= 2 cm
u4
0.02
t, sec
1
1.5
2
j= 3
2.5
-0.02
-0.04
= 0.0486 = 4.86%
-0.06
t, sec
28
29
0.06
u1
u(t), m
0.04
5cm
u2
u3
0.02
P = 15 t
u4
0
0
0.5
1
1.5
2
2.5
-0.02
k
k
-0.04
-0.06
𝑇𝐷 =
𝑇𝐷 =
t, sec
𝑡4 − 𝑡1 1 − 0
=
= 0.33sec
3
3
𝑇𝑛
P=k∆
𝑘=
1 − 𝜁2
𝑇𝑛 = 𝑇𝐷 1 − 𝜁 2 = 0.33 1 − (0.0486)2
15
= 300 𝑡/𝑚
0.05
𝑇𝑛 = 0.3296 sec
30
31
5
3/7/2020
5cm
P = 15t
𝑇𝑛 =
2𝜋
𝜔𝑛
n2 =
k
m
𝜔𝑛 =
𝑚=
𝑐
= 𝜁 = 4.86 %
𝑐𝑐𝑟
2𝜋
2𝜋
=
= 19.03 𝑟𝑎𝑑/sec
𝑇𝑛 0.33
𝑘
300
=
= 0.828 𝑡. sec 2 / 𝑚
𝜔𝑛2 (19.03)2
𝑐𝑐𝑟 = 2𝑚𝜔𝑛 = 2 × 0.828 × 19.03 = 31.51 𝑡. sec/ 𝑚
𝑐 = 𝜁𝑐𝑐 = 0.0486 × 31.51 = 1.53 𝑡. sec/ 𝑚
𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑎𝑠𝑠 × 𝑔 = 0.828 × 10 = 8.28 𝑡
32
33
0.06
u(t), m
0.04
0.02
0
0
0.5
1
1.5
2
2.5
-0.02
-0.04
-0.06
 =
u
1
ln i
2j ui + j
t, sec
0.0486 =
1
0.05
ln
2𝜋 × 𝑗 0.005
j= 7.54 cycles
34
6