ELASTIC STRETCH
Elastic stretch results from recoverable deformation
of
the steel itself. Here, again, a quantity cannot be precisely
calculated. However, the following equation can provide a reasonable approximation for a good many situations.
Changes in length
(ft)-
Change in load (lb) x Length (ft)
Area (inches
2
)x
Modulus
of
Elasticity (psi)
The modulus
of
elasticity is given in Table 17, and the area can be found in Table 18.
TABLE 17 APPROXIMATE MODULUS
OF
ELASTICITY PSI*
Rope Classification Zero through 20% Loading
21
% to 65% Loading
6 x 7 with fiber core
11,700,000
13,000,000
6 x
19
with fiber core 10,800,000
12,000,000
6 x 36 with fiber core 9,900,000 11,000,000
8 x
19
with fiber core 8,100,000 9,000,000
6 x 19 with IWRC 13,500,000 15,000,000
6 x 36 with IWRC 12,600,000
14,000,000
8 x
19
with IWRC 12,000,000
13,500,000
8 x 36 with IWRC 11,500,000
13,000,000
* Applicable to new rope with constructional stretch removed.
EXAMPLE: How much elastic stretch is expected to occur in 200 ft
of
112
inch 6 x 25
FW
ElP
IWRC rope when
loaded to
20%
of
its minimum breaking force?
Minimum Breaking Force = 13.3tons
(26,600 Ib)
20%
of
which = 5,320 lb.
Area
of
112
inch is found by squaring the diameter and multiplying it by the area
of
1 inch rope given
in
Table
18
under the "IWRC" heading and opposite 6 x
25
FW, i.e.,
112
x
112
x .483=.121.
The modulus
of
elasticity
is
found in Table
17
opposite the 6 x 19 IWRC (because 6 x 25FW is a member
of
this
classification) and under the
"Zero through 20% Loading." i.e. 13,500,000 psi.
Substituting these values, the equation reads as follows:
Change
in
length =(5320 x 200) / (.121 x 13,500,000) = 0.65 feet
or
7.8 inches.
5320 x 200
Change in length =
.121
x 13,500,000
=
.65
Ft (7.8 inches)
A word
of
caution concerning the use
of
Table
17:
the higher modulus given under the "21 % to 65% Loading" is
based on the assumption that both the initial and the final load fall within this range.
If
the above example were
restated to the effect that the load was 35% (or
9,310
Ib)
of
the minimum breaking force, it would be incorrect to
rework the problem simply by making two substitutions: the new load and the higher modulus
of
15,000,000 psi. To
do
so would ignore the greater stretch that occurs at the lower modulus during the initial loading.
90 • Wire Rope Technical Board - Wire Rope Users Manual, Fourth Edition
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