APPENDIX A
SAMPLE CALCULATIONS
47
Example Calculation of Predicted Natural Frequency
Given:
Calculate:
Spring constant, k = 395 lb/in.
Effective Vertical Floor Mass Weight = Weff, vert
Weff, vert = Wblocks + Wslab
Weff , vert = 9(60 lb) + 150
lb
ft 3
(12 ft )(12 ft ) 55. in.12
Weff, vert = 3015 lb
in.
ft
  1 air spring 
  4 total air springs 
m = Weff, vert / g
Predicted Natural Frequency = fn,pred
1 k
f n , pred =
m
2π
1
395 lb in.
f n , pred =
2π 3015 lb
386.4 in. s2
fn, pred = 1.13 hz
Example Calculation of Predicted Natural Frequency for Corner Rotation Mode of
Oscillation (Approximate Method)
Given:
Axis of rotation illustrated in Figure 2.7 and discussion of Equations (2.2)(2.4) in Chapter 2.
Calculate:
wLs 2
Weff ,vert = Wb +
2
Wb = 9(60 lb.) = 540 lb
 55
. in. 
w = 150 lb ft 3  in.  = 68.75 psf
 12 ft 
Ls = 12.0 ft
(68.75 psf )(12.0 ft )
2
Weff , vert = 540 lb +
Weff , vert = 5490 lb
2
L
Weff ,rotate = Wb + wL ∫ y 2 dx
0
48
x
= modal displacement of effective section
αL
x = distance from axis of rotation to spring location (corner of slab)
α =(distance from axis of rotation to air spring) / (length of
effective section)
L = length of effective section
L
2
 x 
Weff ,rotate = Wb + wL ∫   dx
 αL 
0
y=
L
3
wL2
 wL   x 
Weff ,rotate = Wb +  2 2    = Wb + 2
3α
α L   3  0
94 in.
α =
= 0.922
102 in.
L = 8.5 ft
(68.75 psf )(8.5 ft )
2
Weff , rotate = 540 lb +
3 (0.922)
2
Weff , rotate = 2488 lb
f n ,rotate = f n ,vert
Weff ,vert
Weff ,rotate
f n , vert = 113
. hz
f n , rotate = (113
. hz ) 5490 lb 2488 lb
f n , rotate = 168
. hz
Example Calculation of Predicted Natural Frequency for Side Rotation Mode of
Oscillation (Approximate Method)
Given:
Axis of rotation illustrated in Figure 2.8 and discussion of equations (2.2)(2.4) in Chapter 2.
Calculate:
Weff ,vert = Wb + wLs 2
Wb = 9(60 lb) = 540 lb
 55
. in. 
w = 150 lb ft 3  in.  = 68.75 psf
 12 ft 
Ls = 6.0 ft
Weff , vert = 540 lb + (68.75 psf )(6.0 ft )
Weff , vert = 3015 lb
2
49
L
Weff ,rotate = Wb + wL ∫ y 2 dx
0
x
y=
αL
x = distance from axis of rotation to spring location (side of slab)
α=(distance from axis of rotation to air spring) / (length of
effective section)
L = length of effective section
L
2
 x 
Weff ,rotate = Wb + wL ∫   dx
 αL 
0
L
3
wL2
 wL   x 
Weff ,rotate = Wb +  2 2    = Wb + 2
3α
α L   3  0
65 in.
α =
= 0.903
72 in.
L = 6.0 ft
(68.75 psf )(6.0 ft )
2
Weff , rotate = 540 lb +
3 (0.922)
2
Weff , rotate = 1552 lb
f n ,rotate = f n ,vert
Weff ,vert
Weff ,rotate
f n , vert = 113
. hz
f n , rotate = (113
. hz ) 3015 lb 1552 lb
f n , rotate = 157
. hz
Example Calculation of Predicted Natural Frequency for Corner Rotation Mode of
Oscillation (Exact Method)
Given:
Calculate:
Axis of rotation illustrated in Figure 2.7 and Equations (2.5) - (2.9) in
Chapter 2.
Using the parallel axis theorem, calculate the mass moment of inertia of the
blocks added to the corner of the floor slab:
I blocks = I + md 2
I = 0 (relative to total Iblocks)
2 (9) (60 lb)
lb - s2
m =
= 33.54
32.2 ft s2
ft
50
9.9 in.
= 7.66 ft
12 in. ft
lb - s2
(7.66 ft )2
= 33.54
ft
= 1968 lb - ft - s2
d = 8.485 ft I blocks
I blocks
I slab = I + md 2
d = 0 (slab rotates about centroidal axis)
I slab =
∫y
2
dm = ρ t
∫y
2
dA = ρ t I x
 1 s2 


ft 3  32.2 ft 
5.5 in.
t =
= 0.458 ft
12 in. ft
1
Ix =
bh 3
12
1

I x = 2
(2)(8.485 ft)(8.485 ft) 3  = 1728 ft 4
 12

ρ = 150 lb
(
)

 1 s2  
I slab =  150 lb 3 
 (0.458 ft )(1728 ft 4 )
ft
32.2
ft



I slab = 3689 lb - ft - s2
I total = I blocks + I slab = (1968 + 3689) lb - ft - s2
Itotal = 5657 lb - ft - s2
M = 2 k (d)(d) Θ
M = 2 395 lb
(
)(
)
7.66 ft x 12 in. ft (7.66 ft ) Θ
in.3
M = 566,244.7 Θ (lb - ft)
Substitute into Equation (2.5):
M + I (d2Θ/dt2) = 0
566,244.7 Θ + 5657 d2Θ/dt2 = 0
ω =
f n , rotate
f n , rotate
2 k d2
566,244.7 lb - ft
=
= 9.92 rad s
I
5657 lb - ft - s2
1
1
=
ω =
9.92 rad s
2π
2π
= 1.58 hz
(
)
51
Example Calculation of Predicted Natural Frequency for Side Rotation Mode of
Oscillation (Exact Method)
Given:
Calculate:
Axis of rotation illustrated in Figure and Equations (2.5) - (2.9) in
Chapter 2.
Using the parallel axis theorem, calculate the mass moment of inertia of the
blocks added to the side of the floor slab:
I blocks = I + md 2
I = 0 (relative to total Iblocks)
4 (9) (60 lb)
lb - s2
m =
= 67.08
32.2 ft s2
ft
7.0 in.
d = 6.0 ft = 5.42 ft
12 in. ft
lb - s2
(5.42 ft )2
I blocks = 67.08
ft
I blocks = 1968 lb - ft - s2
I slab = I + md 2
d = 0 (slab rotates about centroidal axis)
I slab =
∫y
2
ρ = 150 lb
dm = ρ t
ft 3
∫y
2
dA = ρ t I x
 1 s2 


 32.2 ft 
5.5 in.
= 0.458 ft
12 in. ft
1
Ix =
bh 3
12
1
(12 ft)(12 ft)3 = 1728 ft 4
Ix =
12

 1 s2  
I slab =  150 lb 3 
 (0.458 ft )(1728 ft 4 )
ft
32.2
ft



t =
(
)
I slab = 3689 lb - ft - s2
I total = I blocks + I slab = (1968 + 3689) lb - ft - s2
52
Itotal = 5657 lb - ft - s2
M = 4 k (d)(d) Θ
M = 4 395 lb
(
)(
)
5.42 ft x 12 in. ft (5.42 ft) Θ
in.3
M = 566,292.7 Θ (lb - ft)
Substitute into Equation (2.5):
M + I (d2Θ/dt2) = 0
566,292.7 Θ + 5657 d2Θ/dt2 = 0
2 k d2
566,292.7 lb - ft
=
= 9.92 rad s
I
5657 lb - ft - s2
1
1
=
ω =
9.92 rad s
2π
2π
= 1.58 hz
ω =
(
f n , rotate
f n , rotate
)
Example Calculation of Measured Acceleration Response
Given:
Frequency spectrum of acceleration response due to jumping at 1.5 hz
Valve Configuration = None
Frequency Spectrum
Jumping @ 1.5 Hz - F0204A1
150
5.00
4.00
3.00
2.00
1.00
0.00
-1.00
-2.00
-3.00
-4.00
-5.00
Magnitude
Acceleration (%g)
Acceleration vs. Time
Jumping @ 1.5 Hz - F0204A1
100
50
0
0
4
8
12
16
0
2
4
6
Frequency (Hz)
Time (sec)
Measured Input Average Peak Acceleration = 2.24 %g
Magnitude of Frequency Spectrum at First Harmonic = 113.14
Magnitude of Frequency Spectrum at Second Harmonic = 16.32
Magnitude of Frequency Spectrum at Third Harmonic = 14.81
Calculate:
Ratio of Frequency Spectrum Magnitudes
53
8
10
Second / First Harmonic = 16.32 / 113.14 = 0.14
Third / First Harmonic = 14.81 / 113.14 = 0.13
Measured Input Average Peak Acceleration at Second Harmonic =
(2.24 %g) x (0.14) = 0.32 %g
Measured Input Average Peak Acceleration at Third Harmonic =
(2.24 %g) x (0.13) = 0.29 %g
Example Calculation of Measured Force Transmission
Given:
Frequency response of measured input and output force response due to
jumping at 1.5 hz
Valve Configuration = None
Output Force vs. Time
Jumping @ 1.5 Hz - F0204F1
Frequency Spectrum
Jumping @ 1.5 Hz - F0204F1
15000
3180
Magnitude
Output Force (lbf)
3200
3160
3140
10000
5000
3120
3100
0
0
4
8
12
16
0
2
4
6
8
10
Frequency (Hz)
Time (sec)
Input force initial force plate reading = 205.7 lb
Output force initial force plate reading = 3161.6 lb
The following charts show the peak negative (-) and positive (+) readings from
the input and output force response plots minus initial force plate reading
The average impact value is the average net impact for ten cycles.
For example, for the first cycle of the input force:
Plot (-) peak = 11.48 lbf
Plot (+) peak = 312.52 lbf
54
Chart (-) peak = Plot (-) peak - Initial force plate reading
= 11.48 lbf - 205.7 lbf
= - 194.22 lbf
Chart (+) peak = Plot (+) peak - Initial force plate reading
= 312.52 lbf - 205.7 lbf
= 106.82 lbf
Net Impact = Average (-) and (+) impact
− 194.22 + 106.82
=
2
= 150.52 lbf
These calculations were completed for each jumping frequency and
each valve configuration. The results are listed in Table 3.7.
Input Force Chart
(-)
(lbf)
194.22
196.16
196.16
196.16
194.22
196.16
196.16
194.22
194.22
196.16
(+)
(lbf)
Net
Impact
(lbf)
150.52
153.435
149.55
155.375
151.495
151.49
178.68
153.435
195.19
149.55
106.82
110.71
102.94
114.59
108.77
106.82
161.2
112.65
196.16
102.94
Average
Impact: 158.87
Output Force Chart
(-)
(lbf)
23.25
33.05
25.70
26.92
26.92
30.60
31.82
34.28
42.85
26.92
(+)
(lbf)
Net
Impact
(lbf)
28.19
29.41
26.35
25.12
26.35
26.96
28.19
33.09
34.93
26.35
33.13
25.77
27.00
23.32
25.77
23.32
24.55
31.90
27.00
25.77
Average
Impact: 28.49
The average output force is the force transmitted at one spring. The total output force
transmitted to the ground floor is as follows:
Total Output Force
= Output Force at One Spring x 4
= 28.49 lbf x 4
= 113.96 lbf
Measured Force Transmission at First Harmonic
= Total Output Force / Input Force
= (113.96 lbf / 158.87 lbf) x 100
= 71.87 %
55
Measured Force Transmission at Second and Third Harmonic
In a manner similar to the peak acceleration calculations, the ratios of the peak
magnitudes of the frequency spectrum for the input and output force responses were
calculated. The ratio of the first to second harmonic magnitudes was multiplied by the
measured input and output forces at the first harmonic to determine the measured forces at the
second harmonic. For example:
Input Force
Measured Input Force at First Harmonic = 158.87 lbf
Magnitude of Frequency Spectrum at First Harmonic = 77657
Magnitude of Frequency Spectrum at Second Harmonic = 15153
Magnitude of Frequency Spectrum at Third Harmonic = 8818
Ratio of Frequency Spectrum Magnitudes
Second / First Harmonic = 15153 / 77657 = 0.195
Third / First Harmonic = 8818 / 77657 = 0.113
Measured Input Force at Second Harmonic =
(158.87 lbf) x (0.195) = 30.94 lbf
Measured Input Force at Third Harmonic =
(158.87 lbf) x (0.113) = 18.01 lbf
Measured Force Transmission at Second Harmonic
= Total Output Force / Input Force
= (4.32 lbf / 30.94 lbf) x 100
= 13.95 %
Measured Force Transmission at Third Harmonic
= Total Output Force / Input Force
= (3.41 lbf / 18.01 lbf) x 100
= 18.95 %
These calculations were completed for each jumping frequency and
each valve configuration. The results are listed in Table 3.8 and 3.9.
Example Calculation of Damping Ratio Using Logarithmic Decrement Method
Given:
The acceleration response peaks due to HDS impact at the center of the
floor slab.
A typical acceleration response and corresponding peak
acceleration magnitudes are shown below.
56
Acceleration vs. Time
HDS Impact @ Center - F0206A41
Acceleration (%g)
2
0
-2
-4
-6
-8
-10
-12
0
1
2
3
4
Time (sec)
Peak (-)
Peak (+) Average Peak
Acceleration Acceleration Acceleration
(%g)
(%g)
(%g)
0.44
0.82
0.63
0.35
0.23
0.29
0.30
0.18
0.24
0.25
0.13
0.19
0.22
0.08
0.15
0.20
0.06
0.13
Calculate:
Determine the logarithmic decrement and damping ratio using Equation (2.10).
δ = logarithmic decrement
ζ = damping ratio
2πζ
1  xo 
δ=
=
ln 
1 − ζ 2 n  xn 
n=5
1  0.63
δ = ln
 = 0.319
5  013
. 
57
ζ =
δ
0.319
=
2
4π + δ
4π 2 + (0.319)
ζ = 0.0507 = 5.07%
2
2
This calculation was repeated for five of the ten HDS impacts at the center of the slab
for each valve configuration. The damping ratios were averaged to determine the reported
damping ratio for each valve configuration, as shown in Table 3.1.
Example Calculation of Predicted Force Transmission
Given:
Equation 2.12
Vibration characteristics for Valve Configuration = None
ζ = 0.0518; fn = 1.0625 hz
Calculate:
First Harmonic of Jumping Frequency f = 1.5 hz
2




f




1 + 2ζ  f  
n 

 x100

tr = 
2 
2 2


 1 −  f   + 2ζ  f   
f


n
 
   f n   
[
)]
(


 x100
2


2

.
1 + 2(0.0518) 15

10625
.
tr = 
2 2
 + 2(0.0518) 15
 1 − 15
.
.
10625
10625
.
.

 
(
)
(
[
)]
t r = 100.56%
Second Harmonic of Jumping Frequency f = 3.0 hz
[
)]
(
2

1 + 2(0.0518) 3.010625

.
tr = 
2 2
 + 2(0.0518) 3.0
 1 − 3.0
10625
.
10625
.

 
(
)
(
[
t r = 19.86%
58


 x100
2


)]
Third Harmonic of Jumping Frequency f = 4.5 hz
[
)]
(
2

1 + 2(0.0518) 4.510625

.
tr = 
2 2
 + 2(0.0518) 4.5
 1 − 4.5
.
.
10625
10625


 
(
)
(
[
t r = 10.44%
59


 x100
2


)]