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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 avail able
static force is used and is reduced by 30% if the full static force
is
used.
Fig
.3
.5
shows the effect
on
t
he ava ilable force
using
t
he
static
centri
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
weight
,
but
how
much
of
the force may be
used
up
in
centr
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 determine how much
of th
e
dead
weight can be taken up by the static force and how much by the flexures, for
a gi
ven
test
.
Examples:
1. In Fig.3 .6
is
shown an arrangement for
vibration
testing
an electronic
instrum ent on
a
4801
Body
and
4
B13 Head
.
The mass of the instrument to be
tested
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 (5
31bf)
is needed to drive the
syst em at 10
m/s
2
.
If t
he
f ull
stati
c force
is used, there w
ill
still be 70% of the full dynamic force availa
ble
,
or 312
N
(70
Ibf).
The rest of the weight
will
be taken
up
by
the
flexures.
The
down ward
force
exerted
on
the
flex
ures by the remainder of the
w eight
is
96
,
5N
(21
,
7Ibf).
From
Fig.3.4
,
it
can
be
seen that the remaining availabl
e
displacement
is
1 ,75
mm (0,07 in). 10
m
/
s
2
accelera
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
manufact
urer
of
electrical
connectors wants
to test
a
set of
connectors w
eighing
500 g
at
100
m
/
s
2 ,
fro
m
20
to 2
000
Hz,
using
a
480
5
Body
and 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 1
3 mm double displacement
, i.e.
the
full
displace
ment
of
the
4812
Head,
Therefore, the m
ass
of fixture
and
test
obj
ect
mu st be
raised
with 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,
22
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