Calculation of Vertical Fall Arrest Forces This equation allows users to apply imperial units, to compute arresting force. The fall factor f, is the ratio of h/L, and no correction factor is needed. Fp = (0.031Wp · g) + (1.012 ! 0.003125Wp 0.003125Wp · f · Kp) a · b · s c For 0.1 < f < 2, where a = fall arrest device reduction factor, Table 1 b = body support reduction factor, 0.8 for body harness c = rigid weight/manikin factor, 1.4 f = fall factor (h/L), ratio g = gravitational acceleration, 32.2 ft/s! Fp = maximum arresting force, lbf. h = ffreefall reefall distance, ft. Kp = lanyard tension modulus, psi figure 1 L = lanyard/lifeline length, ft. S = shock-absorber reduction factor, table 2 Wp = weight of mass, lbs. Table 1. Fall Arrester Device Reduction Factor: a Inertia type, wire rope LL 0.7 Inertia type, synthetic LL 0.9 Friction 0.7 Mechanical lever 1.0 No fall arrester 1.0 Table 2. Shock-Absorber Reduction Factor: s Tear Stitches 0.6 Tear Fabric (synthetic) 0.7 Tear Fabric (wire rope) 0.6 No shock absorber 1.0 h = free-fall free-fall distance. This is usually twice the U-shaped slack of the lanyard. However, for clearance calculations, be sure to add a dd approximately 5 feet for a harness in order to ensure clearance for the worker’s limbs hanging below the harness D-ring. L = active length of the lanyard or combined lifeline/lanyard length. The longer this line, the longer the stretch that has to be included when calculating clearances.
the maximum a = fall arrester device reduction factor. In general, the reduction of the arresting force by a fall arrest device is the result of several phenomena, of which dissipation of the fall energy due to friction between the fall arrest device and the vertical lifeline is a primary primary contributor. The reduction factor, a, is defined defined as the ratio of the the maximum arrest force in a fall arrest system with a fall arrest device, to the maximum arrest force in a fall arrest system without a fall arrest device (under the same conditions, with all other elements of the fall arrest system system being the same). An inertial wire rope grab can have a reduction factor of 0.7, and a synthetic rope grab can have a value as high as 0.9, if there is no slip. Page 1 of 4
When no fall arrest device is included in the system, use a factor of 1.0. These factors have been experimentally determined for each device. When the exact value of (a) is not known, the highest one for the particular type of fall arrest system should be employed. B = body support factor. This is based upon laboratory drop tests and is 0.8 for body harnesses. c = rigid weight to human weight factor. OSHA has adopted a 1.4 factor (a 220 lb. rigid weight is equivalent to a 310 lb person in its 1910.66, appendix C). s = shock-absorber reduction factor. This is used to provide additional shock absorption to that in the lanyard. Most independent shock absorbers reduce the arresting force on the body to a range of 600 – 900 pounds. Within the overall strength of the shock absorber, the values generally range between 0.4 and 0.7. When very long falls are anticipated, the capacity of the absorber must be checked. K = rope modulus (extensional stretch). In this text, the term modulus is used to indicate the stretch measured when rope is subjected to various loads. There is little correlation between “modulus for stretch” and the term “tension modulus” (normally used in engineering calculations), unless the area of the stress member, the angle of twist of the rope, and the length of the lanyard/vertical lifeline are included in the calculation and yield strengths are not exceeded. Rope modulus, K, versus fall factor, f
K N N/mm!
Kp psi 3 7,256,894
50,000 2
6,531,204
45,000
1
odulus ( K )
5,805,515
40,000
5,079,835
35,000 30,000
4,354,136
25,000
3,628,447
0
0.25
0.5
0.75
1.0
1.25
1.5
1.75
2.0
Fall Factor ( f ) NOTE: Modulus for 5/16-inch diameter wi re rope cable 6 x 21 is 13 million psi.
1. "-inch diameter 3 strand, nylon 6.7 lb/100 ft. 2. 5/8 –inch diameter, 3 strand, nylon 9.6 lb./100 ft. 3. 5/8-inch diameter 3 strand, polypropylene 8.5 lb/100 ft.
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Example 1.
A 220-pound welder was exposed to an elevated fall hazard. He had been issued a 6foot, 5/16-inch (6x21) wire rope lanyard with a fiber core and a full body harness with a single, back D-ring. When the welder slipped and free fell 6 feet, the wire rope lanyard arrested his fall. What was the maximum arrest force (in pounds) experienced by the welder? Fp = (0.031WP · g ) + (1.012 ! 0.003125Wp · f ·Kp) · a · b · s C Wp = 220 lbs. h = 6 ft. L = 6 ft. f = (h/L) = 1 Kp = 13,000,000 psi (taken from U. S. Wire Rope Engineering Handbook) a = 1 (no fall arrester) b = 0.8 s = 1 (no shock absorber) c = 1.4 g = 32.2 ft/s! a · b · s = 1 · 0.8 · 1 = 0.57144 c 1.4 Fp = (0.031 · 220 · 32.2 ) + (1.012 ! 0.003125 · 220 · 1 · 13000000) · 0.57144 = 1,948.43 lbf. Lets see how this breaks down. We will perform the above problem in three steps. Step one divide the problem into its separate entities.
1.012 x ! 0.003125 · 220 · 1 · 13000000)
0.031 x 220 x 32.2
219.604
+
219.604
+ 219.604
1.012 x 2989.565 3025.44 +
1 · 0.8 · 1 1.4 x
.57143
x
.57143
1,728.83
1,948.43
The value of 1,948.43 pounds exceeds the 1,800-pound limit required by OSHA. Thus, the welder should be minimally equipped with a shock-absorbing lanyard component to add to the wire rope lanyard.
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Example 2.
Substituting the appropriate values for this condition yields… a b s c
= 1.0 (no fall arrester) = 0.8 = 0.7 (assume synthetic tear fabric shock absorber) = 1.4
a · b · s = 1 · 0.8 · 0.7 = 0.4 c 1.4 Fp = (0.031 · 220 · 32.2) + 1.012 ! 0.003125 · 220 · 1 · 13000000) · 0.400 = 1,429.78 lbf. The addition of the shock absorber brings the maximum arrest force experienced by the welder to within acceptable limits. Example 3.
A worker weights 185 lbs and is using shock absorber with tear stitches.
Answer: _____________________
Example 4.
A worker weights 200 lbs and is using a wire rope fall arrester, and a wire rope shock absorber.
Answer: _____________________
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