Engine Sling Angle on a Hoist
An engine change is in progress. The engine mass is 240 kg and it has two lifting eyes 3.4 m apart. The available two-leg sling has 2.0 m legs and each leg is marked WLL 2.0 kN (working load limit, in-line). With the hook directly above the mid-point the legs take up the geometry the span forces on them. The hoist is to lift off with a controlled acceleration of 0.5 m/s². A spreader beam exists but is in another bay, and the shift ends in two hours. Take g = 9.81 m/s². The sling certificate is current and the legs are undamaged; the only open question is whether the lift can proceed as rigged.
1. Calculate the tension in each leg as rigged, and state the included angle at the hook.
Read the worked answer
Weight: W = 240 × 9.81 = 2354 N.
Geometry per leg: half-span = 3.4/2 = 1.7 m, leg length = 2.0 m, so vertical projection = √(2.0² − 1.7²) = √1.11 = 1.054 m.
sin θ = 1.054/2.0 = 0.527 → θ = 31.8° above the horizontal, i.e. 58.2° from the vertical.
Included angle at the hook = 2 × 58.2° = 116°.
Vertical equilibrium: 2T sin θ = W → T = 2354/(2 × 0.527) = 2234 N per leg.
2. Is the rigging acceptable against the 2.0 kN WLL? What are the options and what is the maximum engine mass for this sling as rigged?
Read the worked answer
2234 N > 2000 N, so each leg is about 12 % over its WLL - the lift must not proceed as rigged.
Two further points support the decision: the WLL is an in-line rating, and any choke or edge contact reduces it; and 116° is close to the practical 120° ceiling for multi-leg slings, above which leg tension rises very steeply.
Options:
- Fit the spreader beam so the legs hang vertically: T = W/2 = 1177 N - comfortable margin.
- Use longer legs to reduce the included angle.
- Use a four-leg sling, remembering that load is never shared equally between four legs - such slings are normally rated on two or three legs only.
Maximum mass on this sling as rigged: m = 2 × 2000 × 0.527/9.81 = 215 kg, versus the 240 kg engine.
3. What does the 0.5 m/s² lift acceleration do to the leg tension, and why does that matter for the decision?
Read the worked answer
Apparent weight: W' = m(g + a) = 240 × (9.81 + 0.5) = 2474 N, a 5 % increase.
T = 2474/(2 × 0.527) = 2348 N per leg - about 17 % over the WLL.
The 5 % is a controlled figure. A hoist that snatches, or a sling that comes taut with slack in it, produces dynamic factors of 2 or more rather than 1.05. That is why the static margin has to be positive before the lift starts: there is no reserve for a snatch load, however careful the operator.