Container Lashing Force Calculator

Calculate lashing forces on deck containers using ship motion and stow geometry

Compute transverse, longitudinal, and vertical forces on a deck-stowed container from CSS Code acceleration factors, subtract friction, and find the required lashing tension against the rod MSL. Cargo officers and stowage planners use this to verify lashing adequacy. It runs free in your browser on Gera Tools, with nothing uploaded.

Last updated Source: Gera Tools

What is the CSS Code Annex 13 method?

Annex 13 of the IMO Code of Safe Practice for Cargo Stowage and Securing is the standard simplified method for checking that securing arrangements can resist the forces from ship motion. It uses acceleration factors as multiples of g that already include roll, pitch, heave, and gravity components.

When a ship rolls and pitches, every deck container is thrown sideways, forward, and up by forces that can run to hundreds of kilonewtons. This calculator follows the simplified CSS Code Annex 13 method to work out those forces, credit the friction that resists them, and size the lashing tension that the rods must provide.

How it works

The external forces come from the unit mass and the acceleration factors, then friction is subtracted to find what the lashings must restrain:

W   = m·g                      (weight)
F_y = m·(a_y/g)·g              (transverse, usually governing)
F_x = m·(a_x/g)·g              (longitudinal)
F_z = m·(a_z/g)·g              (vertical)
friction = μ·(W − F_z)
net force = max(0, F_y − friction)
required tension per lashing = net force / (n · cos α)
allowable = CS = MSL / 1.5

If the required tension stays below the calculation strength CS, the arrangement is adequate; if not, you need more lashings, a higher-MSL rod, or better friction.

Worked example

A 20-tonne container at a transverse acceleration factor of 0.7 g sees:

  • Weight W = 20 × 9.81 = 196.2 kN
  • Transverse force F_y = 20 × 0.7 × 9.81 = 137.3 kN
  • Assume no upward vertical force; friction coefficient μ = 0.3
  • Friction restraint = 0.3 × 196.2 = 58.9 kN
  • Net force requiring lashings = 137.3 − 58.9 = 78.4 kN

With two lashing rods at 45° to horizontal, each rod must provide: 78.4 / (2 × cos 45°) ≈ 55.4 kN

Against a 100 kN MSL rod (CS = 100/1.5 = 66.7 kN), the arrangement passes. Reduce the number of lashings to one rod or lower the friction coefficient (a wet or icy deck can halve it) and the same stow fails.

Why the transverse case almost always governs

Ship rolling generates the largest sideways accelerations in normal sea states, and the gravity component adds to them when the vessel heels. Annex 13 tables typically assign the highest acceleration factor to the transverse direction, especially for containers stowed at mid-ship. At ship’s ends, pitch-driven longitudinal forces can become comparable, which is why positions near the bow or stern sometimes produce an interesting interaction between longitudinal and transverse cases.

Key variables that change the result

  • Stow position: Acceleration factors depend on the container’s position along and across the ship. Containers stowed high on the stack or far outboard see higher transverse accelerations.
  • Lashing angle: A rod angled too close to vertical contributes little horizontal restraint. The effective horizontal component is force / cos α, so a shallow rod (large α) is far more efficient than a steep one.
  • Friction coefficient: Values range from about 0.2 to 0.4 depending on the surface condition and any anti-skid material. Never assume maximum friction without confirming actual conditions.
  • Stack weight limits: Corner castings and twist locks have their own structural limits; a calculation that passes on lashing force can still fail the stow if cumulative stack weight exceeds the deck or lower container’s rating.

Always verify against the vessel’s Cargo Securing Manual before securing real cargo. The Annex 13 simplified method is a good first-pass check; the CSM’s ship-specific acceleration tables are the authoritative input.