How do you calculate sling tension?

In two steps, per leg. First find that leg's vertical share, which is a statics problem about the load: sum of vertical forces equals the weight and moments about the centre of gravity balance. Then divide that share by the sine of that leg's own angle from horizontal, because a leg can only pull along its own length. The commonest error is doing the second step once for the whole arrangement instead of once per leg.

How do you calculate sling tension for an off-centre load?

Take moments to split the vertical load between the lift points, then apply each leg's own angle factor to its own share. In this article's worked case a 10 t load picked 1.2 m from one point on a 4.0 m span puts 68.7 kN of vertical load on the near point and 29.4 kN on the far one, and because the hook sits over the centre of gravity the legs also have different angles, 68.2 and 47.0 degrees. Those shares become leg tensions of 74.0 and 40.3 kN, so the near leg carries 1.8 times the far leg's tension - a smaller ratio than the shares, because the steeper leg has the better angle factor.

How is the load shared between three sling legs?

By area. Three supports on a rigid body is statically determinate, and each leg's share of the weight is proportional to the area of the sub-triangle formed by the other two lift points and the centre of gravity. Sum the three shares as a check: they must add to the whole load. Then apply each leg's own angle factor to its own share.

Why is a four-leg sling different from a three-leg one?

Because four supports on a rigid body is statically indeterminate. Equilibrium gives three equations and there are four unknown leg forces, so the split depends on leg stiffness, on manufacturing tolerance in leg lengths and on how level the hook is, none of which is a design input. The conventional treatment is to design as though two diagonally opposite legs carry everything, or to resolve the four points to three attachments at the hook and recover a determinate answer.

Should I use the sling angle from the drawing?

Only as a starting point. The angle that matters is measured where the force actually acts, and pin eccentricities move that point: a spreader beam's sling pin sits above the beam axis, a padeye's pin above the load surface, and a shackle adds its own length. In the worked spreader, an arrangement laid out as a nominal 60 degree triangle solves at 59.3 degrees. The error is small, systematic, and always on the unconservative side, so check at the flattest angle the geometry can actually reach.

Sling tension calculation for 2, 3 and 4 legs

The short answer

Every sling tension is two multiplications: the leg's vertical share, then one over the sine of that leg's own angle. Doing them in the wrong order, or once for the whole arrangement, is the usual error.
Two legs, symmetric, is the only case where the share is obvious. Everything else needs moments about the centre of gravity first.
On an asymmetric two-leg pick both the share and the angle change. In the worked case the near leg takes 2.33 times the far leg's vertical share but only 1.84 times its tension, because the steeper leg gets the better angle factor back.
Three legs is determinate and worth arranging deliberately. Four is not, and gets designed as two.
Solve the geometry rather than reading it off the drawing. The worked spreader is set out as a nominal 60 degree triangle and the solver reports 59.3 degrees, because the sling pin sits above the beam axis.

The method, in two steps

Share, then angle. Both steps are ordinary statics, and the whole subject is knowing which one you are doing.

Step one: find each leg's vertical share. This is a statics problem about the load, not about the slings. Sum of vertical forces equals the weight; sum of moments about the centre of gravity is zero. The answer depends on where the lift points are and where the centre of gravity is, and on nothing else.

Step two: divide each share by the sine of that leg's own angle from horizontal. This is a statics problem about the leg. A leg can only pull along its own length, so to deliver a vertical share $V$ at an angle $θ$ it must carry $V / sin θ$ .

Written together for one leg:

Leg tension

T = \frac{V}{sin θ}

$V$: is this leg's own vertical share
$θ$: is this leg's own angle from the horizontal

Two things about that expression cause most of the errors in this subject.

The angle is per leg. Legs of different lengths, or legs to lift points at different distances from the hook, have different angles. Each leg divides its own share by its own sine.

The share is per leg too. The angle factor is not applied to the total load and then divided up. Divide first, then apply each leg's factor.

10 t on two legs at 60 degrees: each leg carries 49.0 kN of vertical share, 56.6 kN of tension, and pulls inwards with 28.3 kN.

Two legs, symmetric

The only genuinely easy case, and worth doing on paper once so the harder ones have something to be compared with.

A 10 t load, two lift points, centre of gravity midway between them. Each leg carries half the weight vertically, whatever the angle.

Angle from horizontal	Vertical share	Leg tension	Inward pull at the lift point
75°	49.0 kN	50.8 kN	13.1 kN
60°	49.0 kN	56.6 kN	28.3 kN
45°	49.0 kN	69.4 kN	49.1 kN
30°	49.0 kN	98.1 kN	85.0 kN

Read the last column as carefully as the third. At 45 degrees the inward pull equals the vertical share exactly, and at 30 degrees it exceeds it. That force squeezes whatever is between the two lift points, and on a long or slender load it is a real compression check that a tension-only calculation never performs.

Two legs, asymmetric

Both the share and the angle move, and they move the same way: the leg nearer the centre of gravity carries more and is steeper, so it gets hit twice.

The same 10 t, picked at two points 4.0 m apart, with the centre of gravity 1.2 m from the left point instead of at midspan.

Step one, the shares. Take moments about the right lift point. The left leg carries the weight multiplied by the distance from the centre of gravity to the right point, divided by the span:

V_{left} V_{right} = \frac{98.1 \times 2.8}{4.0} = 68.7 kN = 98.1 - 68.7 = 29.4 kN

The left leg carries 2.33 times what the right leg does, before any angle is considered.

Step two, the angles. The hook hangs over the centre of gravity, so with both legs reaching a common hook height of 3.0 m the two legs have different angles. The left leg spans 1.2 m horizontally and the right leg 2.8 m:

θ_{left} θ_{right} = arctan \frac{3.0}{1.2} = 68. 2^{\circ} = arctan \frac{3.0}{2.8} = 47. 0^{\circ}

Step three, the tensions.

T_{left} T_{right} = \frac{68.7}{sin 68. 2 ^{\circ}} = 74.0 kN = \frac{29.4}{sin 47. 0 ^{\circ}} = 40.3 kN

The left leg carries 1.84 times the right leg's tension. Notice that the ratio of the tensions is smaller than the ratio of the shares, because the steeper leg has the better angle factor. That is the one piece of good news in an asymmetric pick, and it is not enough to matter: size both lift points for the worse leg unless you are certain which way round the load will be rigged.

Three legs

Determinate, exactly solvable, and worth arranging on purpose. Three points on a rigid body is the largest number that gives a unique answer.

Three supports define a plane, so a rigid load on three lift points has one distribution and statics finds it.

The clean way to compute it is by areas. Put the three lift points at the corners of a triangle and the centre of gravity somewhere inside it. Each leg's share of the weight is proportional to the area of the sub-triangle opposite it - the one formed by the other two points and the centre of gravity.

The worked case: a 6 t load, lift points at (0.0, 0.0), (3.0, 0.0) and (1.2, 2.4) metres, centre of gravity at (1.5, 0.7).

Lift point	Vertical share	Share of total	Leg angle	Leg tension
(0.0, 0.0)	19.1 kN	32.5%	58°	22.6 kN
(3.0, 0.0)	22.6 kN	38.3%	62°	25.6 kN
(1.2, 2.4)	17.2 kN	29.2%	55°	21.0 kN

The shares sum to 58.9 kN, which is the whole 6 t, as they must.

Four legs is a different problem and it is not a harder version of this one. Four supports on a rigid body is statically indeterminate: three equilibrium equations, four unknowns. It has no unique solution and the conventional treatment is to design as though two diagonally opposite legs carry everything. If you want an answer rather than an assumption, resolve four points to three attachments at the hook, or accept the two-leg design case.

Solving the geometry rather than reading it

A hand calculation uses the angle on the drawing. A solver uses the angle the geometry produces, and on any real arrangement those differ.

Every calculation above starts from an angle. Where does that angle come from?

On a hand calculation it comes from the drawing: a nominal triangle, a nominal hook height, a nominal half-span. On a real arrangement the force does not act at the nominal point. A spreader beam's sling pin sits above the beam axis. A padeye's pin sits above the load surface. A shackle adds its own length. Each of those moves the point the leg actually pulls from, and therefore the angle.

Spreader Beam Design Calculator · computed at page render

A 10 t lift on a 6 m spreader, solved

The top slings are set out as a nominal 60 degree triangle over the half-span. The solver reports the angle the geometry actually produces, measured where the force acts.

Hook loadpayload plus rigging and beam self-weight	103.4kN
Tension in each top sling	60.1kN
Angle from horizontal, solvedagainst a nominal 60 degrees on the layout	59.3deg
Vertical component per leg	51.7kN
Inward horizontal component per leg	30.7kN

Governing check: Sling angle within declared limits75.9% utilisationPass

Less than a degree, on this arrangement. It matters because the error is systematic and always in the same direction: the solved angle is flatter than the nominal one, so a hand calculation on nominal dimensions is always slightly unconservative, and the gap grows as the arrangement flattens.

Open this example in the calculator

The practical rule: check the arrangement at the flattest angle the geometry can actually reach, including the pin eccentricities and any realistic sling length variation, not at the angle written on the layout.

Four checks on any tension calculation

Run these before the result leaves your desk. Three of them are arithmetic and take a minute.

Do the vertical components sum to the load? Add every leg's tension multiplied by the sine of its own angle. The total should be the hook load. This catches a wrong share, a wrong angle and a factor applied to the wrong quantity, all at once.

Do the horizontal components cancel? On a symmetric arrangement they must. On an asymmetric one they must still balance, which usually means the load hangs at a tilt that somebody should have thought about.

Is the hook over the centre of gravity? It always is, once the load is free. If your arrangement puts the hook somewhere else, the load will rotate until it is not, and every angle in the calculation changes.

Is the worst leg the one you sized for? On any asymmetric arrangement, identify the worst leg explicitly and say so in the output. "Leg tension 74.0 kN" is not a result. "Left leg 74.0 kN, governing" is.

Seven ways a tension calculation goes wrong

Six are about which quantity the factor was applied to. The seventh is about which angle was used.

1. The angle factor applied to the total load. It applies per leg, to that leg's share.

2. An equal share assumed on an asymmetric load. The share follows from moments about the centre of gravity, and on the worked case one leg carried 2.3 times the other.

3. One angle used for legs that have different angles. On an asymmetric pick with equal-length legs the angles differ substantially: 68 degrees and 47 in the worked case.

4. A four-leg arrangement divided by four. It is indeterminate. Two diagonally opposite legs is the design case.

5. The nominal angle used instead of the solved one. Pin eccentricity and sling length tolerance both flatten the real angle, and both push the answer the same way.

6. The horizontal component ignored. At 45 degrees it equals the vertical share, and it is a real compression in the load.

7. The centre of gravity taken as a point. It has a tolerance. The governing case is at the edge of that tolerance, not in the middle of it.

Common questions

How do you calculate sling tension?: In two steps, per leg. First find that leg's vertical share, which is a statics problem about the load: sum of vertical forces equals the weight and moments about the centre of gravity balance. Then divide that share by the sine of that leg's own angle from horizontal, because a leg can only pull along its own length. The commonest error is doing the second step once for the whole arrangement instead of once per leg.
How do you calculate sling tension for an off-centre load?: Take moments to split the vertical load between the lift points, then apply each leg's own angle factor to its own share. In this article's worked case a 10 t load picked 1.2 m from one point on a 4.0 m span puts 68.7 kN of vertical load on the near point and 29.4 kN on the far one, and because the hook sits over the centre of gravity the legs also have different angles, 68.2 and 47.0 degrees. Those shares become leg tensions of 74.0 and 40.3 kN, so the near leg carries 1.8 times the far leg's tension - a smaller ratio than the shares, because the steeper leg has the better angle factor.
How is the load shared between three sling legs?: By area. Three supports on a rigid body is statically determinate, and each leg's share of the weight is proportional to the area of the sub-triangle formed by the other two lift points and the centre of gravity. Sum the three shares as a check: they must add to the whole load. Then apply each leg's own angle factor to its own share.
Why is a four-leg sling different from a three-leg one?: Because four supports on a rigid body is statically indeterminate. Equilibrium gives three equations and there are four unknown leg forces, so the split depends on leg stiffness, on manufacturing tolerance in leg lengths and on how level the hook is, none of which is a design input. The conventional treatment is to design as though two diagonally opposite legs carry everything, or to resolve the four points to three attachments at the hook and recover a determinate answer.
Should I use the sling angle from the drawing?: Only as a starting point. The angle that matters is measured where the force actually acts, and pin eccentricities move that point: a spreader beam's sling pin sits above the beam axis, a padeye's pin above the load surface, and a shackle adds its own length. In the worked spreader, an arrangement laid out as a nominal 60 degree triangle solves at 59.3 degrees. The error is small, systematic, and always on the unconservative side, so check at the flattest angle the geometry can actually reach.

Sources

Every document below is linked at its publisher or regulator. Xarpis reproduces no standard text; where a clause is named, the identifier is given so you can find it in your own copy.

ASME B30.9Slings
ASME · paid document
The US volume covering alloy steel chain, wire rope, metal mesh, synthetic rope, synthetic webbing and synthetic round slings: rated loads, marking, inspection, and the removal criteria that decide when a sling leaves service. Where published sling rated loads and angle reductions come from.
29 CFR 1926.251Rigging equipment for material handling
US Occupational Safety and Health Administration · free to read
Inspection and safe-use requirements for chain, wire rope, fibre rope, synthetic webbing, shackles and hooks on US construction sites, including the requirement that rigging be inspected before each shift.
LOLER 1998Lifting Operations and Lifting Equipment Regulations
UK Health and Safety Executive · free to read
The UK duty framework for lifting operations: planning by a competent person, supervision, and thorough examination of lifting equipment and accessories. Like OSHA's rules it governs the process, not the arithmetic.
ASME BTH-1Design of Below-the-Hook Lifting Devices
ASME · paid document
Structural, mechanical and electrical design criteria for below-the-hook lifting devices, used alongside ASME B30.20 which carries the safety requirements. The current edition is BTH-1-2023; Xarpis implements the 2020 edition and says so on every result.

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