Updated 4-20-07
Shock and Acceleration Theory
In this lab, you will measure a dynamic cushioning curve for 2 sample foam thicknesses,
and optimize a cushioning system to minimize the effect of an impact on the floppy disk
drive.
Cushioning design and testing related to pr oduct shipping is usually conducted after a
product is ready to be shipped to customers. In addition to protecting against impact
and vibration, well-designed packaging must be able to withstand a wide range of
temperatures and environmental conditions wit hout degradation. Also, the impacts
which a packaging system designed for electronic equipment experiences are usually
low in number, but with potentially high peak accelerations. Thus, the packaging is
designed to withstand a small number of catastrophic impacts, as it is assumed that the
object being protected can handle low accelerations without damage.
Theory
Impact and Shock
A large force applied to a system for a short time interval is often called a shock, or impact, and this
large force can produce correspondingly large accelerations. If a large acceleration is applied to a system
containing electro-mechanical components, such as in a floppy or hard disk drive, failure of one (or
more) of the components can occur. Newton’s second law shows that the acceleration of a body is
related to the forces applied to the body:
F  ma
(1)
where F is the force applied to the body, m is the mass of the body, and a is the acceleration of the
body. The force, and thus the acceleration, may be functions of time, and in the case of a shock or
impact, the force and acceleration are rapidly changing functions of time. If the mass is constant, the
right side of Equation 1 can be written as the derivative with respect to time of the momentum of the
body:
d 
d
d
F  ma  m v 
(mv) 
P
dt  dt
dt
(2)
If a force is applied during a time interval t = t2 - t1, the change in momentum of the body during this
time interval can be calculated by integrating both sides of this equation with respect to time (from t 1 to
t2).
t2
t2
d 
I   F dt    P dt  P2  P1   P
dt 
t1
t1 
1 of 7
(3)
Updated 4-20-07
This force applied over a time interval is called impulse, I, and equals the change in momentum during
that time interval. Since impulse only depends on velocity and mass, and is independent of the impact
surface, the impulse (area under the force-time curve) will be constant (for a given kinetic energy and
mass) for any impact on any surface. Figure 1 shows two theoretical impact response curves for an
object dropped from the same height on two surfaces with different physical properties. In these curves,
A1 will be equal to A2, because the area under each curve is related to the impulse, which is constant for
a constant drop height and object mass.
Figure 1: Curves for Different Impact Surfaces (Constant Mass and Kinetic Energy)
The mechanical and physical properties of the surface that an object strikes (thickness of the material,
modulus of elasticity and surface area), and the amount of kinetic energy contained by the object at the
time of impact determine the profile of the acceleration pulse experienced by the object.
As in a coiled spring, the stiffness or spring rate of a material being deformed elastically can be
represented by a material spring constant k. From Hooke's Law and the elastic stress-strain relationship:
F  kx
(4)

E

(5)
where x is the deformation of the material, F is the total applied force, k is the material spring constant,
E is the modulus of elasticity,  is the applied stress, and  is the strain resulting from the applied stress.
For a material with a cross-sectional area A and thickness L as shown below in Figure 2,  and  can be
related to F and x,

F
A
(6)

x
L
(7)
Substituting these equations into Equation 5 above, we get
2 of 7
Updated 4-20-07
FL
E
xA
(8)
If we rearrange this equation and compare it to Equation 4,
EA 
F   x  k x
 L 
we see that k for a bulk material is
(9)
EA
.
L
Figure 2: Dimensions of a Material Sample with Spring Constant k = AE/L
To understand how representing a bulk material as a spring can allow us to understand the effects of A,
L, and E on accelerations experienced during an impact, we will think about what happens when we
drop the same object, from the same height, on two different materials. Since the drop height is
equivalent in both test, the kinetic energy at the moment before impact will be equal. One material will
be stiff and will have a spring constant klarge, while the other material will be soft and will have a spring
constant ksmall.
When an object experiences an impact with a material acting as a cushioning system, the material
deforms to absorb the kinetic energy contained in the object, Ekinetic. As in all springs, the elastic
deformation of the material allows the material to absorb the kinetic energy and convert it to potential
energy Epotential. At the point when the object has lost all of its kinetic energy, Ekinetic can be related to the
elastic deformation of the material by equation 10:
1
E kinetic  kx 2
2
(10)
where x is the elastic deformation of the material at the point when the object has lost all of the kinetic
energy, and k is the spring constant of the material sample. Since Ekinetic is equal for both drops, we can
write the equation for each impact:
1
1
2
2
E kinetic  k l arg e x l arg e  k small x small
2
2
3 of 7
(11)
Updated 4-20-07
Rearranging this equation gives us
kl arg e .5 x
sma ll



 
x l arg e
k sma ll 
(12)
For each separate material, we can write the Hooke's Law equation as follows:
Fl arg e  k l arg e x l arg e
(13)
Fsmall  k small x small
(14)
and dividing Equation 13 by Equation 14 gives us the ratio of the maximum forces in each impact:
Fl arg e
Fsmall
Substituting the value for
x small
x l arg e

k l arg e x l arg e
k small x small
(15)
from Equation 12 into Equation 15 gives
1
k l arge 2

 


Fsmall ksmall 
Fl arg e
(16)
kl arg e .5

By our definition of klarge and ksmall, the quantity, 
k
  1 , so Flarge > Fsmall. From this equation, we
 small 
can see that for a constant kinetic energy and mass, a material with a large k (k=klarge, AE >>L), acts as a
stiffer spring and applies a larger maximum force during an impact than a material with a small k (k =
ksmall, AE <<L). For absorption of the same amount of kinetic energy, an impact with a material with a
high spring constant will produce a narrower pulse and higher peak acceleration than an impact with a
material with a low spring constant.
Materials that are normally used for cushioning and packaging, such as foam and rubber, have low
values of k and produce wider acceleration pulses and correspondingly lower peak accelerations when
dissipating the same amount of kinetic energy. The following graphs show examples of acceleration data
from impacts with materials with either a high or low spring constant.
4 of 7
Updated 4-20-07
9
8
7
6
5
Ac c e le ra tio n (g )
4
3
2
1
0
-1
-2
0.000
0.050
0.100
0.150
0.200
Tim e (s)
Figure 3: Plot of Acceleration Data for a Material with High Spring Constant
8
7
6
5
4
3
Ac c e le ra tio n (g )
2
1
0
-1
-2
0.000
0.050
0.100
0.150
0.200
Tim e (s)
Figure 4: Plot of Acceleration Data for a Material with Low Spring Constant
Dynamic Cushioning Curve
Once a company has determined the maximum acceleration that a mechanical system can withstand by
testing the system with shock, and vibration, a cushioning system must be designed to prevent
accelerations above this value.
Often a dynamic cushioning curve, a plot of peak acceleration versus static loading, for a given material
thickness and drop height of the object that is being cushioned, is used. Static loading is defined as the
weight of the object that is experiencing the impact divided by the area of the cushioning material
supporting the object. For a given drop height, a series of curves can be generated for different material
5 of 7
Updated 4-20-07
thicknesses, and used to select the appropriate thickness and area of the cushioning material to
minimize the peak acceleration.
Examples of dynamic cushioning curves are shown in Figure 5 below. The different curves correspond to
different material thicknesses. Notice that the minimum shock is in middle range the static loading,
reverse of common intuition that a larger area of cushioning would continually decrease the peak
accelerations experienced during an impact.
Figure 5: An Example of a Dynamic Cushioning Curve
Note that in the curve in Figure 5, a low value of static loading ( static loading 
weight
) corresponds
area
to a larger area supporting the object, while a high value of static loading corresponds to a smaller area
supporting the object.
In the case of large areas ( static loading  0 ), k 
AE
, and k increases linearly with A. As the area
L
increases, the spring constant of the cushioning material increases proportionally, and the cushioning
material behaves as a stiffer and stiffer spring, thus the peak acceleration is greater.
At smaller values of area ( A  0 ), the spring constant k decreases. As the material spring constant
decreases to very small values, the thickness of the material starts to have an effect on the physics of
the dynamic force. For small A, and correspondingly low k values, the deflection of the material will be
large, and if the deflection is too large, the compression of the material causes the force-deflection
curve to deviate from the linear region. Essentially you have compressed the foam as much as it will
6 of 7
Updated 4-20-07
physically compress and have lost all air space in the closed-cell foam and so it no longer behaves as a
linear spring (k is a rapidly increasing function of deflection).
Assignment Questions
1. For each sample plot the dynamic cushioning curve (peak acceleration vs. static loading).
2. Determine the foam area (Amin) you would specify to minimize the peak acceleration (adyn,min) for
each material. Which material works better?
3. Describe the measurement uncertainties associated with the data you collected. You may have
to look up the specifications on your accelerometer. What are possible errors in your
experiment?
4. Calculate and plot the 95% confidence value for the measured maximum acceleration at each
value of foam area.
5. Plot acceleration vs. time for one of your more interesting foam configurations. Note on the
graph what is happening at critical points.
6. Using the information contained in your acceleration vs. time plots, calculate the maximum
displacement of the foam for a few interesting examples. One way to do this is to integrate
acceleration using the trapezoidal rule. You will need to do this twice to generate a
displacement curve. Don’t forget about initial conditions and be sure to note on your graphs
what physically happening at each of the critical points.
7. Calculate the modulus of elasticity, E, and the spring constant, k, by plotting Force vs.
Compression. Does the foam behave like a linear spring? What about with other larger and
smaller foam areas? How are modules of elasticity and the spring constant physically related to
each other?
8. NOTE: The mass of the disk drive assembly is 1470 grams.
7 of 7