Biomedical Engineering Reference
In-Depth Information
proof test in this case is 25,000 psi. If we assume a 10:1 safety margin, the
maximum allowable stress is 2500 psi. The elastic modulus is 10.5 × 10
6
psi
and the unjacketed diameter is 80 μm.
Knowing the allowable stress and the desired strain, we can calculate the
total length of fiber. The elastic modulus
M
is defined as
stress
strain
1
i
M
=
=
stress
×
=
stress
×
(4.12)
δ
l
/
i
δ
l
where
l
is the total length
δ
l
is the total elongation
Then
6
−
6
δ
stress
M l
(
10 6 10
.
×
)(
16 56 10
.
×
)(
39 37
.
)
l
=
=
=
2 74
.
in. mini
mum
(4.13)
2500
The tensile force is the stress multiplied by the cross-sectional area.
2
⎡
⎢
−
6
39 37
⎤
⎥
(80
× 0
1
)
.
F
T
=
2500
π
=
0.019 lb
(4.14)
4
We based these calculations on the assumption we want to minimize the
amount of fiber. The length we calculated is unreasonably short. It will be
more convenient to use a longer fiber. This can be accomplished by simply
reducing the maximum stress. We conclude that there is no basic problem in
achieving the necessary sensitivity.
4.18.4 System Sensitivity
To this point, we related a pressure scale to fiber strain, stress, and ten-
sile force, but did not show the transduction from hydrostatic pressure to
fiber tensile force/stress. As mentioned previously, two methods have been
employed. One is to wind the fiber around a compliant cylinder with the
pressure differential applied between the inside and the outside of the cyl-
inder. The second makes use of the pressure deformation of an elastic jacket
applied directly to the fiber. We will examine both methods.
The density of sea water at 15°C is 63.99 lb ft
−3
. One inch of water is equiva-
lent to a pressure of
P
= 63.99/12
3
= 37(10
−3
) lb in.
−2
. If the hydrostatic pres-
sure operates on area
A
, the force is
F = PA
. From the above data,
A
= 0.01
9 × 12
3
/63.99 = 0.513 in.
2
. Again, this figure corresponds to an impractical
minimum length of fiber. If the fiber length is increased and the maximum
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