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BRUEL & KJAER System V - Page 23

BRUEL & KJAER System V
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However,
it
should
be recognized
that
this
current
passing
through
the
drive
coil is a part
of
the
total
allowable
current
and
therefore
reduces
the
dynamie
force
rating
of
the
shaker
by
reducing
the
amount
of
alternating
current
which
may
be
run
through
the
coil.
At
frequencies
above 1
50
Hz
the
dynamie
rated
force
i.s reduced by
2,5%
if
half
of the
available
static
force
is used and is reduced by
10%
if
the
fu
ll
static
force
is used.
At
fre-
quencies
below
150
Hz
the
dynamie
rated
force
is
reduced
by
15%
if
half
of
the
available
static
force
is used
and
is
reduced
by
30%
if
the
full
static
force
is used.
Fi
g
.3
.5
shows
the
effect
on t he ava
ilab
le
force
us
ing
the
static
cen
tri
ng
force.
It
is
here
that
the
third
controlling
factor
enters
into
the
allowable
dead
weight
calcula-
tion,
i.
e.,
the
test
conditions
. ,
The
test
object
will
be
run
at
a
certain
acceleration
level,
which
limits
not
onl
y
the
weigh
t ,
but
h
ow
much
of
the
for
ce
may
be used up in c
entr-
ing
the
table.
Also,
the
test
will
be
conducted
over a
certain
frequency
range,
and
the
dis-
placement
at
the
low
end
of
this
test
range
wi
ll
affect
the
available
amplitude.
The
two
plots
in
Figs.3.4
and
3.5
can be used
to
det
ermine
how
much
of
th
e dead
weight
can
be
taken
up by
the
static
force
and
how
much
by
the
flexures,
for
a given
test
.
Examples:
1.
In
Fig.3
.6 is
shown
an
arrangement
for
vibration
testing
an
electronic
ins
trum
e
nt
on
a
4801
Body
and
4
B13
Head. The
mass
of
the
instrument
to
be
test
ed
is
16
,8 kg
(37Ib).
and
that
of
the
fixture
is
6,7
kg
(14,
7I
b). The
total
mass
to
be
driven,
inc!ud-
ing
the
mass
of
the
moving
element,
is
therefore
24
,2 kg
(53Ib).
Question
:
Can
this
arrangement
be
driven
at
10
m / s 2
in
the
frequency
range
20
Hz
to
1 kHz?
At
least
237
N (531bf) is needed
to
drive
the
system
at
10
m/
s
2
.
If t
he
f u
ll
st
a
ti
c
forc
e
is used,
there
w
ill
still
be
70%
of
the
full
dynamic
f
orce
av
aila
ble
,
or
312
N (
70
Ibf).
The rest
of
the
weight
will
be
taken
up by
the
flexures.
The
do
wn
war
d
fo
rce
ex
erted
on
the
flex
ures
by
the
remainder
of
the
w e
ight
is
96
,
5N
(21 ,
7Ibf
).
Fr
om
Fig.3.4
,
it
can
be
seen
that
the
remaining
availabl
e displ
acement
is 1,
75
mm
(
0,07
in
).
10
m/ s
2
accel
era-
tion
equals
1,7
5
mm
at
12
Hz, so
the
20
Hz
lower
limit
of
the
test
is
weil
within
the
cap-
ability
of
this
arrangement.
2. A
manu
fact
urer
of
elec
tri
cal c
onn
ec
to
rs
wants
to
test
a
set
of
connectors
we
ighing
500
g at
100
m/ s
2
,
fro
m
20
to
2
000
Hz,
using
a
480
5 Body
an
d a 4 8
12
Head.
Question:
What
is
the
largest
posslble
mass
which
the
fixture
can
have?
100
m/ s
2
at
20
Hz co
rre
spo
nds
to
13
mm
double
displ
acem
ent
,
i.e.
th
e
full
di
splace-
ment
of
the
4812
Head, T
here
for
e,
the
mass
of
fixture
and
t
est
object
mu
st be
rais
ed
wi
th
th
e statiC
centring.
A
study
of
the
lin
es in
Fig.3
.5
will
show
that
they
have
a slope
of
-1,
i. e.,
one
may
say
that:
Force
Available
=
Maximum
Specified
Force-Lifting
Force. The
lifting
force
will
be
the
force
necessary
to
lift
the
test
object
and
fixture,
in
this
case,
..,..,."..
- .\
;1
22

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