Civil, Structural & Geotechnical Engineering
Broms Lateral Pile Capacity Formula: Short Rigid Free-Head Piles in Sand (Cohesionless)
Calculate ultimate lateral soil resistance H_ult (kN) and maximum pile bending moment M_max (kN·m) for short rigid free-head piles in cohesionless sand using Bengt B. Broms’s (1964, ASCE) 3 * K_p passive wedge solution.
Why 3D passive arching triples Rankine K_p in Broms’s lateral pile design!
When wind, wave, bollard, or sign-bridge lateral load H_ult pushes a short rigid free-head pile of diameter B (m) and embedment length L (m) at eccentricity e (m) above ground in cohesionless sand (effective unit weight gamma, friction angle phi, Rankine passive coefficient K_p = (1 + SIN(phi)) / (1 - SIN(phi))), 3D soil arching around the circular shaft mobilizes THREE TIMES classical 2D Rankine passive pressure (p_u = 3 * gamma * z * K_p)!
Assuming the concentrated toe kick-back reaction acts at tip z = L, Bengt B. Broms’s (1964, J. Soil Mech. Found. Div., ASCE) moment equilibrium about the pile tip yields the clean closed-form Ultimate Lateral Capacity: H_ult = (0.5 * gamma * B * (L^3) * K_p) / (e + L)! Zero shear (and peak pile shaft bending moment) occurs at depth f = 0.82 * SQRT(H_ult / (gamma * B * K_p)), producing Maximum Pile Bending Moment M_max = H_ult * (e + (2 / 3) * f) (requiring structural plastic moment capacity M_yield > M_max so soil failure governs as a short rigid pile)!
| Broms Cohesionless Pile Quantity (B = 0.60 m, L = 4.5 m, e = 1.5 m, gamma = 18 kN/m^3, phi = 34°) | Formula | Evaluated Value |
|---|---|---|
| Rankine Passive Coefficient K_p | (1 + SIN(34°)) / (1 - SIN(34°)) | 3.5371 |
| Ultimate Lateral Soil Load H_ult | 0.5 * gamma * B * L^3 * K_p / (e + L) | 290.09 kN |
| Depth to Zero Shear / Peak Moment f | 0.82 * SQRT(H_ult / (gamma * B * K_p)) | 2.26 m (< L = 4.5 m) |
| Maximum Shaft Bending Moment M_max | H_ult * (e + (2/3) * f) | 870.3 kN·m |
Broms (1964) Short Rigid Free-Head Pile Ultimate Lateral Load H_ult (kN) in Sand
Evaluates (0.5 * gamma * B * L^3 * K_p) / (e + L) across diameter B (B2 m), embedment L (B3 m), load eccentricity e >= 0 (B4 m), effective unit weight gamma (B5 kN/m^3), and sand friction angle phi (B6 deg).
=IF(OR(B2 <= 0, B3 <= 0, B4 < 0, B5 <= 0, B6 <= 15, B6 >= 50), "Invalid Pile Geometry, Soil Unit Weight, or Friction Angle phi (15–50 deg)", ROUND((0.5 * B5 * B2 * (B3 ^ 3) * ((1 + SIN(RADIANS(B6))) / (1 - SIN(RADIANS(B6))))) / (B4 + B3), 2))Maximum Pile Bending Moment M_max (kN·m) = H_ult * (e + (2/3) * f)
Computes peak bending moment inside the pile shaft at zero-shear depth f = SQRT(2 * H_ult / (3 * gamma * B * K_p)) (~= 0.8165 * SQRT(H_ult / (gamma * B * K_p))).
=IF(OR(B2 <= 0, B3 <= 0, B4 < 0, B5 <= 0, B6 <= 15, B6 >= 50), "Invalid Pile Geometry, Soil Unit Weight, or Friction Angle phi (15–50 deg)", ROUND(((0.5 * B5 * B2 * (B3 ^ 3) * ((1 + SIN(RADIANS(B6))) / (1 - SIN(RADIANS(B6))))) / (B4 + B3)) * (B4 + (2 / 3) * SQRT((2 * ((0.5 * B5 * B2 * (B3 ^ 3) * ((1 + SIN(RADIANS(B6))) / (1 - SIN(RADIANS(B6))))) / (B4 + B3))) / (3 * B5 * B2 * ((1 + SIN(RADIANS(B6))) / (1 - SIN(RADIANS(B6))))))), 1))Try the example
Use the original values or change them to test your understanding.