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Negative skin friction on single pile and group

Short answer

Single pile: Q_NSF = perimeter × length × α c_u. For a group: equal to the smaller of total individual drag or the weight of the settling block plus its perimeter shear: Q_ng = min( n Q_nsf, γ_f H_f B_g L_g + c_u P_g L_n).

How is negative skin friction calculated for a single pile and a group of piles in clay?

Single pile

The negative skin friction acts over the length LnL_n of the pile that lies in the settling clay (down to the neutral point).

Unit negative skin friction: fn=αcuf_n = \alpha c_u (total stress method, α≈1\alpha \approx 1 for soft clay), or fn=βσv′f_n = \beta\sigma'_v (effective stress method, Bjerrum and Burland, β=0.2\beta = 0.2 to 0.250.25 for clay).

Qnsf=P×Ln×fn=πD Ln cu(α=1)Q_{nsf} = P \times L_n \times f_n = \pi D\,L_n\,c_u \quad (\alpha = 1)

where PP is the perimeter of the pile and DD the diameter.

Group of piles

A group in clay moves with the soil mass between the piles, so the drag is the lesser of:

  1. The sum of the drag on all piles, Qng=n×QnsfQ_{ng} = n \times Q_{nsf}.
  2. The weight of the fill carried by the group, together with the shear along the perimeter of the block of settling clay:
Qng=γfHfBgLg+cu Pg LnQ_{ng} = \gamma_f H_f B_g L_g + c_u\,P_g\,L_n

where Bg×LgB_g \times L_g is the plan of the group, Pg=2(Bg+Lg)P_g = 2(B_g + L_g) its perimeter, HfH_f and γf\gamma_f the thickness and unit weight of the fill (surcharge) that causes the settlement, cuc_u the undrained cohesion of the clay and LnL_n the thickness of the settling clay along the piles.

The calculated drag is added to the structural load when checking the pile capacity, since it acts in the same direction.

✓ Answer checked by Claude Sonnet 5.5

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