Lalanne C. Mechanical Vibrations and Shocks: Mechanical Shock Volume II
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44
Mechanical
shock
Figure
2.16. Standardized primary
and
residual relative displacement
SRS
of
a
TPS
pulse
Figure
2.17. Standardized
positive
and
negative relative displacement
SRS
of
a
TPS
pulse
with zero decay time
Shock
response
spectra
domains
45
Initial
peak
saw
tooth
pulse
Figure
2.19. Standardized primary
and
residual relative displacement
SRS
of
an IPS
pulse
Figure
2.18. Standardized
positive
and
negative relative displacements
SRS
of
an
IPS
pulse
46
Mechanical shock
Figure 2.20.
Standardized positive
and
negative relative displacement
SRS
of
an IPS
with
zero
rise time
Rectangular pulse
Figure 2.21.
Standardized positive
and
negative relative displacement
SRS
of
a
rectangular pulse
Shock response
spectra
domains
47
Trapezoidal
pulse
Figure
2.22.
Standardized
positive
and
negative
relative
displacement
SRS
of
a
trapezoidal
pulse
2.5. Difference between shock response spectrum (SRS)
and
extreme response
spectrum
(ERS)
A
spectrum known
as of
extreme
response
spectrum
(ERS)
and
comparable with
the
shock response spectrum (SRS)
is
often
used
for the
study
of
vibrations (Volume
5).
This spectrum gives
the
largest response
of a
linear single-degree-of-freedom
system
according
to its
natural
frequency, for a
given
Q
factor,
when
it is
subjected
to the
vibration under investigation.
In the
case
of the
vibrations,
of
long duration,
this
response takes place during
the
vibration:
the ERS is
thus
a
primary spectrum.
In
the
case
of
shocks,
we in
general calculate
the
highest
response,
which
takes place
during
or
after
the
shock.
2.6. Algorithms
for
calculation
of the
shock response spectrum
Various algorithms have been developed
to
solve
the
second order
differential
equation [2.9] ([COL 90], [COX 83], [DOK 89], [GAB 80], [GRI 96], [HAL 91],
[HUG
83a], [IRV 86], [MER 91], [MER 93], [OHA 62], [SEI
91] and
[SMA 81]).
One
which leads
to the
most reliable results
is
that
of F. W. Cox
[COX
83]
(Section
2.7.).
Although
these calculations
are a
priori relatively simple,
the
round robins that
were
carried
out
([BOZ
97]
[CHA 94]) showed
differences
in the
results, ascribable
48
Mechanical shock
sometimes
to the
algorithms themselves,
but
also
to the use or
programming errors
of the
software.
2.7.
Subroutine
for the
calculation
of the
shock
response
spectrum
The
following procedure
is
used
to
calculate
the
response
of a
linear single-
degree-of-freedom
system
as
well
as the
largest
and
smallest values
after
the
shock
(points
of the
positive
and
negative SRS, primary
and
residual, displacements
relative
and
absolute accelerations).
The
parameters transmitted
to the
procedure
are
the
number
of
points defining
the
shock,
the
natural pulsation
of the
system
and its
Q
factor,
the
temporal step (presumably constant)
of the
signal
and the
array
of the
amplitudes
of the
signal. This procedure
can be
also used
to
calculate
the
response
of
a
one-degree-of-freedom system
to an
arbitrary excitation,
and in
particular
to a
random
vibration (where
one is
only interested
in the
primary response).
Shock
response
spectra
domains
49
Procedure
for the
calculation
of a
point
of the SRS at
frequency
f
0
(GFA-
BASIC)
From
F. W. Cox
[COX
83]
PROCEDURE S_R_S(npts_signal%,wO,Q_factor,dt,VAR xppO)
LOCAL
i%,a,a1 ,a2,b,b1,b2,c,c1,c2,d,d2,e,s,u,v,wdt,w02,w02dt
LOCAL
p1d,p2d,p1a,p2a,pd,pa,wtd,wta,sd,cd,ud,vd,ed,sa,ca,ua,va,ea
'
npts_signal%
=
Number
of
points
of
definition
of the
shock versus time
'
xpp(npts_signal%)
=
Array
of the
amplitudes
of the
shock pulse
' dt=
Temporal step
'
wO=
Undamped natural pulsation
(2*PI*fO)
'
Initialization
and
preparation
of
calculations
psi=l/2/Q_factor
//
Damping ratio
w=wO*SQR(l-psi
A
2)
//
Damped natural pulsation
d=2*psi*wO
d2=d/2
wdt=w*dt
e=EXP(-d2*dt)
s=e*SIN(wdt)
c=e*COS(wdt)
u=w*c-d2*s
v=-w*s-d2*c
w02=wO
A
2
w02dt=w02*dt
1
Calculation
of the
primary
SRS
'
Initialization
of
the
parameters
srcajprim_min=lE100
//
Negative primary
SRS
(absolute acceleration)
srca_prim_max=-srcajprim_mm
//
Positive primary
SRS
(absolute acceleration)
srcd_prim_min=srca_prim_min
//
Negative primary
SRS
(relative displacement)
srcdjprim_max=-srcd_prim_min
//
Positive primary
SRS
(relative displacement)
displacement_z=0
//
Relative displacement
of the
mass under
the
shock
velocity_zp=0
//
Relative velocity
of the
mass
'
Calculation
of
the
sup.
and
inf.
responses
during
the
shock
at the frequency fO
FOR
i%=2
TO
npts_signal%
a=(xpp(i%-1
)-xpp(i%))/w02dt
b=(-xpp(i%-1
)-d*a)/w02
c2=displacement_z-b
c 1
=(d2*c2-t-velocity_zp-a)/w
displacement_z=s*cl+c*c2+a*dt+b
velocity_zp=u
* c1 +
v*
c2+a
responsedjprim=-displacement_z*w02
//
Relative displac. during shock
x
square
of the
pulsation
50
Mechanical shock
i
responsea_prim=-d*velocity_zp-displacement_z*w02
//
Absolute response
accel.
during
the
shock
'
Positive primary
SRS
of
absolute accelerations
srcaj3rim_max=ABS(MAX(srca_prim_max,responsea_prim))
'
Negative primary
SRS
of
absolute
accelerations
srca_prim_min=MIN(srca_prim_min,responsea_prim)
'
Positive primary
SRS
of
the
relative displacements
srcd_prim_max=ABS(MAX(srcd_prim_max,responsed_prim))
'
Negative primary
SRS
of
the
relative displacements
srcdjprim_min=MrN(srcd_prim_min,responsed_prim)
NEXT
i%
'
Calculation
of the
residual
SRS
1
Initial conditions
for the
residual response
=
Conditions
at the end of the
shock
srca_res_max=responsea_prim
//
Positive residual
SRS of
absolute accelerations
srcajres_min=responseajprim
//Negative residual
SRS of
absolute accelerations
srcd_res_max=responsedjprim
//
Positive residual
SRS of the
relative
displacements
srcd_res_min=responsed_prim
//
Negative residual
SRS of the
relative
displacements
'
Calculation
of
the
phase angle
of
the
first peak
of
the
residual relative
displacement
c 1
=(d2 *displacement_z+velocity_zp)/w
c2=displacement_z
al=-w*c2-d2*cl
a2=w*cl-d2*c2
pld=-al
p2d=a2
IFpld=0
pd=PI/2*SGN(p2d)
ELSE
pd=ATN(p2d/pld)
ENDIF
IF
pd>=0
wtd=pd
ELSE
wtd=PI+pd
ENDIF
'
Calculation
of
the
phase
angle
of
the
first peak
of
residual
absolute
acceleration
bla=-w*a2-d2*al
b2a=w*al-d2*a2
pla=-d*bla-al*w02
p2a=d*b2a+a2*w02
IFpla=0
pa=PI/2*SGN(p2a)
Shock
response
spectra
domains
51
ELSE
pa=ATN(p2a/pla)
ENDIF
IFpa>=0
wta^a
ELSE
wta=PI+pa
ENDIF
FOR
i%=l
TO 2 //
Calculation
of
the
sup.
and
inf.
values
after
the
shock
at the
frequency fO
'
Residual relative
displacement
sd=SIN(wtd)
cd=COS(wtd)
ud=w*cd-d2*sd
vd=-w*sd-d2*cd
ed=EXP(-d2*wtd/w)
displacementd_z=ed*(sd*c
1
+cd*c2)
velocityd_zp=ed*(ud*c
1
+vd*c2)
'
Residual absolute acceleration
sa=SIN(wta)
ca=COS(wta)
ua=w*ca-d2*sa
va=-w*sa-d2*ca
ea=EXP(-d2*wta/w)
displacementa_z=ea*(sa*c
1
+ca*c2)
velocitya_zp=ea*(ua*c
1
+va*c2)
1
Residual
SRS
srcd_res=-displacementd_z*w02
// SRS of the
relative displacements
srca_res=-d*velocitya_zp-displacementa_z*w02
// SRS of
absolute
accelerations
srcd_res_max=MAX(srcd_res_max,srcd_res)
//
Positive
residual
SRS of the
relative
displacements
srcd_res_min=MIN(srcd_res_min,srcd_res)//Negative residual
SRS of the
relative
displacements
srca_res_max=MAX(srca_res_max,srca_res)
//
Positive residual
SRS of the
absolute
accelerations
srca_res_min=MIN(srca_res_min,srca_res)
//
Negative residual
SRS of the
absolute accelerations
wtd=wtd+PI
wta=\vta+PI
NEXT
i%
srcdj)os=MAX(srcd_prim_max,srcd_res_max)
//
Positive
SRS of the
relative
displacements
srcd_neg=MrN(srcdjprim_min,srcd_res_min)
//
Negative
SRS of the
relative
displacements
52
Mechanical
shock
srcd_maximax=MAX(srcd_pos,ABS(srcd_neg))
//
Maximax
SRS of the
relative
displacements
srcajpos=MAX(srca_prim_max,srca_res_max)
//
Positive
SRS of
absolute
accelerations
srca_neg=MIN(srca_prim_min,srca_res_min) //Negative
SRS of
absolute
accelerations
srca_maximax=MAX(srcajpos,ABS(srca_neg))
//
Maximax
SRS of
absolute
accelerations
RETURN
2.8. Choice
of the
digitization
frequency
of the
signal
The
frequency of
digitalization
of the
signal
has an
influence
on the
calculated
response spectrum.
If
this
frequency is too
small:
-The
spectrum
of a
shock
with
zero
velocity change
can be
false
at low
frequency,
digitalization leading
artificially
to a
difference
between
the
positive
and
negative
areas under
the
shock pulse,
i.e.
to an
apparent velocity change that
is not
zero
and
thus leading
to an
incorrect slope
in
this range. Correct restitution
of the
velocity change (error
of
about
1% for
example)
can
require, according
to the
shape
of
the
shock,
up to 70
points
per
cycle.
- The
spectrum
can be
erroneous
at
high
frequencies. The
error
is
here related
to
the
detection
of the
largest peak
of the
response, which occurs throughout shock
(primary spectrum). Figure
2.23
shows
the
error made
in the
stringent case more
when
the
points
surrounding
the
peak
are
symmetrical with
respect
to the
peak.
If
we set
it
can be
shown that,
in
this case,
the
error made according
to the
sampling
factor
S
F
is
equal
to
[SIN
81]
[WIS
83]
Shock
response
spectra
domains
53
Figure 2.23.
Error
made
in
measuring
the
Figure 2.24.
Error
made
in
measuring
the
amplitude
of
the
peak
amplitude
of
the
peak
plotted
against
sampling
factor
The
sampling
frequency
must
be
higher than
16
times
the
maximum
frequency
of
the
spectrum
so
that
the
error made
at
high
frequency is
lower than
2% (23
times
the
maximum
frequency for an
error lower than 1%).
The
rule
of
thumb
often
used
to
specify
a
sampling factor equal
10 can
lead
to an
error
of
about
5%.
The
method proposing
a
parabolic interpolation between
the
points
to
evaluate
the
value
of the
maximum does
not
lead
to
better results.
2.9. Example
of use of
shock response spectra
Let
us
consider
as an
example
the
case
of a
package intended
to
limit
to 100
m/s
2
acceleration
on the
transported
equipment
of
mass
m
when
the
package
itself
is
subjected
to a
half-sine shock
of
amplitude
300
m/s
2
and of
duration
6 ms. One in
addition
imposes
a
maximum displacement
of the
equipment
in the
package (under
the
effect
of the
shock) equal
to e = 4 cm (to
prevent that
the
equipment coming
into contact with
the
wall
of the
package).
It
is
supposed that
the
system made
up by the
mass
m of the
equipment
and the
suspension
is
comparable
to a
one-degree-of-freedom system with
a Q
factor equal
to Q = 5. We
want
to
determine
the
stiffness
k of the
suspension
to
satisfy these
requirements when
the
mass
m is
equal
to 50 kg.