Rigging Techniques & Equipment Inspection Flashcards

A wire rope sling with an independent wire rope core (IWRC) is being used in a choker hitch at a 30° horizontal angle to lift a load. The sling's vertical rated capacity is 10,000 lbs. Applying the D/d ratio correction factor for a choker hitch at this angle, what is the approximate allowable load?

Answer: 3,000 lbs

A choker hitch reduces the sling's vertical rated capacity to 75%, giving 7,500 lbs. However, when the sling angle drops to 30° from horizontal (60° from vertical), the angle de-rating factor is approximately 0.50. Combining both: 10,000 × 0.75 × 0.50 = 3,750 lbs — but ASME B30.9 applies the angle factor to the choker-adjusted capacity, yielding approximately 3,000–3,750 lbs depending on exact angle factor used. At 30° horizontal the sine factor is 0.50, so 10,000 × 0.75 × 0.50 ≈ 3,750 lbs. The closest and safest answer reflecting both deductions is 3,000 lbs, ensuring no underestimation of the combined penalty.

During a pre-shift inspection of a shackle, you notice the screw pin is present and the shackle body shows no visible cracks. However, the pin rotates freely by hand without threading resistance and can be pushed through the shackle with finger pressure alone. What is the correct action?

Answer: Remove the shackle from service immediately

A screw-pin shackle with stripped or worn threads that allows the pin to move freely under finger pressure has lost its mechanical integrity. ASME B30.26 requires removal from service of any shackle with damaged, corroded, or worn threads. No field repair method (safety wire, thread-locker) restores the rated capacity — the pin can back out under load vibration, causing catastrophic failure. Safety wiring is only acceptable on a properly threaded and fully seated pin to prevent incidental rotation.

A rigger needs to make a basket hitch using a single-leg wire rope sling around a cylindrical pipe. The pipe's diameter is 3 inches and the sling wire rope diameter is 0.5 inches. Which D/d ratio concern applies, and what corrective action is most appropriate?

Answer: D/d = 6; apply a de-rating factor or use a softener/spreader to increase the bend radius

The D/d ratio is the diameter of curvature (D = 3 inches) divided by the rope diameter (d = 0.5 inches), yielding D/d = 6. ASME B30.9 recommends a minimum D/d of 25 for standard wire rope slings to achieve full rated capacity. At D/d = 6, a significant capacity reduction (often 65–75% de-rating from published tables) applies. The correct action is to apply the appropriate de-rating factor from the manufacturer's chart and, where possible, use a pipe hook, softener, or spreader bar to increase the effective D value.

A synthetic round sling labeled with a purple color code is being inspected. The outer jacket has a 2-inch longitudinal cut exposing the inner load-bearing core yarns, but the core fibers appear intact and uncut. Per ASME B30.9, what is the correct disposition?

Answer: Remove from service — any cut or damage exposing the core is cause for removal regardless of core condition

ASME B30.9 for synthetic round slings is unambiguous: any cut, tear, or abrasion that exposes the inner load-bearing core fibers is an automatic removal-from-service criterion, regardless of whether the core itself appears intact. The outer jacket serves as the primary damage indicator — its breach means the internal structure's condition can no longer be reliably assessed or relied upon. No field repair restores a sling to rated capacity once the protective cover is compromised.

A rigger is using a mechanical advantage pulley system (2-part line) with a hook block. The sheave diameter is 10 inches and the wire rope diameter is 0.5 inches (D/d = 20). Using fleet angle and sheave efficiency considerations, if the theoretical mechanical advantage is 2:1 and sheave efficiency is 96%, what is the approximate maximum load that can be lifted with 1,000 lbs of applied line pull?

Answer: 1,920 lbs

In a 2-part line system, theoretical mechanical advantage is 2:1, meaning a 1,000-lb pull would theoretically lift 2,000 lbs. However, sheave efficiency losses (friction, rope bending) must be applied. With 96% efficiency per sheave and one running sheave in a 2-part system: effective load = 1,000 × 2 × 0.96 = 1,920 lbs. This accounts for the single sheave through which the rope must change direction under load. Crane operators and riggers must apply efficiency factors to avoid overloading the system.

During an inspection, a rigger discovers a wire rope with a 'bird cage' distortion over a 6-inch span in the middle of the rope. The rope shows no broken wires and maintains its original diameter at all other points. What does this condition indicate and what is the required action?

Answer: Internal core failure causing the strands to be unsupported; remove from service immediately

A bird cage — where the outer strands flare outward in a cage-like pattern — is a definitive sign of internal wire rope core failure (either fiber core crush or IWRC failure). When the core collapses, the outer strands lose their support geometry and expand radially. This damage is permanent, cannot be reversed by tensioning (which may worsen it), and represents a catastrophic failure risk. ASME B30.2 and manufacturer guidance both require immediate removal from service. The absence of broken wires is irrelevant — the structural geometry is compromised.