Theatre Design & Technology - Spring 2013 - 50

Construction of Wire Rope

Area Factor

7x7 GAC

0.471

7x19 GAC

0.472

6x19W with fiber core

0.416

6x19W with IWRC

0.482

6x36WS with fiber core

0.419

6x36WS with IWRC

0.485

8x19W with fiber core

0.366

8x19W with IWRC

0.497

ExamplE: a 200-pound object is connected to a 10-foot length of used
7x19 GAC, 1/4 inch in diameter and free-falls 1 foot (12 inches). What is
the shock load on the wire rope and the beam to which it is attached?
First, calculate the area of the wire rope using the area factor for
7x19 GAC in the table.
Area = 0.25 × 0.25 × 0.472 = 0.0295 square inches

Next, plug the other variables into the shock load equation.

( √
( √

Shock Load = Load × 1 +

1+

Shock Load = Load × 1 +

1+

2 × 12 × 0.0295 × 15,000,000
200 × 10 × 12
10,620,000

24,000
Shock Load = Load × (1+ √ 1 + 442.5 )

)

)

Shock Load = Load × (1+ √ 443.5 )
Shock Load = Load × (1+ 21.059 )

Accurately calculating the shock load of fiber ropes can be more difficult
than calculating the shock load on wire rope for three reasons: 1) because humidity can change the ability of some materials to stretch, and
some natural fibers lose elasticity with age; 2) because there are many
different blends of materials; and 3) because different manufacturers of
rope supply different data about the energy-absorbing capacities of their
ropes. Despite this, it is possible to get a good idea of the shock load of a
falling object being caught by a fiber rope.
As stated earlier, different companies supply different data on the
elasticity of their ropes. We will look at the data provided by two rope
companies, New England Ropes and Yale Cordage, and show how to calculate shock loads on each.
method #1: Ron Reese, an arborist in Chattanooga, Tennessee, developed the first method we will look at. His equation is:

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Rope stretch is a percentage specified by the manufacturer. For
example, New England Ropes KMIII 7/16˝ rope stretches 5.1 percent. Use
"5.1" in the equation for calculating a above.
Rope length is the distance, in feet, between the load and the termination point.
Rope load is the force, in pounds, required to achieve the manufacturer's rope stretch. For example, the rope stretch of New England Ropes
KMIII is 5.1 percent at 10 percent of the minimum breaking strength
(MBS). The MBS is 7,083 pounds, so the rope load = 0.1 × 7,083 pounds
= 708 pounds.
load is weight of object that falls.
Fall is the distance, in feet, that the load free falls.
Let's use the same load, rope length, and fall distance we used in our
example calculation for wire rope. We'll use 1/2-inch Multiline II, a fairly
stretchy rope, instead of 1/4-inch 7x9 GAC.
ExamplE: A 200-pound object is connected to a 10-foot length of 1/2inch diameter Multiline II and free-falls 1 foot. What is the shock load on
the rope and the beam to which it is attached?
According to New England Ropes, Multiline II has an elongation
rate of 1.7 percent at 10 percent of its tensile strength. Since the tensile
strength of 1/2-inch diameter Multiline II is 6,000 pounds, 10 percent
would be 600 pounds.

B = -2 × 0.000141667 × 200
B = -0.056666667

Shock loadS on fiber rope

48

a = (0.005 × Rope Stretch × Rope Length)/ Rope Load
B = -2 × A × Load
C = -Load × Fall Distance (in feet)

0.005 × 1.7 x 10
600
A = 0.000141667

Shock load = 4,411.88 pounds

-B + √B² - (4 × A × C)
4×A

where . . .

A=

Shock Load = 200 × 22.059

Shock Load =

n

C = -200 × 1
C = -200

Plugging these numbers into the shock load equation for fiber rope,
we get:
Shock Load =

0.056666667 + √-0.056666667 ² - (4 × 0.000141667 × -200)
4 × 0.000141667

Shock Load =

0.056666667 + √-0.003211111 - (-0.1133336)
0.000566668

Shock Load =

0.056666667 + √0.116544711
0.000566668

Shock Load =

0.056666667 + 0.341386454
0.000566668

Shock Load =

0.398053254
0.000566668

Shock load = 702.45 pounds

s p r i n g 2013
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