How to Calculate Pile Capacity in Cohesive Soils? Explained in 1-minute

Mon 28th Jul 2025 by ilyas

In this post we will go through how to quickly estimate the pile capacity for a pile in cohesive soils such as clay. In clay soils, the pile resistance is provided by two main components: end bearing; and skin friction. The calculation method for these, and a method to get the allowable capacity of a pile is summarised below.

For information on pile capacities in sand, refer to the link below.

※ How to Calculate Pile Capacity in Granular Soils? Explained in 1-minute

End Bearing Resistance

The end bearing resistance of a pile in clay is based on the base area of the pile and the strength of the soil strata it bears onto. It can be calculated as follows:

Q subscript b equals N subscript c cross times A subscript b cross times C subscript b a s e end subscript

(kN)

Where Nc is typically taken as 9, Ab is the base area of the pile ( $bevelled fraction numerator pi D squared over denominator 4 end fraction$ ), and $C subscript b a s e end subscript$ is the undrained cohesive strength of the clay at the bottom of the pile.

Shaft Resistance

The shaft resistance comes from skin friction between the pile and the surrounding soil along the contact length of the pile. It can be calculated as follows:

Q subscript s equals alpha cross times c with bar on top cross times A subscript s

(kN)

Where α is taken as between 0.4 to 0.5, $c with bar on top$ is the mean cohesive strength of the soil along the pile length, and As is the shaft area ( $pi D L$ ). D is the pile diameter and L is the length.

Ultimate Capacity

The ultimate capacity of the piles can be expressed simply as the sum of the shaft and end bearing resistances:

Qu = Qb + Qs

Pile Allowable Capacity

The allowable capacity of the piles as required to limit deflections can be calculated as the minimum of two different approaches:

Q subscript a equals m i n open curly brackets bevelled fraction numerator Q subscript u over denominator 2 end fraction space comma space fraction numerator Q subscript b over denominator 3 end fraction plus fraction numerator Q subscript s over denominator 1.5 end fraction close curly brackets

Calculation Example

A pile cap with two piles supports a column with a working load (1.0*dead + 1.0*live) of 3000 kN. Each pile is 0.75 m in diameter and 20 m long bearing in clay. The average undrained cohesive strength of the clay is 100 kPa. At the bottom of the pile, the undrained cohesive strength is 250 kPa. Calculate the capacity of a single pile.

Pile base area, Ab = π(0.75)^2 / 4 = 0.44 m2
End bearing resistance, Qb = 9(250)(0.44) = 990 kN

Pile shaft area, As = π(0.75)(20) = 47.1 m2
Shaft resistance, Qs = 0.5(100)(47.1) = 2356 kN

Qu = Qb + Qs = 3,346 kN

Qa = min { Qu/2 , Qb/3 + Qs/1.5 }
= min { 3346/2 , 990/3 + 2356/1.5 }
= min { 1,673 kN , 1,900 kN }
= 1,673 kN

Since the applied load per pile is 3000 kN / 2 piles = 1,500 kN, the pile capacities are OK.

Last Update 28/07/25 07:47 JST

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