Dynamics of Structures

Chapter 3. Response of SDOF to Harmonic Loading

3.5 Examples

Figure 3.5.1     Example 3.5.1

Example 3.5.1

The shown frame is subjected to a harmonic load defined as F(t)=5sin⁡12t kipsF(t) = 5 \sin 12t \space \text{kips} , applied at the top girder that was idealized as a linear distributed mass of 2 kip/ft2 \space \text{kip/ft}. Columns are W10x33 and assume that the gireder is infinitely rigid. Determine the steady-state amplitude for the horizontal motion of the frame. use E =  29000 ksi\text{E = } \space 29000 \space \text{ksi}, I =  170 in4\text{I = } \space 170 \space \text{in}^4 and ignore damping.

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

k=2×12EIL3k = 2 \times \frac{12EI}{L^3}
k=2×12×29000×170(15×12)3k = 2 \times \frac{12 \times 29000 \times 170}{(15 \times 12)^3}
k=20.30 kip/ink = 20.30 \text{ kip/in}

The mass of the system m, m,

m=Wg=2×20386m = \frac{W}{g}=\frac{2 \times 20}{386}
m=0.1036 kip. s2/inm = 0.1036 \space \text{kip. s}^2 \text{/in}

Calculate the system natural frequency ω, \omega,

ω=km=20.300.1036\omega = \sqrt{\frac{k}{m}}=\sqrt{\frac{20.30}{0.1036}}
ω=14.00 rad/s\omega = 14.00 \space \text{rad/s}

The frequency ratio can be calculated as,

r=ωˉω=1214.00=0.857r = \frac{\bar{\omega}}{\omega} = \frac{{12}}{14.00} = 0.857

The Response Deformation Factor (also known as the dynamic magnification factor) can be calculated from Eq. 3.3.6 by setting ζ\zeta to zero,

Rd=11−r2R_d = \frac{1}{1-r^2}
Rd=11−(0.857)2=3.77R_d = \frac{1}{1-(0.857)^2} = 3.77

Steady-state amplitude: U U can be calculated as,

U=ustRd=Fok×RdU = u_{st} R_d = \frac{F_o}{k} \times R_d
U=520.30×3.77U = \frac{5}{20.30} \times 3.77
U=0.928 inU = 0.928 \text{ in}

Observations

As an interesting activity, go back to section 3.1.2, adjust the sliders for kk to be equal to 20.30 rad/s, ω\omega to be equal to 14.00 rad/s, ωˉ\bar{\omega} to be equal to 12 rad/s, FoF_o to 5 kips, click on the total response icon to hide it, and confirm that the steady state response is indeed 0.928 in. You can then show the total response (transient plus steady state) to see the effect of the transient part on the total response. Since this system has no damping, the transient part doesn't fade away and keep affecting the total response indefinitely.

Figure 3.5.2     Example 3.5.2

Example 3.5.2

A rotating machine whose weight is equal to 1500 lb is supported at the middle of the two parrallel steel beams arrangement (E = 29000 ksi). Each beam span and moment of inertia are 10' and 14 in4 respectively. The rotating machine speed is 500 rpm producing an unbalanced vertical force of 60 lb at this running speed. Assuming a 5% viscous damping in this system arrangement and neglecting the weight of the beams, determine the amplitudes of the steady state deflection and acceleration at mid span resulting from the unbalanced force.

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

k=2×48EIL3k = 2 \times \frac{48EI}{L^3}
k=2×48×29000×14(10×12)3k = 2 \times \frac{48 \times 29000 \times 14}{(10 \times 12)^3}
k=22.56 kip/ink = 22.56 \text{ kip/in}

The mass of the system m, m,

m=Wg=15001000×386m = \frac{W}{g}=\frac{1500}{1000 \times 386 }
m=0.00389 kip. s2/inm = 0.00389 \space \text{kip. s}^2 \text{/in}

Calculate the system natural frequency ω, \omega,

ω=km=22.560.00389\omega = \sqrt{\frac{k}{m}}=\sqrt{\frac{22.56}{0.00389}}
ω=76.2 rad/s\omega = 76.2 \space \text{rad/s}

The forcing frequency,

ωˉ=500×2π60=52.4 rad/s\bar{\omega} = 500 \times \frac{2 \pi}{60} = 52.4 \text{ rad/s}

The frequency ratio can be calculated as,

r=ωˉω=52.476.2=0.688r = \frac{\bar{\omega}}{\omega} = \frac{{52.4}}{76.2} = 0.688

The Response Deformation Factor (also known as the dynamic magnification factor) can be calculated from Eq. 3.3.6,

Rd=1(1−r2)2+(2ζr)2R_d = \frac{1}{\sqrt{(1-r^2)^2 + (2 \zeta r)^2}}
Rd=1(1−0.6882)2+(2×0.05×0.688)2R_d = \frac{1}{\sqrt{(1-0.688^2)^2 + (2 \times 0.05 \times 0.688)^2}}
Rd=1.88R_d = 1.88

Steady-state deflection amplitude: U U can be calculated as,

U=ustRd=Fok×RdU = u_{st} R_d = \frac{F_o}{k} \times R_d
U=60/100022.56×1.88U = \frac{60/1000}{22.56} \times 1.88
U=5×10−3 inU = 5 \times 10^{-3} \text{ in}

Steady-state acceleration amplitude: U¨ \ddot{U} can be calculated as,

U¨=ωˉ2U\ddot{U} = \bar{\omega}^2 U
U¨=52.42×5×10−3\ddot{U} = 52.4^2 \times 5 \times 10^{-3}
U¨=13.73 in/s2\ddot{U} = 13.73 \text{ in/s}^2

Figure 3.5.3     Example 3.5.3

Example 3.5.3

A sensitive equipment is protected from the vibration of the surrounding building by bolting it to an isolation pad. The isolation pad is supported on isolation springs as shown in the figure. The combined weight of the isolation pad and the sensitive equipment is 2500 lbs. The surrounding building has rotating machines that makes it vibrate at 1800 rpm. If it is required to limit the vertical motion of the isolation block and the sensitive machine to only 8% of the vibration of the surrounding building, calculate the require spring stiffness assuming that the damping ratio is 0.05. What would be the answer if we ignore the effect of damping? Comment on your results.

In this example, it is required to limit the Response Deformation Factor Rd R_d to 8% or Rd=0.08 R_d = 0.08

The Response Deformation Factor can be calculated from Eq. 3.3.6,

Rd=1(1−r2)2+(2ζr)2R_d = \frac{1}{\sqrt{(1-r^2)^2 + (2 \zeta r)^2}}
0.08=1(1−r2)2+(2×0.05×r)20.08 = \frac{1}{\sqrt{(1-r^2)^2 + (2 \times 0.05 \times r)^2}}

Rearranging,

10.08=(1−r2)2+(0.1×r)2\frac{1}{0.08} = \sqrt{(1-r^2)^2 + (0.1 \times r)^2}
156.25=(1−r2)2+(0.1×r)2156.25 = (1-r^2)^2 + (0.1 \times r)^2
156.25=1−2r2+r4+0.01r2156.25 = 1-2r^2+r^4 + 0.01 r^2
156.25=1−1.99r2+r4156.25 = 1-1.99r^2+r^4
r4−1.99r2−155.25=0r^4 - 1.99r^2 -155.25 = 0
r2=1.99+(−1.99)2+4×1×155.252×1r^2 = \frac{1.99 + \sqrt{(-1.99)^2 + 4 \times 1 \times 155.25}}{2 \times 1}
r2=13.4946r^2 = 13.4946

The frequency ratio can be calculated as,

r=3.673r = 3.673

Since the surrounding building is vibrating at 1800 rpm, then,

ωˉ=1800×2π60=188.49 rad/s\bar{\omega} = 1800 \times \frac{2 \pi}{60} = 188.49 \text{ rad/s}
r=ωˉω=188.49ω=3.673r = \frac{\bar{\omega}}{\omega} = \frac{{188.49}}{\omega} = 3.673
ω=51.318 rad/s{\omega} = 51.318 \text{ rad/s}

The system mass,

m=Wg=25001000×386m = \frac{W}{g}=\frac{2500}{1000 \times 386 }
m=0.00647 kip. s2/inm = 0.00647 \space \text{kip. s}^2 \text{/in}

Now we calculate the isolation system stiffness,

k=mω2=0.00647×51.3182k = m \omega^2 = 0.00647 \times 51.318^2
k=17.04 kip/ink = 17.04 \text{ kip/in}

Calculating the Response Deformation Factor ignoring damping,

Rd=1(1−r2)2+(2ζr)2R_d = \frac{1}{\sqrt{(1-r^2)^2 + (2 \zeta r)^2}}

The equation is reduced to,

Rd=1(1−r2)2R_d = \frac{1}{\sqrt{(1-r^2)^2 }}
0.08=1±(1−r2)0.08 = \frac{1}{±(1-r^2) }
12.5=±(1−r2)12.5 = ±(1-r^2)

Since r must be positive,

12.5=−(1−r2)12.5 = -(1-r^2)
r2=13.5r^2 = 13.5
r=3.674r = 3.674

Notice that it is almost the same value obtained considering 5% damping. It can be concluded that the effect of damping can be neglected for this case.