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SHOCK & VIBRATION STRESS & STRAIN
AS A FUNCTION OF VELOCITY

REVISION M

BY TOM IRVINE

VIBRATIONDATA

AUGUST 29, 2024

1 Tom Irvine background ..................................................................................................................... 3
2 Acronyms & Variables ....................................................................................................................... 4
3 Introduction ...................................................................................................................................... 5
4 Spring-Mass System Example, Free Vibration .................................................................................. 6
5 Stress-velocity Summary for sample structures ............................................................................... 7
6 Stress in an Infinite Rod in Longitudinal Free Vibration ................................................................. 14
7 Stress in a Finite, Fixed-Free Rod, in Longitudinal Free Vibration .................................................. 18
8 Stress in a Finite, Fixed-Free Rod, Longitudinal Response Excited by Resonant Base Excitation... 21
9 Rod with End Mass, Longitudinal Vibration.................................................................................... 24
10 Tapered Rod, Longitudinal Vibration .............................................................................................. 26
11 Stress in a Shear Beam in Free Vibration ........................................................................................ 31
12 Stress due to Bending in a Bernoulli-Euler Beam in Free Vibration ............................................... 37
13 Stress due to Bending in a Bernoulli-Euler Beam Excited by Base Excitation ................................ 60
14 Cantilever Beam Example ............................................................................................................... 68
15 Beam with Point Mass Examples .................................................................................................... 76
16 Beam with End Springs ................................................................................................................... 80
17 Multispan Beam, subjected to base excitation .............................................................................. 85
18 Multispan Beam, subjected to base excitation, intermediate support point at one-third length . 89
19 Multispan Beam, pinned-pinned-free, subjected to base excitation ............................................. 93
20 Beam Bending with mean stress .................................................................................................... 97
21 Stress in a Rectangular Plate in Free Vibration ............................................................................. 102
22 Ring Mode ..................................................................................................................................... 109
23 MIL-STD-810E & Similar References ............................................................................................. 110
24 Gaberson’s Blower Shock Test Series ........................................................................................... 112
25 Morse Chart .................................................................................................................................. 128
26 MIL-STD-810G, METHOD 516.6 .................................................................................................... 129
27 Pseudo Velocity Shock Response Spectrum ................................................................................. 130
28 Half-Sine Pulse .............................................................................................................................. 131
29 El Centro Earthquake .................................................................................................................... 133
30 Re-entry Vehicle Separation Shock ............................................................................................... 135

1
31 SR-19 Motor Ignition & Pressure Oscillation ................................................................................ 137
32 Space Shuttle Solid Rocket Booster Water Impact ....................................................................... 139
33 V-band/Bolt-Cutter Shock............................................................................................................. 141
34 Maximum Velocity Summary ........................................................................................................ 143
35 Velocity Limits of Materials .......................................................................................................... 144
36 Nonlinear Shock Analysis .............................................................................................................. 148
37 Critical Impact Velocity ................................................................................................................. 153
38 Characteristic Impedance Values.................................................................................................. 155
39 Building Velocity Limits ................................................................................................................. 156
40 Machinery Velocity Limits............................................................................................................. 157
41 Piping Velocity Limits .................................................................................................................... 159
42 Civil Engineering Structural Velocity Limits .................................................................................. 161
43 Tacoma Narrows Bridge Failure.................................................................................................... 163
44 Conclusions ................................................................................................................................... 166
45 Bibliography .................................................................................................................................. 167

2
1 TOM IRVINE BACKGROUND

Tom Irvine earned B.S. and M.S. degrees in

engineering science from Arizona State University,
in 1985 and 1987, respectively. He wandered into
the specialty of vibration serendipitously and
found that vibration’s quirky traits were well-
suited to his unconventional personality.

He worked at Orbital Sciences Corporation in Chandler, Arizona, for 11 years as a full-time

employee, and 9 years as a consultant. He derived the shock and vibration levels and
performed hands-on testing for the first Pegasus launch vehicle, which successfully flew in
1990. He has analyzed flight accelerometer data from Pegasus, numerous launch vehicles, and
data from ground separation tests. He has posted hundreds of tutorials and software programs
on his Vibrationdata website, covering shock, vibration, acoustics, structural dynamics, and
signal processing.

He was later a scientist at Dynamic Concepts, Inc., in Huntsville, Alabama, where he worked on
the NASA Space Launch System (SLS) vehicle. He concurrently served as an industry
representative for the NASA Engineering & Safety Center (NESC) Loads and Dynamics Technical
Disciplines Team. Following his NASA stint, he served as a senior technical fellow at Blue Origin.

He currently works at a large aerospace company. Tom can be contacted via Email:
tom@irvinemail.org

Readers are welcome to visit his blog: https://vibrationdata.wordpress.com/

Tom either performed or verified the calculations in this e-book using his Vibrationdata Matlab
GUI package which is freely available at the above blog link. He hopes this book will help
bridge the gap between shock and vibration analysis as taught in universities and as practiced in
industry.

3
2 ACRONYMS & VARIABLES

2.1 VARIABLES

c Wave speed in the material L Length

h Thickness M Bending moment
Spring stiffness or wave number
k U Total energy
depending on context
u Displacement V Shear force
v Velocity Ẅ Base acceleration
x Longitudinal Displacement ε Strain
z Transverse Displacement σ Normal stress
A Cross section area τ Shear stress
D Plate stiffness factor ρ Mass per volume
E Elastic modulus ρ Mass per length
G Shear modulus ϕ Phase angle
Angular natural
I Area moment of inertia ω
frequency (rad/sec)
K Shear factor

Note that the characteristic impedance is ρc.

2.2 KEY ACRONYMS

Nomenclature Definition
PSD Power Spectral Density
PV Pseudo Velocity
SDOF Single-degree-of-freedom
SRS Shock Response Spectrum
VRS Vibration Response Spectrum

4
3 INTRODUCTION

Howard A. Gaberson (1931-2013) championed the idea

that dynamic stress correlated better with pseudo
velocity than with either acceleration or relative
displacement. Gaberson was a shock and vibration
specialist who was with the U.S. Navy Civil Engineering
Laboratory and later the Facilities Engineering Service
Center from 1968 to 2000, mostly conducting dynamics
research. He specialized in shock and vibration signal
analysis and published more than 100 papers and
articles. Gaberson pointed out that the kinetic energy in a
mechanical system is proportional to velocity squared.
The pseudo velocity is thus a measure of the stored peak
energy in the system at a particular frequency and has a
direct relationship to the survival or failure of the system.

Shock and vibration environments produce dynamic stresses which can cause material failure in
structures. The potential failure modes include fatigue, yielding, and ultimate stress limit.

Rinehart and Pearson identified velocity as the key metric for structural failure in “Behavior of
Metals under Impulsive Loads” in 1954. [1]

F.V. Hunt wrote a seminal paper on this subject, titled “Stress and Stress Limits on the
Attainable Velocity in Mechanical Vibration,” published in 1960. [2] This paper gave the
relationship between stress and velocity for several sample structures.

H. Gaberson continued research on stress and modal velocity with a series of papers and
presentations, as shown in References [3], [4], [5].

The purpose of this document is to explore the work of Hunt, Gaberson, and others.
Derivations are given relating stress and velocity for various structures. Some of these
examples overlap the work of previous sources. Other examples are original. In addition, this
document presents some unique data samples for shock events, with the corresponding
spectra plotted in tripartite format. Note that Gaberson uses the term four coordinate paper
(4CP) in place of tripartite.

Note that the following stress-velocity equations do not require that peak stress and peak
velocity occur at the same location or at the same time. Any stress concentration factor must
be applied separately. The equations are useful as quick estimates before any advanced

5
modeling such as finite element analysis. The equations can also be used to “sanity check” any
finite element results. Another use is for trade-studies in preliminary designs.

4 SPRING-MASS SYSTEM EXAMPLE, FREE VIBRATION

Consider the following system.

Figure 4.1. SDOF System

The energy equation is

1 1 (4.1)
U = m [v ] = k[x ]
2 2

The velocity is thus square root of twice the peak energy per unit mass that is stored in the
oscillator. The relationship does not directly depend on the natural frequency.

v = 2U⁄m (4.2)

6
5 STRESS-VELOCITY SUMMARY FOR SAMPLE STRUCTURES

5.1 ROD, LONGITUDINAL VIBRATION

5.1.1 Infinite Rod in Longitudinal Free Vibration

Figure 5.1. Propagating Wave in an Infinite Rod

Consider an infinitely long longitudinal rod undergoing a traveling wave response. The stress is
proportional to the velocity as follows.

σ(x, t) = −ρc v(x, t) (5.1)

This equation is given in References [1], [2], [3], [4], [5], [6], [7] . The derivation is given in
Section 6.

5.1.2 Finite, Longitudinal Free Vibration

Figure 5.2. Propagating Wave in a Finite Rod

7
Gaberson showed that this principle can be extended to a finite rod with common boundary
conditions where the peak modal stress [σn ]max is

[σ ] =ρcv , = ρcω u , (5.2)

A derivation is given in Section 7.

The index n represents the mode number. The peak stress is mainly due to the fundamental
mode in most cases. But higher modes can contribute significant stress in certain cases, such as
pyrotechnic shock response.
Gaberson stated that modal velocity is equal to the pseudo velocity times the participation
factor for the case of a shock response spectrum base input. The physical velocity calculation
would also require the mass-normalized eigen coefficient.

5.1.3 Finite, Fixed-Free Rod, Longitudinal Response Excited by Resonant Base Excitation

Consider a fixed-free rod subjected to base excitation. Equation (5.2) applies as long as the
excitation is sinusoidal with a frequency equal to the natural frequency of the corresponding
mode. The derivation is given in Section 8.

5.2 BEAM

5.2.1 Shear

Consider the free vibration of a shear beam with a rectangular cross-section. The modal shear
stress is proportional to the modal velocity as follows, from the derivation in Section 11.

ρc
τ , = 1.5 v , (5.3)
K

8
5.2.2 Bernoulli-Euler Bending, Free Vibration

Consider the bending vibration of a simply-supported or cantilever beam. The modal stress due
to bending is proportional to the modal velocity as follows, from the derivation in Section 12.

∂ EAρ (5.4)
σ , =Ec y (x, t) =c v ,
∂x I

The above equation can be simplified as follows:

σ = K ρc v , (5.5)

c EAρ c EA
K= = = c A⁄ I (5.6)
ρc I c ρI

Table 5.1. Beam K Variables by Boundary Conditions

Simply-Supported & Fixed-Pinned Fixed-Fixed Free-Free
Cantilever
4 4
c A⁄ I c A⁄ I 1.26 c A⁄I c A⁄ I
3 5

Note that I/A is the is the radius of gyration.

9
The K constants for typical cross-sections are:

Table 5.2. Beam Bending K Factors

Simply-Supported Fixed-Pinned Fixed-Fixed Free-Free
Cross-section & Cantilever
K K K K
Solid Circular 2 2.67 2.52 1.60
Thin Wall Pipe √2 1.89 1.78 1.13
Solid Rectangular √3 2.31 2.18 1.39
Solid Triangle 2.83 3.77 3.57 2.26
Thin Hollow Square √6 / 2 1.63 1.54 0.98
Wide Flange Beams 1.1 to 2.6 1.5 to 3.5 1.39 to 3.28 0.88 to 2.1

Figure 5.3. Wide Flange Beam

American Wide Flange Beams (ASTM A6) have K factors from 1.1 to 1.4 for bending about the
x-axis, with the W14x873 beam having the highest value. The K factors range from 1.9 to 2.6
for bending about the y-axis, with the W12x14 beam having the highest value.
Consider the bending vibration of a simply-supported beam subjected to resonant base
excitation. The modal stress due to bending is proportional to the modal velocity as shown in
the derivation in Section 13. The resulting stress-velocity equation is the same as equations
(5.4) and (5.5).

10
5.3 PLATE

5.3.1 Bending, Free Vibration

Consider bending in a rectangular plate with all edges simply-supported. The following
intermodal stress equations are derived in Section 98.

3 L +νL
σ , = ρc v ,
(5.7)
1−ν L +L

3 L +νL
σ , = ρc v ,
(5.8)
1−ν L +L

5.4 COMPLEX EQUIPMENT

Bateman wrote in Reference [8]:

Of the three motion parameters (displacement, velocity, and acceleration) describing

a shock spectrum, velocity is the parameter of greatest interest from the viewpoint
of damage potential. This is because the maximum stresses in a structure subjected
to a dynamic load typically are due to the responses of the normal modes of the
structure, that is, the responses at natural frequencies. At any given natural
frequency, stress is proportional to the modal (relative) response velocity.

Bateman then gave a formula equivalent to the following equation for mode n.

[σ ] = K Eρ [V ] (5.9)

where
K is a constant of proportionality dependent upon the geometry of the
structure, often assumed for complex equipment to be 4 < K < 8.

Sweitzer, Hull & Piersol also gave an equivalent form of equation (5.9) in Reference [9], with the
same constant of proportionality limits for complex equipment. They also noted that K ≈ 2 for
all normal modes of homogeneous plates and beams. Values of K > 3 may be used to account

11
for stress concentration per Reference [10]. Gaberson quipped that a high K value represents a
poor design. [11]
The equivalent strain relationship for complex equipment is

[σ ] ρ K
[ε ] = =K [V ] = [V ] (5.10)
E E c

5.5 K FACTOR SUMMARY FOR BASIC STRUCTURES

The following table is based on published references and the numerical experiments in this
document. It is intended for reasonable designs and is a “work-in-progress.”

Table 5.3. K Factors Summary

Structure K
Straight Rod Longitudinal 1
Straight Rod Longitudinal with End Mass 1 to 2.2
Tapered Rod 0.5 to 1
Ring, Breathing Mode 1
Beam Bending 1.0 to 3.8
Cantilever Beam Bending with Applied 1.57 to √3
Static Axial Load (See Note 2)
Beam Bending with Point Mass 1.1 to 4.0
Beam Bending with End Springs 0.15 to 2.0
Multispan Beam Bending 1.4 to 2.7
Plate Bending 1 to 2
Overall Range 0.5 to 4.0

Notes:
1. The beam bending factor in the previous table depends on the cross-section and
boundary conditions.

12
2. The stress-velocity relationship can adequately account for the dynamic stress in a beam
subjected to mean stress. The total stress is then calculated by adding the mean stress
to the dynamic stress.

13
6 STRESS IN AN INFINITE ROD IN LONGITUDINAL FREE VIBRATION

The following derivation is based on Reference [12].

Consider the free longitudinal vibration in a rod. The governing equation is

∂ u 1 ∂ u
= (6.1)
∂x c ∂t

Now consider that the rod has an infinite length.

A proposed traveling wave solution to the equation is

u(x, t) = A sin( kx − ωt − ϕ) (6.2)

Taking derivatives of the proposed traveling-wave solution

∂u
= kA cos( kx − ωt − ϕ) (6.3)
∂x

∂ u
= −k A sin( kx − ωt − φ) (6.4)
∂x

∂u
= −ωA cos( kx − ωt − ϕ) (6.5)
∂t

∂ u
= −ω A sin( kx − ωt − φ) (6.6)
∂t

By substitution,
1
−Ak sin( kx − ωt − ϕ) = [−ω A sin( kx − ωt − ϕ)] (6.7)
c

The following constraint results from the substitution

k = ω ⁄c (6.8)

14
The wavenumber is thus

k = ω⁄c (6.9)
Thus,

x
u(x, t) = A sin ω −t −ϕ (6.10)
c

∂ ω x
u(x, t) = A cos ω − t − ϕ (6.11)
∂x c c

∂ x
u(x, t) = −ω A cos ω − t − ϕ (6.12)
∂t c

Hooke’s law for stress and strain is

σ =Eε (6.13)

The axial strain is

∂ 1∂
ε= u(x, t) = − u(x, t) (6.14)
∂x c ∂t

The axial stress σ(x,t) is

∂ E∂ ωE x
σ(x, t) = E u(x, t) = − u(x, t) = A cos ω − t − ϕ (6.15)
∂x c ∂t c c

Define the velocity as v(x, t).

∂
v(x, t) = u(x, t) (6.16)
∂t

The axial stress and strain can thus be calculated from the velocity.

1 (6.17)
𝜀(x, t) = − v(x, t)
c
15
 E
σ(x, t) = − v(x, t) (6.18)
c

Or equivalently,

σ(x, t) = −ρc v(x, t) (6.19)

where ρc is the characteristic specific impedance of the rod material.

Now consider a hypothetical drop shock test machine using a longitudinal rod.

Avionics Box

Box-shaped Adapter

Guide Rod

Bearing Sleeve

Steel Pipe
Free Fall

Steel Base Plate

Figure 6.1. Longitudinal Rod Shock Test Machine

16
Now consider that the rod is initially at rest. It then undergoes a free-fall from an initial height. It is
then stopped abruptly by a rigid floor, whereupon it becomes a fixed-free beam.1

Calculate the maximum velocity and stress without the adapter and test item mounted at the top, per
Reference [13].

Let V be the initial velocity of the rod at the time of initial contact. Limit the response to the first mode.

The maximum modal response velocity is

4V (6.20)
u̇ , =
π

The maximum modal stress response is

4EV (6.21)
σ , =
πc

σ , = ρc u̇ , (6.22)

Stress is proportional to mass density. Consider two rods with the same length and diameter. One is
steel, and the other is aluminum. They are dropped from the same height and allowed to free fall. Each
rod contacts the ground with the same velocity. But the steel rod has about 2.8x the maximum stress
than aluminum due to the mass density difference.

1
I came up with this concept. Some colleagues of mine actually constructed it but had a problem
suppressing the rebound.

17
7 STRESS IN A FINITE, FIXED-FREE ROD, IN LONGITUDINAL FREE VIBRATION

The following derivation is taken from Reference [14].

Consider a long, slender, free-free rod that is dropped such that it has a uniform initial velocity
of V as it strikes the ground. Further assume that the rod becomes attached to the ground at
impact such that its boundary conditions become fixed-free.
The displacement is

8vL 1 nπx
u(x, t) = sin sin( ω t) (7.1)
π c n 2L
, , ,…

∂ 8vL π 1 nπx
u(x, t) = cos sin( ω t) (7.2)
∂x π c 2L n 2L
, , ,…

∂ 4v 1 nπx
u(x, t) = cos sin( ω t) (7.3)
∂x πc n 2L
, , ,…

where
nπc
ω = (7.4)
2L

∂ 8vL πc 1 nπx
u(x, t) = sin cos( ω t) (7.5)
∂t π c 2L n 2L
, , ,…

∂ 4v 1 nπx
u(x, t) = sin cos( ω t) (7.6)
∂t π n 2L
, , ,…

18
Now let

8vL
u(x, t) = u (x, t) (7.7)
π c
, , ,…

The modal displacement is

1 nπx
u (x, t) = sin sin( ω t) (7.8)
n 2L

∂ π nπx
u (x, t) = cos sin( ω t) (7.9)
∂x 2nL 2L

∂ πc nπx
u (x, t) = sin cos( ω t) (7.10)
∂t 2nL 2L

∂ ∂
u (x, t) = u (x, t) (7.11)
∂x ∂t

The maximum stress per mode n is thus proportional to the maximum modal velocity. Note
that the maximum stress location will vary from that of the maximum velocity.

The peak values of harmonic displacement, velocity and acceleration have their maxima at the
antinodes of motion, whereas the peak values of stress and strain have their maxima at the
nodes, per Reference [2]. This statement applies to a finite rod undergoing standing wave
oscillations.

E ∂ E 8vL πc
[σ ] = u = (7.12)
c ∂t c π c 2nL

19
 E ∂ 4vE
[σ ] = u = (7.13)
c ∂t nπc

Note

∂
u = ω [u ] (7.14)
∂t

The maximum stress is thus

E ∂ E
[σ ] = u = ω [u ] (7.15)
c ∂t c

Or in terms of the characteristic impedance,

∂
[σ ] = ρc u = ρc ω [u ] (7.16)
∂t

The maximum strain is

1 ∂ 1
[𝜀 ] = u = ω [u ] (7.17)
c ∂t c

The advantage of calculating stress from velocity is that the natural frequency term is not
required, as it is for calculating stress from displacement.

Furthermore, the characteristic impedance c plays the role of a transfer impedance which
expresses the ratio of the stress at a node of motion to the velocity at an associated antinode,
per Reference [2]. Again, this applies to a finite rod undergoing standing wave oscillations.

20
8 STRESS IN A FINITE, FIXED-FREE ROD, LONGITUDINAL RESPONSE EXCITED BY
RESONANT BASE EXCITATION

The following derivation is taken from Reference [15].

−Γ U (x)
U(x, ω) = Ẅ(ω) (8.1)
(ω − ω ) + j(2ξωω )

The mode shape is

2 ω x
U (x) = sin (8.2)
mL c

d ω 2 ω x
U (x) = cos (8.3)
dx c mL c

The strain is

d
∂ −Γ U (x)
U(x, ω) = Ẅ(ω) dx (8.4)
∂x (ω − ω ) + j(2ξωω )

ω x
∂ 1 2 −Γ ω cos
U(x, ω) = Ẅ(ω) c (8.5)
∂x c mL (ω − ω ) + j(2ξωω )

On a modal basis,

ω x
∂ 1 2 −Γ ω cos
= Ẅ(ω) c
U (x, ω) (8.6)
∂x c mL (ω − ω ) + j(2ξωω )

The phase can be ignored.

21
 ∂ 1 2 Γω
U (x, ω) = Ẅ(ω) (8.7)
∂x c mL (ω − ω ) + (2ξωω )

The velocity v in the frequency domain is

−Γ U (x)
v(x, ω) = jωẄ(ω) (8.8)
(ω − ω ) + j(2ξωω )

2 ω x
U (x) = sin (8.9)
mL c

ω x
2 −Γ sin
v(x, ω) = jωẄ(ω) c (8.10)
mL (ω − ω ) + j(2ξωω )

2 −Γ
[v (x, ω)] = ω Ẅ(ω) (8.11)
mL (ω − ω ) + j(2ξωω )

The phase can be ignored.

2 Γ
[v (x, ω)] = ωẄ(ω) (8.12)
mL (ω − ω ) + (2ξωω )

A further development of the relationship between velocity and stress for this case requires
that the excitation be at the natural frequency, ω = ω .

∂ 1 2 Γω
U (x, ω ) = Ẅ(ω ) (8.13)
∂x c mL (ω − ω ) + (2ξωω )

22
 ∂ Γ 2
U (x, ω ) = Ẅ(ω ) (8.14)
∂x 2ξω c mL

2 Γ
[v (x, ω )] = ω Ẅ(ω ) (8.15)
mL 2ξ ω

Γ 2
[v (x, ω )] = Ẅ(ω ) (8.16)
2ξ ω mL

∂ 1
U (x, ω ) = [v (x, ω )] (8.17)
∂x c

The maximum stress is

∂ E
[σ ] =E U (x, ω ) = [v (x, ω )] (8.18)
∂x c

23
9 ROD WITH END MASS, LONGITUDINAL VIBRATION

E, A, ρ
F
M

Figure 9.1. Rod with End Mass

Consider a fixed-free rod with a point mass on its free end. The rod itself is aluminum, circular
cross-section, diameter = 2 in, and length =24 in. The point mass value is variable.
Solve for the normal modes.
Next, apply a steady-state sine force at the free end with an excitation frequency equal to the
first natural frequency.
The rod is analyzed using the finite element method in Reference [16]. The model consists of
100 elements.
The response is limited to the first mode with an amplification factor Q=10.
The peak velocity occurs at the free end. The peak stress occurs at the fixed end.

The goal is to solve for the scale factor K.

σ
K= (9.1)
ρc v ,

24
 Fundamental Mode, Mass-Normalized Displacements
8
0.5
1
7
2
3
6 4

0
0 5 10 15 20 25
x (inch)

Figure 9.2. Rod with Variable End Mass, Fundamental Mode Shape

The legend value is the ratio of the end mass to the rod mass. The mode
shape tends to become linear for increasing mass ratios.

The forced response results are shown in the following table.

Table 9.1. Rod with End Mass, K Values

Ratio of End Fundamental
Rod Mass End Mass
Mass to Rod Frequency K
(lbm) (lbm)
Mass (Hz)
7.54 3.77 0.5 1406 1.13
7.54 7.54 1 1122 1.32
7.54 15.1 2 851 1.65
7.54 22.6 3 714 1.92
7.54 30.2 4 625 2.17

25
10 TAPERED ROD, LONGITUDINAL VIBRATION

Figure 10.1. Tapered Rod, Circular Cross-section, Aluminum, Fixed-Free

The rod is analyzed using the finite element method in Reference [16]. The model consists of
100 elements.
The response is limited to the first mode with amplification factor Q=10.
Several geometrical configurations are analyzed by varying the length and the diameters.
Each configuration is excited at its fundamental modal frequency with a steady-state sine force.
The peak velocity occurs at the free end. The results are shown in the following table.

The goal is to solve for the scale factor K.

σ
K= (10.1)
ρc v ,

26
 Table 10.1. Tapered Rod K Factors, Part I
Length Diameter at Diameter Fundamental Peak Stress
(in) Fixed End at Free fn (Hz) Location (in) K
(in) End (in)
16 1.33 0.67 3910 3.68 0.67
24 2 1 2607 5.52 0.67
32 2.67 1.33 1955 7.36 0.67

Notes for Table 10.1.

1. The free end diameter is always one-half of the fixed end diameter.
2. The fixed end diameter is always 1/12th of the length.

Table 10.2. Tapered Rod K Factors, Part 2

Length Diameter at Diameter Fundamental Peak Stress
(in) Fixed End at Free fn (Hz) Location (in) K
(in) End (in)
24 1 0.5 3910 5.52 0.67
24 2 1 2607 5.52 0.67
24 3 1.5 1955 5.52 0.67

Notes for Table 10.2.

1. The free end diameter is always one-half of the fixed end diameter.
2. The fixed end diameter to length ratio is variable

27
 Table 10.3. Tapered Rod K Factors, Part 3
Length Diameter at Diameter Fundamental Peak Stress
(in) Fixed End at Free fn (Hz) Location (in) K
(in) End (in)
24 2 0.25 3085 5.76 0.58
24 2 0.5 2958 5.76 0.60
24 2 1 2607 5.52 0.67
24 2 1.5 2288 3.60 0.80
24 2 2 2047 0 1.00

Notes for Table 10.3.

1. The free end to the fixed end diameter ratio is variable
2. The fixed end diameter to length ratio is variable

28
 10 -3 Displacement Maximum per Length
1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0
0 5 10 15 20 25
x (in)

Figure 10.2. Displacement, Length=24 in, Fixed End Diameter=2 in, Free End Diameter = 1 in

The excitation force was 100 lbf at 2607 Hz. The displacement plot is essentially the
fundamental mode shape. All points are in phase for the fundamental mode. The
corresponding stress is shown in the next plot.

29
 Stress Maximum per Length
600

500

400

300

200

100

0
0 5 10 15 20 25
Time (sec)

Figure 10.3. Stress, Length=24 in, Fixed End Diameter=2 in, Free End Diameter = 1 in

30
 11 STRESS IN A SHEAR BEAM IN FREE VIBRATION

The following derivation is based on Reference [17].

Consider a beam which undergoes shear displacement only. Note that a shear beam model is
used for seismic analysis of certain civil engineering structures.

x u (x, t)

Figure 11.1. Beam Shear

Assume a uniform cross-section and mass density.
The transverse shear displacement u(x, t) is governed by the equation

∂ u ∂ u
ρA = KGA (11.1)
∂t ∂x

∂ u KG ∂ u
= (11.2)
∂t ρ ∂x

By substitution,

(11.3)
KG ∂ ∂
[U(x)T(t)] = [U(x)T(t)]
ρ ∂x ∂t

Perform the partial differentiation.

KG
U (x)T(t) = U(x)T̈(t) (11.4)
ρ

Divide through by U(x)T(t).

31
 KG U ″ (x) T̈(t)
= (11.5)
ρ U(x) T(t)

Each side of the above equation must equal a constant. Let ω be a constant.

KG U (x) T̈(t)
= = −ω (11.6)
ρ U(x) T(t)

The time equation is

T̈(t)⁄T(t) = −ω (11.7)

T̈(t) = −ω T(t) (11.8)

T̈(t) + ω T(t) = 0 (11.9)

Propose a solution

T(t) = a sin(ωt) + b cos(ωt) (11.10)

Ṫ(t) = aω cos(ωt) − b ω sin(ωt) (11.11)

T̈(t) = −a ω sin(ωt) − bω cos(ωt) (11.12)

Verify the proposed solution. Substitute into equation (11.6).

−a ω sin(ωt) − bω cos(ωt) + ω [a sin(ωt) + b cos(ωt)] = 0 (11.13)

0=0 (11.14)

Equation (D-10) is thus verified as a solution.

There is not a unique ω, however, in equation (11.6). This is demonstrated later in the
derivation. Thus, a subscript n must be added as follows.

32
 T (t) = a sin(ω t) + b cos(ω t) (11.15)

The spatial equation is

KG U ″ (x)
= −ω (11.16)
ρ U(x)

The shear wave speed is

c= KG⁄ρ (11.17)

ω
U ″ (x) = − U(x) (11.18)
c

ω
U ″ (x) + U(x) = 0 (11.19)
c

The displacement solution is

ωx ωx
U(x) = d sin + e cos (11.20)
c c

The slope equation is

ω ωx ωx
U′(x) = d cos − e sin (11.21)
c c c

Now consider that the shear beam is fixed-free. The left boundary conditions is

u (0, t) = 0 (zero displacement) (11.22)

U(0)T(t)=0 (11.23)

U(0)=0 (11.24)

33
The right boundary condition is

∂
u(x, t)| =0 (zero stress) (11.25)
∂x

U’(L)T(t)=0 (11.26)

U’(L)=0 (11.27)

Substitute equation (11.22) into(11.20).

e=0 (11.28)

Thus, the displacement equation becomes

ωx
U(x) = d sin (11.29)
c

ω ωx
U′(x) = d cos (11.30)
c c

Substitute equation (11.27) into equation (11.30).

ωL
d cos =0 (11.31)
c

The constant d must be non-zero for a non-trivial solution. Thus,

ω L 2n − 1
= π, n = 1, 2, 3, … (11.32)
c 2

2n − 1 c
ω = π , n = 1, 2, 3 … (11.33)
2 L

The ω term is given a subscript n because there are multiple roots.

34
 2n − 1 c
ω = π , n = 1, 2, 3, … (11.34)
2 L

The displacement function for the fixed-free beam is

ω x
U (x) = d sin (11.35)
c

(2n − 1)πx
U (x) = d sin (11.36)
2L

T (t) = a sin(ω t) + b cos(ω t) (11.37)

u(x, t) = U (x)T (t) (11.38)

(2n − 1)πx
u(x, t) = sin a sin(ω t) + b cos(ω t) (11.39)
2L

∂ (2n − 1)πx
u(x, t) = ω sin a cos(ω t) − b sin(ω t) (11.40)
∂t 2L

2n − 1 c
ω = π , n = 1, 2, 3, … (11.41)
2 L

∂ 2n − 1 c
u(x, t) = π (11.42)
∂t 2 L

∂ (2n − 1)π (2n − 1)πx

u(x, t) = cos a sin(ω t) + b cos(ω t) (11.43)
∂x 2L 2L

35
 ∂ 2n − 1
u(x, t) = π (11.44)
∂x 2L

∂u 1 ∂u
= (11.45)
∂x c ∂t

∂
V = GA u(x, t)
∂x
(2n − 1)π (2n − 1)πx (11.46)
= GA cos a sin(ω t) + b cos(ω t)
2L 2L

The modal shear stress for a rectangular beam is

V ∂
τ , = 1.5 = 1.5 G u (x, t) (11.47)
A ∂x

V 1.5 G ∂
τ , = 1.5 = u (x, t) (11.48)
A c ∂t

V 1.5 ρc ∂
τ , = 1.5 = u (x, t) (11.49)
A k ∂t

The shear factors for two cross-sections are given in the following table, as taken from
Reference [18].

Table 11.1. Thomson’s Shear Factors

Shear Factor
Cross-Section
K
Rectangular 2/3
Circular 3/4

36
12 STRESS DUE TO BENDING IN A BERNOULLI-EULER BEAM IN FREE VIBRATION

12.1 SIMPLY-SUPPORTED BEAM

Consider the following simply-supported beam.

EI, ˆ

Figure 12.1. Beam Bending

Recall that the governing differential equation is

∂ y ∂ y
−EI =ρ (12.1)
∂x ∂t

The spatial solution from Reference [19] is

Y(x) = a sinh(βx) + a cosh(βx) + a sin(βx) + a cos(βx) (12.2)

d
Y(x) = a β sinh(βx) + a β cosh(βx) − a β sin(βx) − a β cos(βx) (12.3)
dx

The boundary conditions at the left end x = 0 are

Y(0) = 0 (zero displacement) (12.4)

d Y
=0 (zero bending moment) (12.5)
dx

37
The boundary conditions at the free end x = L are

Y(L) = 0 (zero displacement) (12.6)

d Y
=0 (zero bending moment) (12.7)
dx

Apply boundary condition (E-4) to (E-2).

a +a =0 (12.8)

a = −a (12.9)

Apply boundary condition (E-5) to (E-3).

a −a =0 (12.10)

a =a (12.11)

Equations (12.8) and (12.10) can only be satisfied if

a =0 (12.12)

a =0 (12.13)

The spatial equations thus simplify to

Y(x) = a sinh(βx) + a sin(βx) (12.14)

d
Y(x) = a β sinh(βx) − a β sin(βx) (12.15)
dx

Apply boundary condition (12.6) to equation (12.14).

a sinh(βL) + a sin(βL) = 0 (12.16)

38
Apply boundary condition(12.7) to equation (12.15).

a β sinh(βL) − a β sin(βL) = 0 (12.17)

a sinh(βL) − a sin(βL) = 0 (12.18)

sinh(βL) sin(βL) a 0
= (12.19)
sinh(βL) −sin(βL) a 0

By inspection the above equation can only be satisfied if a1 = 0 and a3 = 0. Set the determinant
to zero in order to obtain a nontrivial solution.

−sin(βL)sinh(βL) − sin(βL)sinh(βL) = 0 (12.20)

−2 sin(βL)sinh(βL) = 0 (12.21)

sin(βL)sinh(βL) = 0 (12.22)

The previous equation is satisfied if

β L = nπ, n = 1, 2, 3, … (12.23)

nπ
β = , n = 1, 2, 3, … (12.24)
L

The natural frequency term ω is

ω =β EI⁄ρ (12.25)

nπ (12.26)
ω = EI⁄ρ , n = 1,2,3, …
L

39
 1 nπ (12.27)
f = EI⁄ρ , n = 1,2,3, …
2π L

1 nπ
f = EI⁄ρ , n = 1,2,3, … (12.28)
2π L

The eigenvector and its second derivative at this point are

Y(x) = a sinh(βx) + a sin(βx) (12.29)

d Y(x)
= a β sinh(βx) − a β sin(βx) (12.30)
dx

The eigenvector derivation requires some creativity. Recall

Y(L) = 0 (zero displacement) (12.31)

d Y
=0 (zero bending moment) (12.32)
dx

Thus,

d Y
+Y = 0 for x = L and  n L  n, n = 1,2,3, … (12.33)
dx

nπ nπ
1− a sinh(nπ) + 1 − a sin(nπ) = 0, n = 1,2,3 … (12.34)
L L

The sin(nπ) term is always zero. Thus a1 = 0.

The eigenvector for all n modes is

Y (x) = a sin(nπx/L) (12.35)

40
Mass-normalize the eigenvectors as follows

ρY (x)dx = 1 (12.36)

ρa sin (nπx/L) dx = 1 (12.37)

ρa
[1 − cos( 2nπx/L)] = 1 (12.38)
2

ρa 1
x− sin( 2nπx/L) =1 (12.39)
2 2β

ρa L
=1 (12.40)
2

2
a = (12.41)
ρL

2
a = (12.42)
ρL

2
Y (x) = sin(nπx/L) (12.43)
ρL

The complete solution is thus

2
y(x, t) = sin(nπx/L) a sin(ω t) + b cos(ω t) (12.44)
ρL

41
 ∂ nπ 2
y(x, t) = cos(nπx/L) a sin(ω t) + b cos(ω t) (12.45)
∂x L ρL

∂ nπ 2
y(x, t) = − sin(nπx/L) a sin(ω t) + b cos(ω t) (12.46)
∂x L ρL

nπ
ω = EI⁄ρ , n = 1,2,3, … (12.47)
L

∂ ρ 2
y(x, t) = −ω sin(nπx/L) a sin(ω t) + b cos(ω t) (12.48)
∂x EI ρL

Assume a + b =1 (12.49)

∂ ρ 2 (12.50)
y (x, t) =ω
∂x EI ρL

∂ 2
y (x, t) =ω (12.51)
∂t ρL

∂ ρ ∂
y (x, t) = y (x, t) (12.52)
∂x EI ∂t

∂ ρA ∂
y (x, t) = y (x, t) (12.53)
∂x EI ∂t

Let c be the longitudinal wave speed in the material.

42
 c = E⁄ρ (12.54)

∂ 1 A ∂
y (x, t) = y (x, t) (12.55)
∂x c I ∂t

∂ EI A ∂
M = EI y (x, t) = y (x, t) (12.56)
∂x c I ∂t

M c
σ = (12.57)
I

∂ Ec A ∂
𝜎 =Ec y (x, t) = y (x, t) (12.58)
∂x c I ∂t

(12.59)
∂ EAρ ∂
𝜎 = Ec y (x, t) =c y (x, t)
∂x I ∂t

The maximum strain is

∂ Aρ ∂
ε =c y (x, t) =c y (x, t) (12.60)
∂x EI ∂t

43
12.2 FIXED-FREE BEAM

EI, 

Figure 12.2. Fixed-Free Beam

The eigenvalues from Reference [19] are

n β L
1 1.87510
2 4.69409
3 (2n-1)/2

The mode shape equation is

1 cos(βL) + cosh(βL)
Y(x) = [cosh(βx) − cos(βx)] − [sinh(βx) − sin(βx)]
ρL sin(βL) + sinh(βL)
(12.61)

d β cos(βL) + cosh(βL)
Y(x) = [sinh(βx) + sin(βx)] − [cosh(βx) − cos(βx)]
dx ρL sin(βL) + sinh(βL)

(12.62)

44
d β cos(βL) + cosh(βL)
Y(x) = [cosh(βx) + cos(βx)] − [sinh(βx) + sin(βx)]
dx ρL sin(βL) + sinh(βL)

(12.63)

Fixed-Pree Beam, Fundamental Mode

2

1.8

1.6

1.4

1.2

0.8

0.6

0.4

0.2

0
0 10 20 30 40 50 60 70 80 90 100
Percent Length

Figure 12.3. Fixed-Free Beam, Bending Mode Shape

For the fundamental mode, the peak stress occurs at the fixed end, and the peak velocity at the
free end.

45
The complete solution is

y(x, t) = Y (x) a sin(ω t) + b cos(ω t) (12.64)

∂ d
y(x, t) = Y (x) a sin(ω t) + b cos(ω t) (12.65)
∂x dx

Assume a + b =1 (12.66)

∂ 2β (12.67)
y (x, t) =
∂x ρL

∂ 2ω
y (x, t) = (12.68)
∂t ρL

ω =β EI⁄ρ (12.69)

∂ 2β √EI
y (x, t) = (12.70)
∂t ρL ρ

∂ 2β √EI
y (x, t) = (12.71)
∂t ρL ρA

∂ ρA ∂
y (x, t) = y (x, t) (12.72)
∂x EI ∂t

∂ ρA ∂
y (x, t) = y (x, t) (12.73)
∂x EI ∂t

46
Let c be the longitudinal wave speed in the material.

c = E⁄ρ (12.74)

∂ 1 A ∂
y (x, t) = y (x, t) (12.75)
∂x c I ∂t

∂ EI A ∂
M = EI y (x, t) = y (x, t) (12.76)
∂x c I ∂t

M c
σ = (12.77)
I

∂ Ec A ∂
𝜎 =Ec y (x, t) = y (x, t) (12.78)
∂x c I ∂t

∂ EAρ ∂
𝜎 = Ec y (x, t) = c y (x, t) (12.79)
∂x I ∂t

47
12.3 FIXED-PINNED BEAM

EI,

Figure 12.4. Fixed-Pinned Beam

The eigenvalues from Reference [19] are

n β L
1 3.9266
2 7.0686
3 10.2102
4 13.3518
5 16.4934

The mode shape equation is

1 [sinh(βL) + sin(βL)]
Y(x) = [sinh(βx) − sin(βx)] − [cosh(βx) − cos(βx)]
ρL [cosh(βL) + cos(βL)]
(12.80)

d β [sinh(βL) + sin(βL)]
Y(x) = [cosh(βx) − cos(βx)] − [sinh(βx) + sin(βx)]
dx ρL [cosh(βL) + cos(βL)]

(12.81)

48
d β [sinh(βL) + sin(βL)]
Y(x) = [sinh(βx) + sin(βx)] − [cosh(βx) + cos(βx)]
dx ρL [cosh(βL) + cos(βL)]

(12.82)

Fixed-Pinned Beam, Fundamental Mode

1.6

1.4

1.2

0.8

0.6

0.4

0.2

0
0 10 20 30 40 50 60 70 80 90 100
Percent Length

Figure 12.5. Fixed-Pinned Beam, Fundamental Bending Mode Shape

The peak stress occurs at the fixed end. A numerical analysis shows the peak displacement for
the fundamental mode occurs at 0.581L relative to the fixed end with an amplitude of
1.5 ω ρL .

49
Here is the Matlab script used to determine the point of maximum deflection for the
fundamental mode:
L = 1;
BL = 3.9266;
B = BL / L;

num = sinh(BL) + sin(BL);

den = cosh(BL) + cos(BL);
A = num / den;

x0 = 0.5; % initial point

z = fzero(@(x) cosh(B*x) - cos(B*x) - A*(sinh(B*x) + sin(B*x)), x0);

yy = sinh(B*z) - sin(B*z) - A*(cosh(B*z) - cos(B*z));

fprintf('\n Point of maximum deflection: %7.3g L \n',z);

fprintf(' Amplitude: %7.3g \n',abs(yy));

The complete displacement solution is

y(x, t) = Y (x) a sin(ω t) + b cos(ω t) (12.83)

∂ d
y(x, t) = Y (x) a sin(ω t) + b cos(ω t) (12.84)
∂x dx

Assume a + b =1 (12.85)

∂ 2β (12.86)
y (x, t) =
∂x ρL

∂ 1.5 ω
y (x, t) = (12.87)
∂t ρL

50
 ω =β EI⁄ρ (12.88)

∂ 1.5 β √EI
y (x, t) = (12.89)
∂t ρL ρ

∂ 1.5 β √EI
y (x, t) = (12.90)
∂t ρL ρA

∂ 2 ρA ∂
y (x, t) = y (x, t) (12.91)
∂x 1.5 EI ∂t

∂ 4 ρA ∂
y (x, t) = y (x, t) (12.92)
∂x 3 EI ∂t

Let c be the longitudinal wave speed in the material.

c = E⁄ρ (12.93)

∂ 4 A ∂
y (x, t) = y (x, t) (12.94)
∂x 3c I ∂t

∂ 4 EI A ∂
M = EI y (x, t) = y (x, t) (12.95)
∂x 3c I ∂t

M c
σ = (12.96)
I

51
 ∂ 4Ec A ∂
𝜎 =Ec y (x, t) = y (x, t) (12.97)
∂x 3c I ∂t

∂ 4 c EAρ ∂
𝜎 = Ec y (x, t) = y (x, t) (12.98)
∂x 3 I ∂t

12.4 FREE-FREE BEAM

EI, 

Figure 12.6. Free-Free Beam

The eigenvalues from Reference [19] are

n β L
1 0
2 4.73004
3 7.85320
4 10.9956

52
 Free-Free Beam, Fundamental Mode
1.5

0.5

-0.5

-1

-1.5

-2

-2.5
0 10 20 30 40 50 60 70 80 90 100
Percent Length

Figure 12.7. Fee-Free Beam, Fundamental Bending Mode Shape

The mode shape equation is

1 − cosh(βL) + cos(βL)
Y(x) = [sinh(βx) + sin(βx)] + [cosh(βx) + cos(βx)]
ρL sinh(βL) + sin(βL)
(12.99)

d β − cosh(βL) + cos(βL)
Y(x) = [cosh(βx) + cos(βx)] + [sinh(βx) − sin(βx)]
dx ρL sinh(βL) + sin(βL)

(12.100)

53
d β − cosh(βL) + cos(βL)
Y(x) = [sinh(βx) − sin(βx)] + [cosh(βx) − cos(βx)]
dx ρL sinh(βL) + sin(βL)

(12.101)

For the first elastic mode, the peak stress is at the midpoint, and the peak velocity is at each of
the two free ends. The complete solution is

y(x, t) = Y (x) a sin(ω t) + b cos(ω t) (12.102)

∂ d
y(x, t) = Y (x) a sin(ω t) + b cos(ω t) (12.103)
∂x dx

Assume a + b =1 (12.104)

∂ 1.62 β (12.105)
y (x, t) =
∂x ρL

∂ 2ω
y (x, t) = (12.106)
∂t ρL

ω =β EI⁄ρ (12.107)

∂ 2β √EI
y (x, t) = (12.108)
∂t ρL ρ

∂ 2β √EI
y (x, t) = (12.109)
∂t ρL ρA

54
 ∂ ρA ∂
y (x, t) = y (x, t) (12.110)
∂x EI ∂t

∂ 1.61 ρA ∂
y (x, t) = y (x, t) (12.111)
∂x 2 EI ∂t

∂ ρA ∂
y (x, t) = 0.8 y (x, t) (12.112)
∂x EI ∂t

Let c be the longitudinal wave speed in the material.

c = E⁄ρ (12.113)

∂ 0.8 A ∂
y (x, t) = y (x, t) (12.114)
∂x c I ∂t

∂ 0.8 EI A ∂
M = EI y (x, t) = y (x, t) (12.115)
∂x c I ∂t

M c
σ = (12.116)
I

∂ 0.8 E c A ∂
𝜎 =Ec y (x, t) = y (x, t) (12.117)
∂x c I ∂t

55
 ∂ EAρ ∂
𝜎 = Ec y (x, t) = 0.8 c y (x, t) (12.118)
∂x I ∂t

12.5 FIXED-FIXED BEAM

EI, ρ

Figure 12.8. Fixed-Fixed Beam

The eigenvalues from Reference [19] are

n β L
1 4.73004
2 7.85321
3 10.9956
4 14.13717
5 17.27876

56
 Fixed-Fixed Beam, Fundamental Mode
1.6

1.4

1.2

0.8

0.6

0.4

0.2

0
0 10 20 30 40 50 60 70 80 90 100
Percent Length

Figure 12.9. Fixed-Fixed Beam, Bending Mode Shape

The mode shape equation is

1 sinh(βL) + sin(βL)
Y(x) = [cosh(βx) − cos(βx)] − [sinh(βx) − sin(βx)]
ρL cosh(βL) − cos(βL)

(12.119)

d β [sinh(βL) + sin(βL)]
Y(x) = [sinh(βx) + sin(βx)] − [cosh(βx) − cos(βx)]
dx ρL [cosh(βL) − cos(βL)]

(12.120)

57
d β [sinh(βL) + sin(βL)]
Y(x) = [cosh(βx) + cos(βx)] − [sinh(βx) + sin(βx)]
dx ρL [cosh(βL) − cos(βL)]

(12.121)

For the fundamental mode, the peak stress occurs at each fixed end, and the peak velocity
occurs in the middle. A numerical analysis shows the maximum peak displacement for the
fundamental mode has an amplitude of 1.59 ω ρL . The complete solution is

y(x, t) = Y (x) a sin(ω t) + b cos(ω t) (12.122)

∂ d
y(x, t) = Y (x) a sin(ω t) + b cos(ω t) (12.123)
∂x dx

Assume a + b =1 (12.124)

∂ 2β (12.125)
y (x, t) =
∂x ρL

∂ 1.59 ω
y (x, t) = (12.126)
∂t ρL

ω =β EI⁄ρ (12.127)

∂ 1.59 β √EI
y (x, t) = (12.128)
∂t ρL ρ

∂ 1.59 β √EI
y (x, t) = (12.129)
∂t ρL ρA

58
 ∂ 2 ρA ∂
y (x, t) = y (x, t) (12.130)
∂x 1.59 EI ∂t

∂ ρA ∂
y (x, t) = 1.26 y (x, t) (12.131)
∂x EI ∂t

Let c be the longitudinal wave speed in the material.

c = E⁄ρ (12.132)

∂ 1.26 A ∂
y (x, t) = y (x, t) (12.133)
∂x c I ∂t

∂ 1.26 EI A ∂
M = EI y (x, t) = y (x, t) (12.134)
∂x c I ∂t

M c
σ = (12.135)
I

∂ 1.26 E c A ∂
𝜎 =Ec y (x, t) = y (x, t) (12.136)
∂x c I ∂t

∂ EAρ ∂
𝜎 = Ec y (x, t) = 1.26 c y (x, t) (12.137)
∂x I ∂t

59
13 STRESS DUE TO BENDING IN A BERNOULLI-EULER BEAM EXCITED BY BASE
EXCITATION

13.1 SIMPLY-SUPPORTED BEAM

Consider a beam simply-supported at each end subjected to base excitation.

y(x,t)
EI, ρ, L

w(t)

Figure 13.1. Beam, Base Excitation

The following is taken from Reference [20]. The governing differential equation is

∂ y ∂ y ∂ w
EI +ρ = −ρ (13.1)
∂x ∂t ∂t

The mass-normalized mode shapes are

2
Y (x) = sin β x (13.2)
ρL

d 2
Y (x) = − β sin(β x) (13.3)
dx ρL

d
Y (x) = − β Y (x) (13.4)
dx

60
 The eigenvalues are

nπ
β = , n = 1,2,3, … (13.5)
L

The natural frequencies are

ω =β EI/ρ (13.6)

The relative displacement response Y(x, ω) to base acceleration is

−Γ Y (x)
Y(x, ω) = Ẅ(ω) (13.7)
(ω − ω ) + j2ξ ωω

d
∂ −Γ Y (x)
Y(x, ω) = Ẅ(ω) dx (13.8)
∂x (ω − ω ) + j2ξ ωω

∂ Γ β Y (x)
Y(x, ω) = Ẅ(ω) (13.9)
∂x (ω − ω ) + j2ξ ωω

On a modal basis,

∂ Γ β
Y (x, ω) = Ẅ(ω) [Y (x)] (13.10)
∂x (ω − ω ) + j2ξ ωω

The phase can be ignored.

∂ Γ β
Y (x, ω) = Ẅ(ω) [Y (x)] (13.11)
∂x (ω − ω ) + (2ξ ωω )

61
The relative velocity v in the frequency domain is

−Γ Y (x)
v(x, ω) = jωẄ(ω) (13.12)
(ω − ω ) + j2ξ ωω

On a modal basis,

−Γ
[v (x, ω)] = jωẄ(ω) [Y (x)] (13.13)
(ω − ω ) + j2ξ ωω

The phase can be ignored.

Γ
[v (x, ω)] = ωẄ(ω) [Y (x)] (13.14)
(ω − ω ) + (2ξ ωω )

The second derivative and velocity are related by

∂ β
Y (x, ω) = [v (x, ω)] (13.15)
∂x ω

ω =β EI/ ρ (13.16)

∂ ω ρ
Y (x, ω) = [v (x, ω)] (13.17)
∂x ω EI

Note consider the case where ω = ω .

∂ ρ
Y (x, ω) = [v (x, ω)] (13.18)
∂x EI

∂ 1
Y (x, ω) = [v (x, ω)] (13.19)
∂x c √I

62
The bending moment is

∂ E√I
M = EI Y (x, ω) = [v (x, ω)] (13.20)
∂x c

∂
M = EI Y (x, ω) = EAIρ[v (x, ω)] (13.21)
∂x

The bending stress is

∂ EAρ
σ =cE Y (x, ω ) =c [v (x, ω)] (13.22)
∂x I

This concept can be extended to the multi-modal response for the special case of a simply-
supported beam.
Again,

∂ Γ β Y (x)
Y(x, ω) = Ẅ(ω) (13.23)
∂x (ω − ω ) + j2ξ ωω

−Γ Y (x)
v(x, ω) = jωẄ(ω) (13.24)
(ω − ω ) + j2ξ ωω

Thus

∂ −β
Y(x, ω) = v(x, ω) (13.25)
∂x jω

∂ jβ
Y(x, ω) = v(x, ω) (13.26)
∂x ω

63
13.2 CANTILEVER BEAM

Now consider a fixed-free beam subjected to base excitation.

y(x, t)

w(t)

Figure 13.2. Cantilever Beam, Base Excitation

The eigenvalues are

n n L
1 1.875104
2 4.69409
3 7.85476
4 10.99554
5 (2n-1)/2

Note that the root value formula for n > 5 is approximate.

The natural frequencies are

ω =β EI/ρ (13.27)

64
The mass-normalized mode shapes can be represented as
1
Y (x) = {[cosh(β x) − cos(β x)] − D [sinh(β x) − sin(β x)] } (13.28)
ρL

where

cos β L + cosh β L
D = (13.29)
sin β L + sinh β L

β
Y (x) = {[sinh(β x) + sin(β x)] − D [cosh(β x) − cos(β x)] } (13.30)
ρL

β
Y (x) = {[cosh(β x) + cos(β x)] − D [sinh(β x) + sin(β x)] } (13.31)
ρL

β
Y (x) = {[sinh(β x) − sin(β x)] − D [cosh(β x) + cos(β x)] } (13.32)
ρL

Note that

1 cos β L + cosh β L
Y (L) = cosh β L − cos β L − sinh β L − sin β L
ρL sin β L + sinh β L

(13.33)

Table 13.1. Cantilever Beam Eigenvalues

n β L Y (L)
1 1.875104 2 ρL
2 4.69409 -2 ρL
3 7.85476 2 ρL
4 10.99554 -2 ρL

65
Thus,
Y (0) = β Y (L) (13.34)

The relative displacement response Y(x, ω) to base acceleration is

−Γ Y (x)
Y(x, ω) = Ẅ(ω) (13.35)
(ω − ω ) + j2ξ ωω

d
∂ −Γ Y (x)
Y(x, ω) = Ẅ(ω) dx (13.36)
∂x (ω − ω ) + j2ξ ωω

On a modal basis,

d
∂ −Γ Y (x)
̈ dx
Y (x, ω) = W(ω) (13.37)
∂x (ω − ω ) + j2ξ ωω

The maximum occurs at x=0. Solve for the first mode.

d
∂ −Γ Y (0)
Y (0, ω) = Ẅ(ω) dx (13.38)
∂x (ω − ω ) + j2ξ ωω

∂ −Γ β
Y (0, ω) = Ẅ(ω) [Y (L)] (13.39)
∂x (ω − ω ) + j2ξ ωω

The phase can be ignored.

66
 ∂ Γ β
Y (0, ω) = Ẅ(ω) [Y (L)] (13.40)
∂x (ω − ω ) + (2ξ ωω )

The relative velocity v in the frequency domain is

−Γ Y (x)
v(x, ω) = jωẄ(ω) (13.41)
(ω − ω ) + j2ξ ωω

The first modal velocity is

−Γ
[v (x, ω)] = jωẄ(ω) [Y (x, ω)] (13.42)
(ω − ω ) + j2ξ ωω

The phase can be ignored.

Γ
[v (x, ω)] = ωẄ(ω) [Y (x, ω)] (13.43)
(ω − ω ) + (2ξ ωω )

The maximum velocity occurs at x=L.

Γ
v (L, ω) = ωẄ(ω) Y (L) (13.44)
(ω − ω ) + (2ξ ωω )

∂ β
Y (0, ω) = v (L, ω) (13.45)
∂x ω

The bending moment is

∂ β
M(0, ω) = EI Y (0, ω) = EI v (L, ω) (13.46)
∂x ω

The bending stress is

∂ β
σ =cE Y (0, ω) = c E v (L, ω) (13.47)
∂x ω

67
14 CANTILEVER BEAM EXAMPLE

The objectives of this section are to

1. Demonstrate that the stress-velocity proportionality factor for beams is uniform for all modes.
2. Perform a base input PSD case to determine whether the stress-velocity relationship can be
extended to multi-mode response (yes, but more research is needed).

The analysis is performed using the classical method.

Parameters
Aluminum
Solid Cylinder, diameter = 0.25 inch
Length = 12 inch
Q=10 for all modes

Figure 14.1. Cantilever Beam, Base Excitation

Mode 1 fn= 47.72 Hz

1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

0 2 4 6 8 10 12
x (inch)

Figure 14.2. Cantilever Beam, Bending Mode 1

68
 Mode 2 fn= 299 Hz
1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

0 2 4 6 8 10 12
x (inch)

Figure 14.3. Cantilever Beam, Bending Mode 2

Mode 3 fn= 837.3 Hz

1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

0 2 4 6 8 10 12
x (inch)

Figure 14.4. Cantilever Beam, Bending Mode 3

69
 Mode 4 fn= 1641 Hz
1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

0 2 4 6 8 10 12
x (inch)

Figure 14.5. Cantilever Beam, Bending Mode 4

Mode 5 fn= 2712 Hz

1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

0 2 4 6 8 10 12
x (inch)

Figure 14.6. Cantilever Beam, Bending Mode 5

70
 Bending Stress Frequency Response Function x=0 inch
10 4

10 3

10 2

10 1

10 0

10 -1
0
10

00

00
10

10

40
Frequency (Hz)

Figure 14.7. Cantilever Beam Stress

Relative Velocity Frequency Response Function x=12 inch

10 2

10 1

10 0

10-1

10-2

10-3
0
10

00

00
10

10

40

Frequency (Hz)

Figure 14.8. Cantilever Beam Velocity

71
 Frequency Response Function Maxima Across Length
10 4 10 2
Stress
Relative Velocity

10 3 10 1

10 2 10 0

10 1 10 -1

10 0 10 -2
0
10

00

00
10

10

40
Frequency (Hz)

Figure 14.9. Cantilever Beam Stress & Velocity FRFs

The characteristic impedance for aluminum is ρc = 51.8 lbf sec/in^3. The proportionality factor
for the first mode assuming 1 G input is

σ ,
K= (14.1)
ρc v ,

2057 lbf⁄in
K= (14.2)
(51.8 lbf sec/in^3 )(20.1 in⁄sec)

K ≈2 (14.3)

This is the expected value for a solid cylindrical beam. By inspection, the higher modes also
have the same K value.

72
 Acceleration Power Spectral Density 42.1 GRMS Overall
10

0.1

0.01

0.001

0.0001

1e-05

00

00
10

0
10

10

20
Frequency (Hz)

Figure 14.10. Base Input, Acceleration PSD

The base input acceleration PSD is designed to provoke a somewhat uniform stress response
among the first four beam modes given that its corresponding velocity PSD is flat.

Velocity Power Spectral Density 2.24 in/sec RMS Overall

0.01

0.001
00

00
10

0
10

10

20

Frequency (Hz)

Figure 14.11. Base Input, Velocity PSD

73
 Relative Velocity Response PSD x=12 in, 5.94 in/sec RMS
10 0

10 -1

10 -2

10 -3

10 -4

10 -5

10 -6 0
10

00
00
10

10

20
Frequency (Hz)

Figure 14.12. Velocity Response at Free End

Stress Response PSD x=0 in, 525 psi RMS

104

103

102

101

100

10-1

10-2
0
10

0
10

0
10

20

Frequency (Hz)

Figure 14.13. Stress Response at Fixed End

74
 PSD Relative Velocity Mode 1 4.11 in/sec rms
100 Mode 2 3.52 in/sec rms
Mode 3 3.04 in/sec rms
Mode 4 2.74 in/sec rms

10-1

10-2

10-3

10-4
10

00
00
0
10

10

20
Frequency (Hz)

Figure 14.14. Velocity Response at Free End by Mode

The Modal Relative Velocity rms values in the following table are calculated from the square of the area
under each peak over a one-third octave band centered at each peak frequency. The use of the one-
third octave band format needs further consideration.

Table 14.1. Modal Velocity and Stress

Modal Relative Velocity Modal Stress

Mode
(in/sec) rms (psi) rms

1 1.86 193

2 2.58 267

3 2.53 262

4 2.54 263

The overall velocity from the four values in the above table is 4.8 in/sec rms. The overall stress is
496 psi rms which is 5.5% lower than the overall level of 525 psi rms in the stress response PSD in
Figure 14.13. An under-prediction is expected because the stress PSD also contains some non-
resonant response energy.

75
15 BEAM WITH POINT MASS EXAMPLES

15.1 CANTILEVER BEAM

y(x, t)

w(t)

Parameters
Aluminum, Length = 10 inch
Solid Cylinder, diameter = 0.375 inch
Q=10 for all modes
256 elements
6 modes included

Figure 15.1. Cantilever Beam with End Mass

The effect of the end mass on the stress velocity relationship is shown in the following table.

Table 15.1. Cantilever Beam with End Mass, K Values

Ratio of End Fundamental
Rod Mass End Mass
Mass to Beam Frequency K
(lbm) (lbm)
Mass (Hz)
0.110 0 0 103.1 2.0
0.110 0.0275 0.25 72.7 2.6
0.110 0.055 0.5 59.2 3.1
0.110 0.0825 0.75 51.2 3.6
0.110 0.110 1 45.7 4.0

76
 Fundamental Mode, Mass-Normalized Displacement
120
0
0.25
0.50
100
0.75
1.00

80

60

40

20

0
0 1 2 3 4 5 6 7 8 9 10
x (inch)

Figure 15.2. Cantilever Beam with End Mass, Fundamental Mode

The legend value is the ratio of the end mass to the beam mass.

77
15.2 SIMPLY-SUPPORTED BEAM WITH CENTER MASS

y(x,t)
M
EI, ρ, L

w(t)

Parameters
Aluminum, Length = 20 inch
Solid Cylinder, diameter = 0.375 inch
Q=10 for all modes
256 elements
6 modes included

Figure 15.3. Simply-Supported Beam with Center Mass

The effect of the center mass on the stress velocity relationship is shown in the following table.

Table 15.2. Simply-Supported Beam with Center Mass, K Values

Center Ratio of Center Fundamental
Rod Mass
Mass Mass to Beam Frequency K
(lbm)
(lbm) Mass (Hz)
0.220 0 0 72.3 2.0
0.220 0.055 0.25 59.1 2.6
0.220 0.110 0.5 51.1 3.1
0.220 0.165 0.75 45.7 3.6
0.220 0.220 1 41.7 4.0

The point mass yields the same K value for the same mass ratio in for both the cantilever and
simply-supported beams as shown in the previous two tables.

78
 Fundamental Mode, Mass-Normalized Displacement
60
0
0.25
50 0.5
0.75
1

40

30

20

10

0
0 2 4 6 8 10 12 14 16 18 20
x (inch)

Figure 15.4. Simply-Supported Beam with Center Mass, Fundamental Mode

The legend value is the ratio of the center mass to the beam mass.

79
16 BEAM WITH END SPRINGS

The following examples may be useful for structures mounted via isolation springs.

E, I, 

k1 k2

Figure 16.1. Beam Bending with End Springs

Parameters
Aluminum, Length = 24 inch
Solid Cylinder, diameter = 0.5 inch
Q=10 for all modes
256 elements
6 modes included

Calculate the frequency response functions for a point force applied at the center of the beam in the
vertical axis. Use these functions for determining the stress-velocity relationship coefficient.

80
 Table 16.1. Solid Cylindrical Beam, Normal Modes Analysis Results

k1 & k2 Peak Velocity Peak Stress

f1 (Hz) f2 (Hz) f3 (Hz) K
(lbf/in) (in/sec/lbf) (psi/lbf)
50,000 132 506 1052 4.9 493 2.0
5000 118 334 559 4.9 463 1.9
500 64.5 123 337 6.2 357 1.1
50 22.5 39.4 307 14.8 314 0.42
5 7.2 12.5 304 45.4 309 0.15

The peak velocity and stress each occur at the fundamental modal frequency.

Mode 1 fn= 22.5 Hz

4

-1

-2

-3

-4
0 5 10 15 20 25
x (in)

Figure 16.2. Mode 1, Solid Cylindrical Beam Bending with End Springs, k1=k2=50 lbf/in

81
 Mode 2 fn= 39.4 Hz
4

-1

-2

-3

-4
0 5 10 15 20 25
x (in)

Figure 16.3. Mode 2, Solid Cylindrical Beam Bending with End Springs, k1=k2=50 lbf/in

Mode 3 fn= 307 Hz

4

-1

-2

-3

-4
0 5 10 15 20 25
x (in)

Figure 16.4. Mode 3, Solid Cylindrical Beam Bending with End Springs, k1=k2=50 lbf/in

82
 Mobility FRF H 257 257
180

-180

10

00

00
10

10

20
10 2
X 22.4492
Y 14.831

10 1

10 0

10 -1

10 -2
0
10

00

00
10

10

20
Frequency (Hz)

Figure 16.5. Velocity FRF, Solid Cylindrical Beam Bending with End Springs, k1=k2=50 lbf/in
10

00

00
0
10

10

20
0
10

00

00
10

10

20

Figure 16.6. Stress FRF, Solid Cylindrical Beam Bending with End Springs, k1=k2=50 lbf/in

83
Repeat the analysis for the same beam but with a rectangular cross-section, 2 inch wide, 0.5
inch thick.

Table 16.2. Rectangular Beam, Normal Modes Analysis Results

k1 & k2 Peak Velocity Peak Stress

f1 (Hz) f2 (Hz) f3 (Hz) K
(lbf/in) (in/sec/lbf) (psi/lbf)
50,000 77 302 658 6.6 583 1.74
5000 73 245 430 6.7 568 1.67
500 50.5 106 219 7.2 467 1.27
50 19.6 34.8 180 13.8 380 0.54
5 6.4 11.1 176 40.4 366 0.18

The peak velocity and stress each occur at the fundamental modal frequency.

84
17 MULTISPAN BEAM, SUBJECTED TO BASE EXCITATION

y(x, t)
EI, L
ẅ(t)

Figure 17.1. Multispan Beam, Pinned-Pinned-Pined

The beam is aluminum, with a circular cross-section, 0.25 inch diameter. The length is 36
inches. The beam is pinned at each end and in the middle.
The FEA model has 256 beam elements with one rotational and one translational degree-of-
freedom per node. The analysis is performed via a Matlab FEA script. [21] Selected mode
shapes are shown in the following figures.

Table 17.1. Normal Modes Analysis Results

Mode Frequency (Hz) Modal Mass Percent

1 59.5 0
2 93.0 74.0
3 238 0
4 301 0.68
5 536 0
6 629 11.2
7 953 0
8 1075 0
9 1488 0.19
10 1641 0
11 2143 4.3
12 2326 0.09

85
 Mode 1 freq= 59.53 Hz
80

60

40

20

-20

-40

-60

-80
0 5 10 15 20 25 30 35
x (inch)

Figure 17.2. Multispan Beam, Pinned-Pinned-Pined, Mode 1

Mode 2 freq= 93 Hz
80

70

60

50

40

30

20

10

0
0 5 10 15 20 25 30 35
x (inch)

Figure 17.3. Multispan Beam, Pinned-Pinned-Pined, Mode 2

86
 Mode 4 freq= 301.4 Hz
80

60

40

20

-20

-40

-60

-80
0 5 10 15 20 25 30 35
x (inch)

Figure 17.4. Multispan Beam, Pinned-Pinned-Pined, Mode 4

Mode 6 freq= 628.8 Hz

80

60

40

20

-20

-40

-60

-80
0 5 10 15 20 25 30 35
x (inch)

Figure 17.5. Multispan Beam, Pinned-Pinned-Pined, Mode 6

87
 Maximum Transmissibility Functions
10 4 10 2
Stress at 18 in
PV at 7.59 in

10 3 10 1

10 2 10 0

10 1 10 -1

10 0 10 -2
0
10

00

00
10

20
10
Frequency (Hz)

Figure 17.6. Multispan Beam, Pinned-Pinned-Pined

A uniform base excitation was applied via the three pinned support locations from 10 to 2000
Hz. The amplification factor was Q=10 for all modes. 12 modes were included in the response.
The maximum stress and velocity are each driven by second mode at 93 Hz.
The maximum stress occurs at the center point. The maximum velocity occurs at 7.59 inch from
left end (also at -7.59 inch from right end).
Recall

σ ,
K= (17.1)
ρc v ,

Now assume that the base excitation is a steady-state sine function at the 93 Hz modal
frequency. The resulting proportionality constant is K = 2.67.

88
18 MULTISPAN BEAM, SUBJECTED TO BASE EXCITATION, INTERMEDIATE SUPPORT
POINT AT ONE-THIRD LENGTH

y(x, t)
EI, L
ẅ(t)

Figure 18.1. Multispan Beam, Pinned-Pinned-Pined, One-Third Length Intermediate Support

The beam is aluminum, with a circular cross-section, 0.25 inch diameter. The length is 36
inches. The beam is pinned at each end and at 12 inches from the left end.
The FEA model has 256 beam elements with one rotational and one translational degree-of-
freedom per node. The analysis is performed via a Matlab FEA script. [21] Selected mode
shapes are shown in the following figures.

Table 18.1. Normal Modes Analysis Results

Mode Frequency (Hz) Modal Mass Percent

1 42.9 42.7
2 134 9.0
3 187 18.3
4 329 11.8
5 536 0.0
6 638 0.27
7 883 2.3
8 1206 1.0
9 1357 2.6
10 1705 2.3
11 2143 0.0
12 2343 0.1

89
 Mode 1 freq= 42.91 Hz
100

80

60

40

20

-20
0 5 10 15 20 25 30 35 40
x (inch)

Figure 18.2. Multispan Beam, Pinned-Pinned-Pined, One-Third Length Support, Mode 1

Mode 2 freq= 133.9 Hz

80

60

40

20

-20

-40

-60

-80
0 5 10 15 20 25 30 35 40
x (inch)

Figure 18.3. Multispan Beam, Pinned-Pinned-Pined, One-Third Length Support, Mode 2

90
 Mode 3 freq= 187.3 Hz
100

80

60

40

20

-20

-40

-60
0 5 10 15 20 25 30 35 40
x (inch)

Figure 18.4. Multispan Beam, Pinned-Pinned-Pined, One-Third Length Support, Mode 3

Mode 4 freq= 329.1 Hz

100

80

60

40

20

-20

-40

-60

-80
0 5 10 15 20 25 30 35 40
x (inch)

Figure 18.5. Multispan Beam, Pinned-Pinned-Pined, One-Third Length Support, Mode 4

91
 Maximum Transmissibility Functions
10 4 10 2
Stress at 25.7 in
PV at 25 in

10 3
10 1

10 2

10 0

10 1

10 -1
0
10

10 -1 10 -2
0
10

00

00
10

20
10
Frequency (Hz)

Figure 18.6. Multispan Beam, Pinned-Pinned-Pined, One-Third Length Intermediate Support

A uniform base excitation was applied via the three pinned support locations from 10 to 2000
Hz. The amplification factor was Q=10 for all modes. 12 modes were included in the response.
The maximum stress and velocity are each driven by first mode at 42.9 Hz.
The maximum velocity and stress occur at 25.0 and 25.7 inch from left end, respectively.
Recall

σ ,
K= (18.1)
ρc v ,

Now assume that the base excitation is a steady-state sine function at the 42.9 Hz modal
frequency. The resulting proportionality constant is K = 1.80.

92
19 MULTISPAN BEAM, PINNED-PINNED-FREE, SUBJECTED TO BASE EXCITATION

y(x, t)
EI, L
ẅ(t)

Figure 19.1. Multispan Beam, Pinned-Pinned-Free

The beam is aluminum, with a circular cross-section, 0.25 inch diameter. The length is 20
inches. The beam is pinned at the left end and at the middle.
The FEA model has 256 beam elements with one rotational and one translational degree-of-
freedom per node. The analysis is performed via a Matlab FEA script. [21] Selected mode
shapes are shown in the following figures.

Table 19.1. Normal Modes Analysis Results

Mode Frequency (Hz) Modal Mass Percent

1 44.3 24.3
2 228 26.9
3 385 25.8
4 837 0
5 1127 3.0
6 1834 3.5
7 2252 4.3
8 3216 0
9 3763 0.9
10 4984 1.3
11 5659 1.7
12 7138 0

93
 Mode 1 freq= 44.32 Hz
180

160

140

120

100

80

60

40

20

-20
0 2 4 6 8 10 12 14 16 18 20
x (inch)

Figure 17.2. Multispan Beam, Pinned-Pinned-Free, Mode 1

Mode 2 freq= 227.7 Hz

120

100

80

60

40

20

-20

-40

-60
0 2 4 6 8 10 12 14 16 18 20
x (inch)

Figure 17.3. Multispan Beam, Pinned-Pinned-Free, Mode 2

94
 Mode 3 freq= 384.8 Hz
150

100

50

-50

-100

-150

-200
0 2 4 6 8 10 12 14 16 18 20
x (inch)

Figure 17.4. Multispan Beam, Pinned-Pinned-Free, Mode 3

Mode 4 freq= 837.1 Hz

150

100

50

-50

-100

-150
0 2 4 6 8 10 12 14 16 18 20
x (inch)

Figure 17.5. Multispan Beam, Pinned-Pinned-Free, Mode 4

95
 Maximum Transmissibility Functions
10 4 10 2
Stress at 10 in
PV at 20 in

10 3 10 1

10 2 10 0

10 1 10 -1

10 0 10 -2
0
10

00

00
10

20
10
Frequency (Hz)

Figure 17.6. Multispan Beam, Pinned-Pinned-Free

A uniform base excitation was applied via the three pinned support locations from 10 to 2000
Hz. The amplification factor was Q=10 for all modes. 12 modes were included in the response.
The maximum stress and velocity are each driven by first mode at 44.3 Hz.
The maximum velocity occurs at the free end. The maximum stress occurs at the midpoint.
Recall

σ ,
K= (19.1)
ρc v ,

Now assume that the base excitation is a steady-state sine function at the 44.3 Hz modal
frequency. The resulting proportionality constant is K = 1.42.

96
20 BEAM BENDING WITH MEAN STRESS

Nonzero mean stress can result from residual stress, thermal stress, applied static force or
inertial load, or pressurization.
The natural frequency of a cantilever beam is affected by the type of axial load applied:
 Tensile load: Increases the natural frequency
 Compressive load: Decreases the natural frequency
 Critical compressive load: Causes the frequency to reach zero and the beam to buckle
The following example shows that the stress-velocity relationship can adequately account for
the dynamic stress in a beam subjected to mean stress with a modified k scale factor. The total
stress is then calculated by adding the mean stress to the dynamic stress.

Fy

Fx

Figure 20.1. Cantilever Beam with Static Axial Tensile Load

Parameter Value

Material AL 6061-T6

Boundary Conditions Fixed-Free

Length 24 in

Width 2 in

Thickness 0.5 in

Amplification Factor Q=20

97
The beam is analyzed using the finite element analysis with 64 beam elements using a Matlab script
based on Reference [22].

The fundamental frequency of the beam without the axial load is 27.55 Hz. The frequency increases to
49.19 Hz when a static tensile load of Fx=2400 lbf is applied as shown in the next figure. The cross-
sectional area is 1 inch^2. The static axial stress is thus 2400 psi.

Mode 1 freq= 49.19 Hz

25

20

15

10

0
0 5 10 15 20 25
x (inch)

Figure 20.2. Cantilever Beam with Static Axial Tensile Load, Fundamental Mode Shape

Next, apply sine force Fy=2.5 lbf at the fundamental frequency along with static Fx for the modal
transient forced response analysis. Include the fundamental mode only.

98
 Stress Node 1
Peaks ( 9397 psi, -4597 psi)
10000

5000

-5000
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (sec)

Figure 20.3. Total Stress at Beam’s Fixed End

The mean stress is 2400 psi. The peak dynamic stress 6997 psi.

99
 Velocity Node 65
Peaks ( 87.7 in/sec, -87.7 in/sec)
100

80

60

40

20

-20

-40

-60

-80

-100
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (sec)

Figure 20.4. Velocity at Beam’s Free End

Recall
σ ,
K= (20.1)
ρc v ,

The scale factor is k = 1.57 for this dynamic stress in this example. This is less than the
k = √3 factor for a cantilever beam without a static axial load.
The experiment was performed for additional cases the results are shown in the following
table. The excitation frequency was the same as the natural frequency in each case.

100
Table 20.1. Normal Modes Analysis Results

Static Axial Fundamental Scale Factor

Load (lbf) Frequency (Hz) k
0 27.55 √3
600 34.84 1.61
1200 40.46 1.57
2400 49.19 1.57

101
21 STRESS IN A RECTANGULAR PLATE IN FREE VIBRATION

21.1 SIMPLY-SUPPORTED PLATE BENDING EQUATION

Consider the following rectangular plate.

Y
a

0 X

Figure 21.1. Plate Bending

The governing equation of motion from Reference [23] is

∂ z ∂ z ∂ z ∂ z
D +2 + + ρh =0 (21.1)
∂x ∂x ∂y ∂y ∂t

Assume a harmonic response.

z(x, y, t) = Z (x, y) exp( jωt) (21.2)

The modal displacement for a plate simply-supported on all four edges is

mπx nπy
Z =A sin sin (21.3)
a b

102
The mass-normalization factor is

2
A = (21.4)
ρabh

The natural frequencies are

D mπ nπ
ω = + (21.5)
ρh a b

The partial derivates are

∂ mπ mπx nπy
Z = A cos sin (21.6)
∂x a a b

∂ mπ mπx nπy
Z =− A sin sin (21.7)
∂x a a b

∂ nπ mπx nπy
Z = A sin cos (21.8)
∂y b a b

∂ nπ mπx nπy
Z =− A sin sin (21.9)
∂y b a b

The modal bending moments are

∂ ∂
M , = −D Z +ν Z (21.10)
∂x ∂y

∂ ∂
M , = −D ν Z + Z (21.11)
∂x ∂y

103
The maximum modal stress at any given cross section is

6 6D ∂ ∂
σ , = M , =− Z +ν Z (21.12)
h h ∂x ∂y

6 6D ∂ ∂
σ , = M , =− Z +ν Z (21.13)
h h ∂y ∂x

The total maximum modal stress for any mode is

6DA mπ nπ
σ , = +ν (21.14)
h a b

6DA nπ mπ
σ , = +ν (21.15)
h b a

Note that
∂ mπx nπy
Z = ωA sin sin (21.16)
∂t a b

For free vibration,   mn

∂ mπx nπy
Z =ω A sin sin (21.17)
∂t a b

D mπ nπ
ω = + (21.18)
ρh a b

The velocity is

∂ D mπ nπ mπx nπy
Z = + A sin sin (21.19)
∂t ρh a b a b

104
 ∂ D mπ nπ
Z = + A (21.20)
∂t ρh a b

21.2 SIMPLY-SUPPORTED PLATE EXAMPLES

Consider the von Mises and Tresca stresses at the center of the plate which is undergoing free vibration
at its fundamental frequency. The plate is steel, 0.125 inch thick, with ν = 0.3.

Recall

σ ,
K= (21.21)
ρc v ,

The following results were obtained via a Matlab script.

Table 21.1. Fundamental Mode K Factor, m=1, n=1

K K
a (in) b (in) a/b
von Mises Tresca
12 12 1.0 1.18 1.18
18 12 1.5 1.25 1.42
24 12 2.0 1.35 1.56
30 12 2.5 1.42 1.64
36 12 3.0 1.47 1.69
42 12 3.5 1.51 1.72
48 12 4.0 1.53 1.74
54 12 4.5 1.55 1.76
60 12 5.0 1.56 1.77

Note that K Tresca is uniform for higher modes for a given a/b ratio, but K von Mises varies by
mode number.

105
21.3 INTERMODAL SEGMENT

Consider a higher mode. The vibration mode for an intermodal segment can be represented
with n=m=1.
Hunt [2] notes:
It is relatively more difficult to establish equally general relations between antinodal
velocity and extensionally strain for a thin plate vibrating transversely, owing to the
more complex boundary conditions and the Poisson coupling between the principal
stresses. One can deal, however, with the velocity strain relations in the interior of
such a plate by invoking again the fact that conditions along an interior nodal line
correspond to those of a simple edge support. Each intermodal segment can,
therefore, be treated as if it were a simply supported rectangular plate of dimensions
Lx, Ly vibrating in its fundamental mode, where Lx and Ly are the nodal separations
along the X and Y axes, as shown in the following figure.

Z(x,y)

Lx
Ly

Y X

Figure 21.2. Rectangular Plate Vibrating in a Higher-order Mode

106
The dashed lines are the nodal lines. An intermodal segment is isolated for analysis as a plate
of dimension Lx, Ly with simply-supported edges vibrating in its intermodal fundamental
mode.
The subscript int is used to denote intermodal segment.

6DA π π
σ , = +ν (21.22)
h L L

6DA π π
σ , = +ν (21.23)
h L L

∂ D π π
Z = + A (21.24)
∂t ρh L L

∂
Z
∂t
A = (21.25)
D π π
+
ρh L L

π π
+ν
6D ∂ L L
σ , = Z (21.26)
h ∂t D π π
+
ρh L L

π π
+ν
6 Dρh ∂ L L
σ , = Z (21.27)
h ∂t π π
+
L L

The plate stiffness factor is

Eh
D= (21.28)
12(1 − ν )

107
By substitution

Eh ρ π π
6 +ν
12(1 − ν ) ∂ L L
σ = Z (21.29)
,
h ∂t π π
+
L L

π π
+ν
Eρ ∂ L L
σ , =6 Z (21.30)
12(1 − ν ) ∂t π π
+
L L

3 Eρ ∂ πL + ν(π L )
σ , = Z (21.31)
1 − ν ∂t πL + (π L )

3 Eρ ∂ L +νL
σ , = Z (21.32)
1 − ν ∂t L +L

3 ∂ L +νL
σ , = ρc Z (21.33)
1−ν ∂t L +L

Similarly,

3 ∂ L +νL
σ , = ρc Z (21.34)
1−ν ∂t L +L

The Hunt equations agree with the Tresca results for the previous simply-supported plate
example.

108
22 RING MODE

Consider a ring’s breathing mode, or first extensional mode. The mode shape is shown below using a
model constructed in Simcenter Nastran. The nodes undergo radial displacement, with all nodes in-
phase. The undeformed ring is the center ring. The fully expanded ring is the outer ring.

Figure 22.1. Ring Breathing Mode

The material is aluminum 6061-T651. The diameter is 48 inch. The height is 1 inch. The
thickness is 0.25 inch. The model consists of 360 plate elements, with 1 degree spacing.
A uniform force of 1 lbf was applied in the radial axis along the outer surface of the ring at each
of the 720 nodes. The force was applied in the frequency domain at 1387 Hz which is the
breathing mode frequency. The maximum velocity was 86.16 in/sec. The maximum stress was
4595 psi, which was higher than the von Mises stress. The results confirmed the following
equation.

[σ ] =ρcv , (22.1)

Stress is proportional to mass density. Consider two rods with the same length and diameter. One is
steel, and the other is aluminum. They are dropped from the same height and allowed to free fall. Each
rod contacts the ground with the same velocity. But the steel rod has about 2.8x the maximum stress
than aluminum due to the mass density difference. Note that the longitudinal wave speed c for both
steel and aluminum is about 200,000 in/sec.

109
23 MIL-STD-810E & SIMILAR REFERENCES

An empirical rule-of-thumb in MIL-STD-810E states that a shock response spectrum is

considered severe only if one of its components exceeds the level

Threshold = [ 0.8 (G/Hz) * Natural Frequency (Hz) ] (23.1)

For example, the severity threshold at 100 Hz would be 80 G.

This rule is effectively a velocity criterion.

MIL-STD-810E states that it is based on unpublished observations that military-quality

equipment does not tend to exhibit shock failures below a shock response spectrum velocity of
100 inches/sec (254 cm/sec). Howard Gaberson credits Allan Piersol for this threshold, but
Gaberson’s blower shock test series may have been a contributing source. [11]

The previous threshold Equation has a built-in 6 dB margin of conservatism.

Note that this rule was not included in MIL-STD-810F or G, however.

SMC-TR-06-11 states that a response velocity to the shock less than 50 inches/second is judged
to be non-damaging. This is the case if the shock response spectrum value in G is less than 0.8
times the frequency in Hz.

ECSS-E-HB-32-25A gives equation (23.1) for “structural and non-sensitive equipment” but
excludes electronic equipment.

110
Gaberson [11] further expanded the severity levels as

Table 23.1. Gaberson Shock Severity Levels, Approximate

Pseudo Pseudo
Velocity Velocity Classification
(ips) (m/sec)
40 1 Mild
80 2 Moderate
100 2.5 Severe
200 5 Very Severe
400 10 Extreme

Pseudo velocity is covered in Section 27.

See also the Morse chart in Section 25.

111
24 GABERSON’S BLOWER SHOCK TEST SERIES

Howard Gaberson performed an insightful series of shock tests on squirrel cage blowers. [11]
The tests were performed at different test facilities with varying pulse types and incremental
amplitudes. The goal was to experimentally characterize shock severity. Each of the tests in
the following table eventually caused failures except for the LW72 test. Note that the first two
letters of each test name are used to identify the test pulse type, such as TP for terminal peak
sawtooth.

Test Description
TP60 60 inch drop to a terminal peak sawtooth
HS54 54 inch drop on rubber pad to get half sine
PB24 24 inch drop on hard phenolic block
LW72 72 inch hammer drop to shelf bracket on MIL-S-901 lightweight shock machine
MW36 36 in hammer drop on the MIL-S-901 medium weight shock machine
HW4 4th shot on the floating shock platform at Hunter’s Point

The failures consisted of

1. Spider members broken or deformed: belt flies off

2. Motor knocked out of its supports: belt flies off or motor cannot turn
3. Self-tapping bolts deformed, loosened, pulled out
4. Feet bolt holes deformed

112
Test blower as originally
configured.
Weak sheet metal
Arrows point to thin spider motor mount
members supporting beam
bearings, and to sheet metal
screws.

Notice sheet metal legs.

Figure 24.1. Squirrel Cage Blower

113
 Rubber Pad for
Half-sine Test

Figure 24.2. HS54, Drop Table Shock Machine, China Lake

114
Mylar covered opening
with orifice to simulate
realistic delta p

Manometer to check
for adequate delta p

Terminal peak
sawtooth programmer

Figure 24.3. TP60, Shock Machine with Terminal Sawtooth Peak Programmer

115
Phenolic block taped to
shock machine anvil for
short duration half sine
test

Figure 24.4. PB24, Phenolic Block High Impact Test Setup

116
Figure 24.5. LW72, Blower Mounted on MIL-S-901 Lightweight Shock Machine

117
 Pendulum Hammer

Figure 24.6. LW72, Blower Mounted on MIL-S-901 Lightweight Shock Machine, Perspective
View

118
Figure 24.7. MW36, Blower Mounted on MIL-S-901 Medium Shock Machine

119
Figure 24.8. HW4, Two Blowers Mounted in the Floating Shock Platform
The blowers are mounted on separate plates with varying stiffness.

120
Spot Weld Failed

Spot weld pulled

out allowing belt
to loosen.

Figure 24.9. HS54, Half-sine Test Failure

121
Pulley side bearing
support spot weld failed
allowing bearing support
to deflect and loosen belt

Figure 24.10. TP60, Terminal Sawtooth Pulse Failure

122
 Spider member pulled out
of housing, loosing belt.

Figure 24.11. PB24, Phenolic Block Test Failure

123
Leg pulled away from
blower body & sheet
metal screws gone

Bearing support bent

Motor mount bent

rubbing pulley

Accelerometer

Leg deformed

Figure 24.12. MW36, Medium Weight Failures

124
Figure 24.13. MW36, Medium Weight Failures

125
Figure 24.14. Base Input Time Histories
All six shocks are plotted to the same scale. Only the LW72 pulse (MIL-S-901 lightweight shock)
did not fail the blower.

126
Figure 24.15. Shock Response Spectra

All six shocks are plotted to the same scale. Again, only the blue LW72 pulse did not fail the
blower. The LW72 curve reaches about 140 ips. This data may have been a contributing source
of the 100 ips severity threshold in MIL-STD-810E even though the blower is not “military-
quality.”

Recall that MIL-STD-810E states that its velocity criterion is based on unpublished observations
that military-quality equipment does not tend to exhibit shock failures below a shock response
spectrum velocity of 100 ips.

127
25 MORSE CHART

SHOCK RESPONSE SPECTRA - DAMAGE THRESHOLDS

5
10
Probable Damage 300 ips
Potential Damage 100 ips
Failure unlikely below this limit 50 ips
4
10
PEAK ACCEL (G)

3
10

2
10

1
10

0
10
10 100 1000 10000

NATURAL FREQUENCY (Hz)

Threshold Formula
300 ips [ 4.8 (G/Hz) * Natural Frequency (Hz) ]
100 ips [ 1.6 (G/Hz) * Natural Frequency (Hz) ]
50 ips [ 0.8 (G/Hz) * Natural Frequency (Hz) ]

Figure 25.1. Morse Chart

The curves in the above figure are taken from Reference [24]. The curves are defined by the
following formulas. The 100 ips threshold is defined in part by the observation that the severe
velocities which cause yield point stresses in mild steel beams turn out to be about 130 ips, per
Reference [25].

128
26 MIL-STD-810G, METHOD 516.6

A more complete description of the shock (potentially more useful for shock damage
assessment, but not widely accepted) can be obtained by determining the maximax
pseudo-velocity response spectrum and plotting this on four-coordinate paper where, in pairs
of orthogonal axes, the maximax pseudo-velocity response spectrum is represented by the
ordinate, with the undamped natural frequency being the abscissa and the maximax
absolute acceleration along with maximax pseudo-displacement plotted in a pair of
orthogonal axes, all plots having the same abscissa.

The maximax pseudo-velocity at a particular SDOF undamped natural frequency is thought

to be more representative of the damage potential for a shock since it correlates with
stress and strain in the elements of a single degree of freedom system (paragraph 6.1,
reference f). If the testing is to be used for laboratory simulation, use 516.6-8 a Q value of ten
and a second Q value of 50 in the processing. Using two Q values, a damped value and a value
corresponding to light damping, provides an analyst with information on the potential
spread of materiel response.

It is recommended that the maximax absolute acceleration SRS be the primary method of
display for the shock, with the maximax pseudo-velocity SRS the secondary method of
display and useful in cases in which it is desirable to be able to correlate damage of simple
systems with the shock.

(End Quote)

Note that Reference f is:

Harris, C., and C. E. Crede, eds., Shock and Vibration Handbook, 5th Edition, NY,
McGraw-Hill, 2001.

129
27 PSEUDO VELOCITY SHOCK RESPONSE SPECTRUM

The Shock Response Spectrum (SRS) models the peak response of a single-degree-of-freedom
(SDOF) system to a base acceleration, where the system’s natural frequency is an independent
variable. The SRS method is thoroughly covered in Reference [26]. The purpose of this section
is to present some additional notes.

The absolute acceleration and the relative displacement of the SDOF system can be readily
calculated.

The velocity, however, is more difficult to calculate accurately. The “pseudo velocity” is an
approximation of the relative velocity.

The following statements are in terms of the respective peak values.

The pseudo velocity is equal to the relative displacement multiplied by the natural frequency
ωn which has units of radians per second. This is the preferred calculation method.

The pseudo velocity is approximately equal to the absolute acceleration divided by ωn.

The pseudo acceleration is equal to the relative displacement multiplied by the natural
frequency ωn2.

The pseudo acceleration is thus equal to the pseudo velocity multiplied by the natural
frequency ωn. There may be little reason if any to calculate pseudo acceleration in practice,
however, because the absolute acceleration can be calculated directly.

Furthermore, one of the advantages of the pseudo velocity SRS is that it tends to produce a
more uniform SRS than either acceleration or relative displacement.
Additional information is given in Reference [27].

130
28 HALF-SINE PULSE

SRS Q=10, Half-Sine Pulse, 10 G, 11 msec

SRS Q=10, Half-Sine Pulse, 10 G, 11 msec, 1 to 1000 Hz

Parameter Max Min Range (dB)
Displacement (inch) 4.0 1.1e-04 91
Velocity (in/sec) 25.2 0.64 32
Acceleration (G) 16.5 0.41 32

Figure 28.1.
The velocity and acceleration spectra have the same range in terms of decibels.

131
Note that the pseudo velocity SRS would converge to the velocity change area under the half-
sine acceleration pulse as the natural frequency approaches zero and as the Q value
approaches infinity. Damping reduces the pseudo velocity plateau level to less than the
velocity change by precluding the build-up of multiple response cycles.

The velocity change for the given pulse is 27 in/sec.

The net velocity change equation is

2ẍ t (28.1)
Δẋ =
π

The velocity change for the sample pulse is

2(10 G)(386 in/sec )(0.011 sec)

Δẋ = = 27 in/sec (28.2)
π

An ideal half-sine pulse is for educational purposes only. Half-sine tests for components must
have zero net velocity whether performed on a shaker or drop tower.

132
29 EL CENTRO EARTHQUAKE

ACCELERATION TIME HISTORY EL CENTRO EARTHQUAKE 1940

NORTH-SOUTH COMPONENT

0.5

0.4

0.3

0.2

0.1
ACCEL (G)

-0.1

-0.2

-0.3

-0.4

-0.5
0 5 10 15 20 25

TIME (SEC)

Figure 29.1. El Centro (Imperial Valley) Earthquake

Nine people were killed by the May 1940 Imperial Valley earthquake. At Imperial, 80
percent of the buildings were damaged to some degree. In the business district of Brawley,
all structures were damaged, and about 50 percent had to be condemned. The shock
caused 40 miles of surface faulting on the Imperial Fault, part of the San Andreas system in
southern California. It was the first strong test of public schools designed to be
earthquake-resistive after the 1933 Long Beach quake. Fifteen such public schools in the
area had no apparent damage. Total damage has been estimated at about $6 million. The
magnitude was 7.1.

The El Centro earthquake was the first major quake in which calibrated, strong-motion
source data was measured which would be useful for engineering purposes. This data has
obvious application to the design of buildings, bridges, and dams in California. It also has
some surprising aerospace uses. Consider a rocket vehicle mounted as a cantilever beam
to a launch pad at Vandenberg AFB on the central California coast. Engineers must verify
that the vehicle can withstand a hypothetical seismic event prior to launch. The El Centro
earthquake data has thus been used in some analyses as a modal transient input to the
rocket vehicle’s finite element model.

133
 SRS Q=10 El Centro Earthquake North-South Component

SRS, El Centro Earthquake, Results for the Domain from 0.1 to 6.4 Hz
Parameter Max Min Range (dB)
Displacement (inch) 15.0 0.2 38
Velocity (in/sec) 31.0 7.4 12
Acceleration (G) 0.93 0.012 38

Figure 29.2.

The SRS of the time history is shown in the above figure in tripartite format.

One of the advantages of the pseudo velocity SRS is that it has a smaller amplitude range
than either the displacement or acceleration. The pseudo velocity SRS is thus less
frequency-dependent.

134
30 RE-ENTRY VEHICLE SEPARATION SHOCK

ACCELERATION TIME HISTORY RV SEPARATION

10000

5000
ACCEL (G)

-5000

-10000
91.462 91.464 91.466 91.468 91.470 91.472 91.474 91.476 91.478

TIME (SEC)

Figure 30.1.

The time history is a near-field, pyrotechnic shock measured in-flight on an unnamed rocket
vehicle. The separation device was linear shape charge.

The SRS of the time history is shown in the following figure in tripartite format.

135
 SRS Q=10 RV Separation Shock

SRS, RV Separation, Results for the Domain from 100 to 10,000 Hz

Parameter Max Min Range (dB)
Displacement (inch) 0.041 0.0012 31
Velocity (in/sec) 526 14 31
Acceleration (G) 20,400 23 59

Figure 30.2. RV Separation, Flight Data

The values in the above table begin at 100 Hz because there is some uncertainty regarding the
accuracy of the data below 100 Hz, which is sometimes the case for near-field pyrotechnic
shock events due to baseline shift, saturation, accelerometer piezoelectric crystal resonance,
etc. The velocity range is 28 dB lower than the acceleration range.

136
31 SR-19 MOTOR IGNITION & PRESSURE OSCILLATION

SR-19 Motor Ignition Static Fire Test Forward Dome

1000

500
ACCEL (G)

-500

-1000
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

TIME (SEC)

Figure 31.1. SR-19 Motor, Test data

The SR-19 is a solid-fuel rocket motor. The combustion cavity has a pressure oscillation at 650
Hz.

137
 SRS Q=10 SR-19 Motor Ignition Forward Dome

SRS, SR-19 Motor Ignition, Forward Dome, Results for the Domain from 10 to
6500 Hz
Parameter Max Min Range (dB)
Displacement (inch) 0.73 1.7e-04 73
Velocity (in/sec) 295 6.8 33
Acceleration (G) 3224 7.1 53

Figure 31.2.
The peak at 680 Hz is due to the SR-19 motor oscillation which results from a standing pressure
wave in the combustion cavity.

138
32 SPACE SHUTTLE SOLID ROCKET BOOSTER WATER IMPACT

ACCELERATION SRB WATER IMPACT FWD IEA

100

50
ACCEL (G)

-50

-100
0 0.05 0.10 0.15 0.20

TIME (SEC)

Figure 32.1.
The data is from the STS-6 mission. Some high-frequency noise was filtered from the data.

139
 SRS Q=10 SRB Water Impact, Forward IEA

SRS, SRB Water Impact, Forward IEA, Results for the Domain from 10 to 750 Hz
Parameter Max Min Range (dB)
Displacement (inch) 0.76 0.002 52
Velocity (in/sec) 209 10 26
Acceleration (G) 259 5.6 33

Figure 32.2.
The velocity SRS has the smallest dynamic range in terms of decibels.

140
33 V-BAND/BOLT-CUTTER SHOCK

ACCELERATION V-BAND/BOLT-CUTTER SEPARATION SOURCE SHOCK

300

200

100
ACCEL (G)

-100

-200

-300
0 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040

TIME (SEC)

Figure 33.1.
The time history was measured during a shroud separation test for a suborbital launch vehicle.

141
 SRS Q=10 V-band/Bolt-Cutter Shock

SRS Q=10, V-band/Bolt-Cutter Shock

Parameter Max Min Range (dB)
Displacement (inch) 0.032 4e-04 38
Velocity (in/sec) 11.4 2.0 15
Acceleration (G) 1069 0.4 69

Figure 33.2.
The velocity SRS has the smallest dynamic range in terms of decibels.

142
34 MAXIMUM VELOCITY SUMMARY

Table 34.1. Maximum Velocity & Dynamic Range of Shock Events

Maximum Velocity
Event Pseudo Velocity Dynamic Range
(in/sec) (dB)
RV Separation, Linear Shape Charge 526 31

SR-19 Motor Ignition, Forward Dome 295 33

SRB Water Impact, Forward IEA 209 26

Half-Sine Pulse, 50 G, 11 msec 125 32

El Centro Earthquake, North-South Component 31 12

Half-Sine Pulse, 10 G, 11 msec 25 32

V-band/Bolt-Cutter Source Shock 11 15

The peak velocity comparison is useful, but a consideration of natural frequency is still needed
because the dynamic range is greater than zero, at least 12 dB, in each case.

143
 35 VELOCITY LIMITS OF MATERIALS

Gaberson gave the following limits in Reference [5].

Table 35.1. Severe Velocities, Fundamental Limits to Modal Velocities in Structures, Part I
Rod Beam Plate
ρ
Material E (psi) σ (psi) Vmax Vmax Vmax
(lbm/in^3)
(in/sec) (in/sec) (in/sec)
Douglas Fir 1.92e+06 6450 0.021 633 366 316
Aluminum
10.0e+06 35,000 0.098 695 402 347
6061-T6
Magnesium
6.5e+06 38,000 0.065 1015 586 507
AZ80A-T5
33,000 226 130 113
Structural Steel,
29e+06 0.283
High Strength
100,000 685 394 342

The following tables are taken from Reference [25]. The original sources are noted. The
velocity terms are “modal velocities at the elastic limit.”

144
(Values from Sloan, 1985, Packaging Electronics)

Table 35.2. Severe Velocities, Fundamental Limits to Modal Velocities in Structures, Part II
Rod Beam
E ρ ult yield
Material μ Vmax Vmax
(1e+06 psi) (lbm/in^3) (ksi) (ksi)
(in/sec) (in/sec)
Aluminum 5052 9.954 0.334 0.098 34 24 477.4 275.9
Aluminum
9.954 0.34 0.098 42 36 716.2 413.9
6061-T6
Aluminum
9.954 0.334 0.1 77 66 1299.8 751.3
7075-T6
Be 42 0.1 0.066 86 58 684.5 395.7
Be-Cu 18.5 0.27 0.297 160 120 1005.9 581.5
Cadmium 9.9 0.3 0.312 11.9 11.9 133.0 76.9
Copper 17.2 0.326 0.322 40 30 250.5 144.8
Gold 11.1 0.41 0.698 29.8 29.8 210.4 121.6
Kovar 19.5 - 0.32 34.4 59.5 468.0 270.5
Magnesium 6.5 0.35 0.065 39.8 28 846.4 489.3
Nickel 29.8 0.3 0.32 71.1 50 318.1 183.9
Silver 10.6 0.37 0.38 41.2 41.2 403.4 233.2
Solder 63/37 2.5 0.4 0.30008 7 7 158.8 91.8
Steel 1010 30 0.292 0.29 70 36 239.8 138.6
Stainless 28.4 0.305 0.29 80 40 273.9 158.3
Alumina al203 54 - 0.13 25 20 148.3 85.7
Beryllia Beo 46 - 0.105 20 20 178.8 103.4
Mira 10 - 0.105 - 5.5 105.5 60.0
Quartz 10.4 0.17 0.094 27.9 27.9 554.5 320.5
Magnesia Mgo 10 - 0.101 12 12 234.6 135.6
EPO GLS G10 X/Y 2.36 0.12 0.071 25 35 1680.1 971.1
EPO GLS G10 Z 2.36 0.12 0.071 25 35 1680.1 971.1
Lexan 0.379 - 0.047 9.7 9.7 1428.1 825.5
Nylon 0.217 - 0.041 11.8 11.5 2395.6 1384.8
Teflon 0.15 - 0.077 - 4 731.3 422.7
Mylar 0.55 - 0.05 25 25 2962.2 1712.3

145
 (Values from Roark, 1965, p 416)

Table 35.3. Severe Velocities, Fundamental Limits to Modal Velocities in Structures, Part II
Rod Beam
E ρ ult yield
Material μ Vmax Vmax
(1e+06 psi) (lbm/in^3) (ksi) (ksi)
(in/sec) (in/sec)
Aluminum Cast Pure 9 0.36 0.0976 11 11 230.6 133.3
Al Cast 220-T4 9.5 0.33 0.093 42 22 459.9 265.8
Al 2014-T6 10.6 0.33 0.101 68 60 1139.4 658.6
Beryllium Cu 19 0.3 0.297 150 140 1158.0 669.4
Cast Iron, Gray 14 0.25 0.251 20 37 357.8 224.2
Mg AZ80A-T5 6.5 0.305 0.065 55 38 1148.7 663.0
Titanium Alloy 17 0.33 0.161 115 110 1306.5 755.2
Steel Shapes 29 0.27 0.283 70 33 226.3 130.8
Concrete 3.5 0.15 0.0868 0.35 0.515 18.4 10.6
Granite 7 0.28 0.0972 - 2.5 59.6 34.4

146
 Figure 35.1. Herring Gull Dropping Clam

https://palmbeachcountynaturally.wordpress.com/

Examples from biological nature are given in the following table. Some bird species drop clams
and nuts on the ground to open them. The shell may not break open if dropped on soft sand.
Rocks and asphalt are much more effective ground surfaces.

Table 35.4. Severe Velocities from Nature, Gaberson [11]

Material Velocity Threshold

Walnut Crow drops walnut 18 ft onto asphalt, 408 ips

Clam Seagull drops clam 10 ft onto asphalt, 304 ips
Definite Crack at 1 inch drop, 27.8 ips
Egg
No crack at ½ inch drop, 19.6 ips

The velocity in the above table is calculated from the drop height.

v= 2gh (35.1)

147
36 NONLINEAR SHOCK ANALYSIS

Figure 36.1. Stress-Strain Curve

The plot shows the applied strain on the horizontal axis and the resisting stress on the vertical
axis. The area inside the triangle represents the maximum energy that can be absorbed
elastically if the member is strained up to its proportional limit stress. The member will return
to its original position when the load is removed. This area is also referred to as resilience or
modulus of resilience.

The yield stress, denoted σy, is the stress needed to induce plastic deformation in the
specimen. The yield stress is taken to be the stress needed to induce a specified amount of
permanent strain, typically 0.2% since it is often difficult to pinpoint the exact stress at which
plastic deformation begins.
The remaining area, which is not cross-hatched, represents the plastic (or inelastic) energy if
the member is strained up to its fracture point. This area is also referred to as the toughness or
modulus of toughness.
A large amount of energy can be absorbed, if permanent, plastic deformation is acceptable.

148
Figure 36.2. Toughness & Facture

Ductile materials undergo observable plastic deformation and absorb significant energy before
fracture. An additional portion of the energy is dissipated as heat during load and unloading
cycles.
Brittle fracture is characterized by very low plastic deformation and low energy absorption
before breaking.
Toughness is the ability of a material to absorb energy and plastically deform without
fracturing. One definition of material toughness is the amount of energy per unit volume that a
material can absorb before rupturing. This measure of toughness is different from that used for
fracture toughness, which describes load bearing capabilities of materials with flaws.
Toughness requires a balance of strength and ductility. A material should withstand both high
stresses and high strains to be tough. Generally, strength indicates how much force the
material can support, while toughness indicates how much energy a material can absorb before
rupturing.

Toughness (or, deformation energy, UT) is measured in units of joule per cubic meter (J/m 3) in
the SI system and inch-pound-force per cubic inch (in·lbf/in 3) in US customary units.

149
 Figure 36.3. Absorbed Energy in Stress-Strain Curve

Figure 36.4. Strain or Work Hardening

The uniaxial stress–strain curve shows typical work hardening plastic behavior of materials in
uniaxial compression. For work hardening materials the yield stress increases with increasing
plastic deformation. The strain can be decomposed into a recoverable elastic strain εp and an
inelastic strain εe. The stress at initial yield is σo.

150
36.1.1 Material Response to Shock

A material can sometimes sustain an important dynamic load without damage, whereas the
same load, statically, would lead to plastic deformation or to failure. [28] Many materials
subjected to short duration loads have ultimate strengths higher than those observed when
they are static.
Hopkinson noted that copper and steel wire can withstand stresses that are higher than their
static elastic limit and are well beyond the static ultimate limit without separating
proportionality between the stresses and the strains. This is provided that the length of time
during which the stress exceeds the yield stress is of the order of 1 millisecond or less.
Tests carried out on steel (annealed steel with a low percentage of carbon) revealed that the
initiation of plastic deformation requires a definite time when stresses greater than the yield
stress are applied. It was observed that this time can vary between 5 milliseconds (under a
stress of approximately 352 MPa) and 6 seconds with approximately 255 MPa; with the static
yield stress being equal to 214 MPa). Other tests carried out on five other materials showed
that this delay exists only for materials for which the curve of static stress deformation presents
a definite yield stress, and the plastic deformation then occurs for the load period.

Table 36.1. Annealed Steel Test Results

Stress Stress
Parameter
(MPa) (ksi)
Static Yield Stress 214 31.1

5 msec for plastic deformation onset 352 51.1

6 sec for plastic deformation onset 255 37.0

151
Reference [29] includes the following table where the data was obtained for uniaxial testing
using an impact method.

Table 36.2. Dynamic Strengthening of Materials

Static Strength Dynamic Strength Impact Speed

Material
(psi) (psi) (ft/sec)
2024 Al (annealed) 65,200 68,600 >200
Magnesium Alloy 43,800 51,400 >200
Annealed Copper 29,900 36,700 >200
302 Stainless Steel 93,300 110,800 >200
SAE 4140 Steel 134,800 151,000 175
SAE 4130 Steel 80,000 440,000 235
Brass 39,000 310,000 216

152
37 CRITICAL IMPACT VELOCITY

Figure 37.1. Tensile Impact Test Method, ISO 8256 method A (Image Courtesy of Instron)
The specimen is clamped in vice supported by the frame of the pendulum, and it is broken by
the impact between pendulum and crosshead member, which is always clamped on the other
extremity of the specimen.

Gaberson et al have used the yield stress to calculate the corresponding maximum allowable
velocity for a given material and geometry. But higher velocities can be endured if some plastic
deformation or work-hardening is allowed.
The following is taken from Reference [1].
Behavior changes occur in test specimens undergoing a series of impact tests of increasing
velocity. Each velocity at which this occurs is called a critical velocity. For example, there is a
critical velocity at which permanent deformation will occur at the specimen’s impacted end.
Another and perhaps the most important critical velocity is that at which a specimen factures at
the impact point, known as the critical impact velocity (CIV).

153
Von Karman and others have assumed in discussing the dynamics of strain propagation that the
plastic strain propagation velocity depends on the slope of the stress-strain curve at that value
of strain, when a specimen is subjected to a tensile impact test.
The velocity of propagation c is given by

dσ
dε (37.1)
c=
ρ

where dσ/dε is the slope of the engineering stress-strain curve at the given strain value.

The impact velocity v that will produce a plastic strain ε is given by

dσ
dε (37.2)
v = dε
ρ

The slope of the engineering tensile stress-stress strain curve dσ/dε becomes zero when the
strain reaches that corresponding to the ultimate stress limit. Theory, predicts that rupture will
occur at the instant of impact at the impacted end of the specimen when the impact velocity is

dσ
dε (37.3)
v ≥ v = dε
ρ

CIV data for common materials is scarce. But a few values were given previously in Table 36.2.

The impact speed for 2024 Al (annealed) rod is > 200 ft/sec (2400 in/sec).

The yield stress-based velocity for this same material is about 925 in/sec.

The impact speed is thus 2.6 times higher than the yield speed. This is an important benefit of
ductile aluminum and steel alloys.

154
 38 CHARACTERISTIC IMPEDANCE VALUES

Note that: 1 Pa sec/m = 1 kg/(m^2 sec)

Table 38.1. Characteristic Impedance, Metric

Characteristic
Elastic Speed
Density Impedance
Modulus (m/sec)
(MPa sec/m)
Material (kg/m^3) (GPa) Bar Bulk Bar Bulk

Aluminum 2700 71 5150 6300 13.9 17.0

Steel 7700 195 5050 6100 39.0 47.0

Table 38.2. Characteristic Impedance, English I

Characteristic
Elastic Speed
Density Impedance
Modulus (in/sec)
[ lbm/(in^2 sec) ]
Material (lbm/in^3) (psi) Bar Bulk Bar Bulk

Aluminum 0.1 1e+07 2.03e+05 2.48e+05 2.03E+04 2.48E+04

Steel 0.285 3e+07 1.99e+05 2.40e+05 5.67E+04 6.84E+04

Table 38.3. Characteristic Impedance,

English II
Characteristic
Impedance
(psi sec/in)
Material Bar Bulk

Aluminum 52.5 64.2

Steel 147 173

155
39 BUILDING VELOCITY LIMITS

Recommended limit values for traffic are given in the following table, as taken from References
[30], [31], [32]. The peak value is taken from the velocity time history.

Table 39.1. Recommended Ground Limit Values for Traffic

Recommended Vertical Velocity
Type of building and foundation
Vmax
(mm/sec) (in/sec)
Especially sensitive buildings and buildings
1 0.039
of cultural and historic value
Newly-built buildings and/or foundations
2 0.079
of a foot plate (spread footings)

Buildings on cohesion piles 3 0.12

Buildings on bearing piles or friction piles 5 0.20

The vertical velocity is with respect to the ground vibration.

The specification implies that building stress correlates more closely with velocity than with
either displacement or acceleration.

External Vibration Sources Internal Vibration Sources

1. Seismic activity 1. HVAC equipment

2. Subway, road and rail systems 2. Elevator and conveyance systems
3. Industrial facilities 3. Fluid pumping equipment
4. Construction equipment 4. Human footfall, dancing, aerobics
5. Wind 5. Exercise equipment

156
40 MACHINERY VELOCITY LIMITS

ISO 2372 (10816) Standards provide guidance for evaluating vibration severity in machines
operating in the 10 to 200Hz (600 to 12,000 RPM) frequency range. Examples of these types of
machines are small, direct-coupled, electric motors and pumps, production motors, medium
motors, generators, steam and gas turbines, turbo-compressors, turbo-pumps and fans. Some
of these machines can be coupled rigidly or flexibly, or connected though gears. The axis of the
rotating shaft may be horizontal, vertical or inclined at any angle.

Table 40.1. Machine Velocity Limits

Class l Individual parts of engines and machines, integrally connected with the complete
machine in its normal operating condition. (Production electrical motors of up to 20 HP (15 kW)
are typical examples of machines in this category.)
Class ll Medium-sized machines typically, electric motors with 20 to 75 HP (15-75 kW) without
special foundations, rigidly mounted engines, or machines on special foundations up to 400 HP
(300 kW).

157
Class lll Large prime movers and other large machines with rotating masses mounted on rigid
and heavy foundations which are relatively stiff in the direction of vibration measurement.
Class lV Large prime movers and other large machines with rotating masses mounted on
foundations which are relatively soft in the direction of vibration measurement (for example,
turbo-generator sets, especially those with lightweight sub-structures).
The above table is for general machines. ISO 2372 (10816) also has specific levels for steam
turbine and generator systems, critical industrial machines over 15 kW, gas turbines, hydro-
turbines, and reciprocating machines.

158
41 PIPING VELOCITY LIMITS

Figure 41.1. Pipework Vibration Limits [33]

The main causes for piping vibrations are

1. Turbulence in the liquid flow
2. Compressors, pumps and other equipment that produce mechanical forces
3. High-frequency acoustic excitations caused by high-pressure
4. Surge, water hammer, and momentum changes due to sudden valve closure
5. Cavitation or vapor bubble collapse due to high pressure drop

Flow induced vibration (FIV) is the result of turbulence in the process fluid, which occurs due to
major flow discontinuities such as bends, tees, partially closed valves, and small bore
connections. The high levels of broadband kinetic energy created downstream of these sources
is concentrated at low frequencies, less than 100 Hz, and can lead to excitation of vibration
modes of the piping and connected equipment. The extent of this problem depends on the
piping design, support configuration and stiffness, valve operation, and other related factors
which determine the severity of the resulting vibration.

159
A relief or control valve on piping systems in gas service, or other pressure reducing devices,
can generate high levels of high frequency acoustic energy, an effect commonly referred to
as acoustic induced vibration (AIV). In addition to high noise levels arising external to the
piping, this excitation can result in high frequency vibration of the pipe wall, with the
potential for high dynamic stresses at welded features such as supports and small bore
connections. This in turn can lead to the possibility of fatigue cracking within a relatively
short period of time (minutes or hours).
Flow induced pulsation (FIP) can be caused by dead leg branches in pipework, which can be
excited as acoustic resonances with discrete frequencies. These resonances can induce large
shaking forces in the pipework, leading to integrity and safety risks.

Figure 41.2. Example of vortex shedding from an object in the flowstream (Souce: acusim.com)

Figure 41.3. Example of Dead Leg Branch

160
42 CIVIL ENGINEERING STRUCTURAL VELOCITY LIMITS

Figure 42.1. Construction-induced Vibration

Buildings, bridges, towers and other large structures may be subjected to excitation from:

 Wind
 Earthquakes
 Automobile traffic
 Subways and railroads
 Nearby construction and demolition
 Internal mechanical equipment
Offshore structures are also exposed to ocean waves.
High vibration levels in buildings can cause cracking in plaster walls and masonry, as well as
broken windows. Extreme vibration can cause cracking or shifting of foundations or bearing
walls.
Construction vibration becomes a legal matter when there is potential damage to nearby
buildings and homes.

161
Beards [34] gives the following limits for these structures.

Table 42.1. Structural Velocity Limits

VRMS VRMS Damage
(mm/sec) (in/sec) Effect
Up to 5 Up to 2 Most unlikely

5 to 10 2 to 4 Unlikely
Possible – check
Over 10 Over 4
dynamic stress

162
43 TACOMA NARROWS BRIDGE FAILURE

Figure 43.1. Tacoma Narrow Bridge Torsional Mode

Figure 43.2. Tacoma Narrows Bridge Collapse

163
The original Tacoma Narrows Bridge known as “Galloping Gertie” was opened to traffic on July
1, 1940. It was located in Washington State’s Puget Sound.
Strong winds caused the bridge to collapse on November 7, 1940. Initially, 35 mile per hour
winds excited the bridge's transverse vibration mode, with an amplitude of 1.5 feet. This
motion lasted 3 hours. [35]
The wind then increased to 42 miles per hour beginning at 9:45 AM. In addition, a support
cable at mid-span snapped, resulting in an unbalanced loading condition. The bridge response
thus changed to a 0.2 Hz torsional vibration mode, with an amplitude up to 28 feet peak-to-
peak. The corresponding velocity was 168 in/sec zero-to-peak.
The stress-velocity relationship gives a stress of 50 ksi assuming a proportionality factor of K=2
per the following equation with the materials properties of steel.

σ = K ρc v , (43.1)

The bridge collapsed at approximately 11:00 AM. The most likely cause was aerodynamic
instability or bridge flutter. But vortex shedding may have also been a factor.
The bridge’s deck was made from concrete and carbon steel, although the specific steel alloy is
not readily available.
The bridge would have undergone about 900 torsional cycles at approximately 50 ksi, plus an
unknown number of bending cycles during its four-month life.
Any fatigue analysis is uncertain because the alloy, heat treatment and stress concentration
factors are all unknown. Furthermore, published S-N curves typically begin well above 1000
cycles. But a “ballpark calculation” is still possible.
The following S-N curve suggests that the stress limit would be roughly 80 ksi for 1000 cycles.
The ultimate tensile strength may have been twice that level.
The Palmgren-Miners cumulative damage index is about (50 ksi/80 ksi) = 0.63 for the bridge’s
torsional oscillation stress. Failure occurs when this index reaches 1 per classical theory, but
some references give a threshold of 0.1 for conservatism.
The cumulative index would have been about 1 for a stress concentration factor of 1.5. Again,
the true cumulative index would be even higher if the bending cycles had been included with
their respective stress levels. Traffic-induced loads and thermal expansion and contraction
would have contributed as well.
The “aerodynamic instability vs. vortex shedding debate” will continue. But fatigue was almost
certainly the failure mode.

164
Figure 43.3. Typical Carbon Steel S-N Curve

165
44 CONCLUSIONS

44.1 STRESS-VELOCITY RELATIONSHIP

Modal stress is directly proportional to modal velocity for both free vibration and resonant
excitation. The proportionality equation does not depend on frequency, although the velocity
itself depends on frequency.

There are limitations to any stress-velocity equation, however. Crandall [36] noted that stress
and velocity may each have concentrations, particularly for nonuniform structures.

44.2 SHOCK RESPONSE SPECTRA DYNAMIC RANGE

Six shock pulses were considered in terms of their respective shock response spectra. One was
an analytical half-sine shock pulse. Another was an earthquake strong-motion time history.
The other four were measured shock events from rocket vehicles or motors. The dynamic
range for the velocity ranged from 12 to 33 dB among these six samples. In each case the
amplitude range for velocity was less than or equal to that for the corresponding displacement
and acceleration.

Thus, the response velocity is frequency-dependent, although less so than either displacement
or acceleration.

The natural frequency of any component attached to the base structure must still be
considered with respect to the velocity spectrum.

Shock response spectra should be plotted in tripartite format so that the effect of
displacement, velocity and acceleration can be considered together on a single plot. This plot
may be in addition to separate plots of each of the three amplitude metrics.

Again, modal stress is directly proportional to modal velocity. But other failure modes may
have better correlation with either displacement or acceleration. Loss of sway space must be
considered with respect to relative displacement, for example. Thus, attention should be given
to each of the three amplitude metrics.

166
45 BIBLIOGRAPHY

[1] S. Rinehart and J. Pearson, Behavior of Metals under Impulsive Loads, Cleveland, Ohio: American
Society of Metals, 1954.

[2] F. Hunt, Stress and Strain Limits on the Attainable Velocity in Mechanical Systems, JASA, 32(9) 1123-
1128, 1960.

[3] Chalmers and Gaberson, Modal Velocity as a Criterion of Shock Severity, Shock and Vibration
Bulletin, Naval Research Lab, December 1969.

[4] H. Gaberson, Using the Velocity Shock Spectrum for Damage Potential, San Diego, CA: SAVIAC, 74th
Symposium, 2003.

[5] H. Gaberson, An Overview of the Use of the Pseudo Velocity Shock Spectrum, El Segundo, CA:
Dynamic Environments Workshop, The Aerospace Corporation, 2007.

[6] K. Graff, Wave Motions in Elastic Solids, New York: Dover, 1991.

[7] R. Clough and J. Penzien, Dynamics of Structures, New York: McGraw-Hill, 1975.

[8] A. Piersol and T. Paez, Harris’ Shock and Vibration Handbook, 6th Edition, Chapter 27 Shock testing
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[9] A. Piersol and T. Paez, Harris’ Shock and Vibration Handbook, 6th Edition Chapter 40, Equipment
Design, New York: McGraw-Hill, 2010.

[10] A. Piersol, Preliminary Design Procedures for Equipment Exposed to Random Vibration
Environments, Journal of the IEST, Winter 2001.

[11] H. Gaberson, Pseudo Velocity Shock Analysis, Short Course Slides, SAVIAC.

[12] T. Irvine, Notes on Mechanical Shock and Vibration Waves, Revision B,, Vibrationdata, 2000.

[13] T. Irvine, Initial Velocity Excitation Of The Longitundinal Modes, Revision C, Vibrationdata, 2024.

[14] T. Irvine, Initial Velocity Excitation of the Longitudinal Modes in a Beam for Pyrotechnic Shock
Simulation, Vibrationdata, 2008.

[15] T. Irvine, Rod Response to Longitudinal Base Excitation, Steady-State and Transient, Revision B,
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[16] T. Irvine, Longitudinal Vibration of a Tapetered Rod via the Finite Element Method, Vibrationdata,
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[17] T. Irvine, Natural Frequencies of a Shear Beam, Vibrationdata, 2009.

[18] W. Thomson, Theory of Vibration with Applications, 2nd Edition, New Jersey: Prentice-Hall, 1981.

[19] T. Irvine, Bending Frequencies of Beams, Rods, and Pipes, Rev M, Vibrationdata, 2010.

[20] T. Irvine, Steady-State Vibration Response of a Simply-Supported Beam Subjected to Base Excitation,
Vibrationdata, 2010.

[21] T. Irvine, Longitudinal Vibration of a Rod via the Finite Element Method, Revision B, 2008:
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[24] R. Morse, Presentation at Spacecraft & Launch Vehicle Dynamics Environments Workshop Program,
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[27] T. Irvine, The Pseudo Velocity Shock Response Spectrum, Revision A, Vibrationdata, 2010.

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168
View publication stats

Defense, 1989.

169