Dynamics of Structures

Chapter 2. Damped Single Degree of Freedom System

Chapter 2
2.1 What Is Damping?2.2 Developing the Eq. of Motion2.3 Logarithmic Decrement2.4 Damped SDOF Animation2.5 Examples

2.5 Examples

Figure 2.5.1     Example 2.5.1

Example 2.5.1

In an experiment to determine the damping ratio of the shown frame, the frame was displaced by 1 in and let to undergo a free vibration. It was observed that the vibration peak of the second cycle was reduced by 5% from the original maximum displacement. If the stiffeness of each column is 1500 kip/in and the weight of the beam is 150 kips. Neglecting the mass of the columns, what is the estimated damping ratio? Also what is the frame damping constant? (g = 386 in/s²)

We first need to calculate the equivalent stiffness of the system k, k,

k=2×1500=3000 kip/ink = 2 \times 1500 = 3000 \space \text{kip/in}

The mass of the system m, m,

m=Wg=150386=0.39 kip. s2/inm = \frac{W}{g}=\frac{150}{386}=0.39 \space \text{kip. s}^2 \text{/in}

Calculate the system natural frequency ω, \omega,

ω=km=30000.39\omega = \sqrt{\frac{k}{m}}=\sqrt{\frac{3000}{0.39}}
ω=87.7 rad/s\omega = 87.7 \space \text{rad/s}

The logarithmic decrement δ, \delta, can be calculated as,

δ=ln⁡(u1u2)\delta = \ln\left(\frac{u_1}{u_2}\right)
δ=ln⁡(1.0 in0.95 in)=0.0513\delta = \ln\left(\frac{1.0 \, \text{in}}{0.95 \, \text{in}}\right) = 0.0513

And the damping ratio ζ, \zeta, can be calculated as,

ζ=δ2π=0.05132π=0.00816\zeta = \frac{\delta}{2\pi} = \frac{0.0513}{2\pi} = 0.00816

And the damping constant c, c, can be calculated as,

c=2mωζc = 2m\omega\zeta
c=2×0.39×87.7×0.00816c = 2 \times 0.39 \times 87.7 \times 0.00816
c=214.7 kip.sec/inc = 214.7 \space \text{kip.sec/in}

Figure 2.5.2     Example 2.5.2

Example 2.5.2

The two-mass spring system shown in the Figure, we are interested in the differential displacement of the two masses u=u2−u1u=u_2 - u_1. Derive the equation of motion of the system in terms of u.

Example 2.5.2 Free Body Diagram

The equation of motion of the first mass,

m1u¨1−cu˙−ku=0     (1)m_1\ddot{u}_1 - c\dot{u} - ku = 0 \ \ \ \ \ (1)

The equation of motion of the second mass,

m2u¨2+cu˙+ku=0     (2)m_2\ddot{u}_2 + c\dot{u} + ku = 0 \ \ \ \ \ (2)

Multiplying equation (1) by m2m_2 and equation (2) by m1m_1,

m1m2u¨1−c m2u˙−km2u=0     (3)m_1 m_2 \ddot{u}_1 - c \ m_2 \dot{u} - k m_2 u = 0 \ \ \ \ \ (3)
m1m2u¨2+c m1u˙+km1u=0     (4)m_1 m_2 \ddot{u}_2 + c \ m_1 \dot{u} + k m_1 u = 0 \ \ \ \ \ (4)

Subtract equation (4) from equation (3),

m1m2(u¨2−u¨1)+cu˙(m1+m2)+ku(m1+m2)=0     (5)m_1 m_2 (\ddot{u}_2 - \ddot{u}_1) + c \dot{u} (m_1 + m_2) + k u (m_1 + m_2) = 0 \ \ \ \ \ (5)

Rearranging,

m1m2u¨+cu˙(m1+m2)+ku(m1+m2)=0     (6)m_1 m_2 \ddot{u} + c \dot{u} (m_1 + m_2) + k u (m_1 + m_2) = 0 \ \ \ \ \ (6)

Equation (6) represents the equation of motion of the two-mass system in terms of uu,

Figure 2.5.3     Example 2.5.3

Example 2.5.3

A 300 lb vibrating machine is supported by a beam supported on two springs and two dampers. It was found that the vertical displacement under the weight of the machine is 1 in. The dampers reduced the amplitude of the vibration to 0.9 times the initial displacment after one complete free vibration cycle. Determine the undamped natural frequency and the damped natural frequency (Neglect the weight of the beam compared to the weight of the machine).

The vertical stiffness can be obtained from knowing the vertical force and the vertical displacement,

k=wΔ=3001.0=300 lb/ink = \frac{w}{\Delta} = \frac{300}{1.0} = 300 \space \text{lb/in}

The mass of the machine is,

m=wg=300386=0.78 lb.s2/inm = \frac{w}{g} = \frac{300}{386} = 0.78 \space \text{lb.} \text{s}^2 \text{/in}

The system natural frequency,

ω=km=3000.78\omega =\sqrt{\frac{k}{m}} = \sqrt{\frac{300}{0.78}}
ω=19.612rad/s\omega = 19.612 \text{rad/s}

Let's assume that the initial displacement u1=uu_1 = u , then the second peak is u2=0.9uu_2 = 0.9u, we can calculate the logarithmic decrement as,

δ=ln(u1u2)=ln(u0.9u)=0.105\delta =ln(\frac{u_1}{u_2}) = ln(\frac{u}{0.9 u}) = 0.105

For small damping ratios,

δ=2πζ\delta =2 \pi \zeta

Or,

δ=2π×0.105=0.0167\delta =2 \pi \times 0.105 = 0.0167

Finally, the damped natural frequency can be calculated as,

ωd=ω1−ζ2 \omega_d = \omega \sqrt{1- \zeta^2}
ωd=19.6121−0.01672 \omega_d = 19.612 \sqrt{1- 0.0167^2}
ωd=19.609 rad/s \omega_d = 19.609 \space \text{rad/s}

It should be noted that the effect of damping on the natural frequency is so small since the damping provided by the dampers is also small.