huangyhg

opinions on accuracy
hi all,
we are using a nylon rigging sling as an arrestor in the event that the suspension of a piece of equipment partially failed. in the event of suspention failure, one end of the equipment would fall 8" before being stopped by the sling. my question is this: how many of you structural guys would be happy using the formula f = w * h / s for results in finding the impact loads on the sling.
where
f = force on impact with the sling
w = portion of equipment weight
h = drop height
s = elongation of sling on impact
there seems to be a great deal of disagreement in our office as to this method being thorough enough.
thanks in advance for any opinions.
assuming that the nylon acts like a true spring (probably reasonable) and using newton's laws to define a more detailed model, i get this comparison:
for s < h: above formula is conservative (overestimates the value of f).
for s = h: above formula gives accurate value of f.
for s > h: above formula underestimates the value of f.
for a simple formula, it is not too bad when s is about equal to h.
hi haggis
when i look at your formula and study the units, i end up
with an impact force equal to the weight of the fallen equipment so i think this is incorrect.
my approach would be :-
w(h+x)= (f^2/(2*e))*al and x=(f*l)/e
w(h+((f*l)/e))= (f^2/(2*e))*al
this will give a quadratic equation in f.
rearrange the formula to make f the subject and solve the quadratic by using the standard formula for solving quadratic's given in most mathmatic books.
the above formula i have given equates the potential energy of the load to the strain energy in the sling after impact.
where w= mass of falling equipment
h= height equipment falls before striking sling
x= maximum extension of sling which can be replaced by
(f*l/e)
where f= max stress in the sling produced by
falling mass
l= length of sling
e= modulus of elasticity of sling material.
regards desertfox
the potential energy of the load = mgh
m= mass of load, h= distance dropped.
the energy stored in the sling = kx/2
k= spring rate of sling, x= deflection of sling
so mgh=kx/2
deflection x = 2mgh/k
you can probably determine k accurately enough by measuring the stretch of the sling caused by a known load. this value can then be used to determine the load in the sling at maximum deflection.
jeff
hi notnats
could i just point out that the energy stored in the sling if acting like a spring should read
stored energy = (k*x^2)/2
and not (k*x)/2
regards desertfox
thanks desertfox, i just logged back in to correct that.
that makes:
deflection x = (2mgh/k)^0.5
jeff
i have a couple of comments:
first, is the sling already engaged with the equipment and that the 8 inch drop is the elongation of the sling (hence the statement that the equipment will fall 8 inches before being stopped)or is the 8 inches the separation between the equipment and sling?

second, i would hesitate to determine the spring constant by applying a static load to the sling. if this is a nylon web material, the fibers of the webbing will slide and stretch before the entire sling comes into tension versus a spring of constant cross-section. the elastic properties of the sling will be non-linear. if you know the manufacturer of the sling, get its load rating.
third, roark mentions that the stress in an elastic material due to sudden loading approaches twice that of static loading due to wave propagation through the material (shock loading).
in light of the above comments, i would recommend a factor of safety of at least 3.
thanks to all for the responses.
here's what i have been using in place of the simple formula:
drop distance = (a * t^2) / 2
ke = 0.5 * m * v^2
v = a * t
m = w / a
where:
a = acceleration due to gravity = 32.17 ft/sec^2
v = velocity in ft/sec
m = mass
w = weight in lbs.
t = time for drop in seconds.
ke = kinetic energy from drop in ft.lbs.
d = distance to arrest fall in ft. (stretch in sling)
drop distance known as 8鈥濃�︹�?solve for time (t)
0.6 = (a * t^2) / 2
(a * t^2) = 0.6 * 2 = 1.2
t^2 = 1.2 / a = 1.2 / 32.17 = 0.037
t = 0.193 secs. to impact.
v. = a * t = 32.17 * 0.193 = 6.21 ft/sec at impact.
ke = .5 * m * v^2 = .5 * 114 * 6.21^2 = 2198 ft.lbs.
force at impact = 2198 / d
to answer vmiratfff"> , (1) the sling is in a slack condition due to required movement of the equipment and if the suspension failed, the equipment would drop 8" before contacting the sling.
(2) the sling is a nylon web material and by applying a static load at 2000 lb. increments, it behaved as you described. the first 2 loads applied gave lots of elongation before it was in tension then each incremental load gave less elongation. yes, the results are non linear.
(3) the load ratings are on the sling as required by the ohsa i believe and is rated at 6000 lbs, 4880 lbs and 12000 lbs. when being used as a lifting sling in straight sling, choker and basket configurations. the manufacturer has no values for impact loading as, from his point of view, and i tend to agree with him, the sling is manufactured for lifting purposes only.
which may beg the question, why are we using it as an arrestor. however, if the equipment suspension failed and the sling was subjected to the drop load, it would be a one off use and be replaced.
thanks again for the responses.
i should have mentioned to vmiratfff"> that the impact loads i am getting are more than twice the static load figures in using the above formulae. using a static load factor of 2:1, the safety factor of the sling is reduced to 1.4:1. from it's rated 5:1 as a lifting device. we pulled a sling to failure at 29500 lbs.
thanks again.
let h be the distance the mass free-falls before the sling begins to go into tension. let s be the amount the sling then stretches to totally arrest the downwards motion of the mass.
potential energy before release equals strain energy at time of arrest:
mg(h+s) = (1/2)(ks^2)
solve this quadratic for s:
s = mg[1+sqrt(1+2kh/(mg))]/k
peak force in sling = ks = mg[1+sqrt(1+2kh/(mg))]
this result explicitly includes the "sudden loading" effect refered to by roark.
hth

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