Crane outrigger pad bearing pressure

Introduction

Most mobile crane overturns start at the ground: an outrigger punches through soft ground, a void or a buried service, and the crane tips. The check is simple in principle: the outrigger load divided by the area that carries it must stay below what the ground can bear. The difficulty is the area, because a float of a few tenths of a square metre puts megapascals on the ground and a mat only helps if it is stiff enough to spread the load. This calculator takes the outrigger load (from the crane maker or a planning estimate), the float and mat sizes, treats the mat either as rigid (with a bending check) or as spreading the load through its thickness, and compares the result with BS 8004 presumed values or a tested allowable pressure.

It is for lift planners, appointed persons, port and site engineers, and students of temporary works.

Formulas

F = manufacturer value,  or ≈ 0.65 × (crane + load + rigging) g,  or (crane + load + rigging) g
q_float = F / A_float
rigid mat:  q = (F + W_mat) / (L_mat B_mat),   bending σ = 3 q c² / t²,  c = largest overhang beyond the float
spread:     L_eff = min(L_mat, L_float + 2 t tan θ),  q = F / (L_eff B_eff) + W_mat / A_mat
required area = F / q_allow,   square side = √(F / q_allow)

BS 8004 presumed allowable values (lower bounds): rock 4 000, dense gravel 600, compact sand 300, very stiff clay 300, medium dense gravel 200, stiff clay 150, medium dense sand 100, firm clay 75 kPa; soft or loose ground must be tested.

Assumptions and limitations

• Uniform pressure under float and mat; the rigid-mat case is only as good as the mat's stiffness and strength.
• Presumed values guard against shear failure, not settlement; they need foundation widths of about 1 m and a low water table.
• No voids, services, slopes, excavations or dynamic effects; marginal cases need a temporary works design.

Worked example

A 60 t crane lifting 10 t: the manufacturer's outrigger load for the lift is 400 kN on a 0.6 × 0.6 m float; 2 × 2 m timber mat 150 mm thick weighing 0.5 t; stiff clay.

1. Under the float: 400/0.36 = 1 111 kPa, far beyond any soil.
2. Rigid mat: (400 + 4.9)/4 = 101 kPa, 67 % of the 150 kPa presumed for stiff clay: acceptable.
3. Bending: overhang 0.7 m, σ = 3 × 101 × 0.49/0.0225 = 6.6 MPa, within 10 MPa for hardwood: the rigid assumption holds.
4. If the mat only spread the load at 45° through its 150 mm thickness, the effective area would be 0.9 × 0.9 m and the pressure 495 kPa: a flexible mat this size would fail the check.
5. Fully effective, the ground needs at least 2.67 m², a 1.63 m square mat.

The planning estimate 0.65 × 70.5 t × g = 449 kN would have given 114 kPa; the upper bound of all weight on one outrigger 691 kN.

FAQ

Related calculators

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• Concrete volume and rebar weight — permanent crane bases.
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• Column buckling — outrigger beams and props.