Bearing Capacity of a Strip Footing Resting on Reinforced Fly Ash Slope: An Analytical Approach

Abstract

This paper presents an analytical approach to determine the ultimate load-bearing capacity of a footing subjected to a vertical load and resting on the top of a fly ash slope without or with single or multi-layers geogrid reinforcement. The load-bearing capacity values obtained by using the proposed analytical approach were compared with those obtained from experimental and numerical approaches reported recently in the literature. After validation of the analytical values of the load bearing capacity, an empirical formula based on a regression analysis was developed between the bearing capacity ratio and geogrid index (α), which physically signifies the mobilized strength of geogrid reinforcement under a given footing load and is a function of the number of reinforcement layers (N) and the embedment ratio (z/B).

Appendices

Appendix 1

Calculation of theoretical bearing capacity of fly ash slopes.

• Step 1: Calculation of F _cc and F _γc [9]

  In the present investigation, E = 8,000 kPa, c = 20 kPa, ϕ = 14°, γ = 13.82 kN/m³, ϕ _μ  = 35°, B = 0.10 m and L = 0.30 m, Substituting the value of ϕ in Eq. (9), we get μ = 0.43. Substituting the value of E and μ in Eq. (8), we get G = 2797.21 kPa.

  On substituting the above values in Eqs. (3) and (4), we get I _r  = 138.66 and I _r(cr)  = 28.18, respectively. Since in the present case, Ir ≥ I_{r(c r)}, F _cc  = F _qc  = F _γc  = 1

• Step 2: Bearing capacity of unreinforced fly ash slopes: For β = 45°, D _e /B = 2.0, ϕ = 14°, N _cq , = 5.0 and N _γq  = 1.15 [13]. On substituting these values in Eq. (2), we get, q _o  = 100.80 kPa

• Step 3: Bearing capacity of reinforced fly ash slopes (Single layer). For β = 45°, D _e /B = 2.0 and z/B = 1.0.

  Using Eqs. (11a) and (10), we get, m = 0.85 and f _e  = 0.595, respectively. Also for z/B = 1.0, From Fig. 6, A ₁ = 0.335, A ₂ = 0.175 and A ₃ = 0.140. From Fig. 3, X ₀ /B = 0.82 and X ₀  = 0.082. From Fig. 8, L ₀ /B = 2.60 and L ₀  = 0.26. Substituting the above values in Eq. (38), we get, q _R  = 154.60 kPa

• Step 4: Bearing capacity of reinforced fly ash slopes (Multiple layer reinforcement). Taking, N = 4, and using Eq. (11a) and from Fig. 6, we get for successive reinforcement layer.

  • For first layer, Z ₁ /B = 0.25, m ₁  = 0.9625, A ₁ = 0.35, A ₂ = 0.31 and ΔH ₁  = 0.25

  • For second layer, Z ₂ /B = 0.50, m ₂  = 0.925, A₁ = 0.345, A₂ = 0.25 and ΔH ₂  = 0.25

  • For third layer, Z ₃ /B = 1.0, m ₃  = 0.85, A ₁ = 0.335, A ₂ = 0.175 and ΔH ₃  = 0.50

  • For fourth layer, Z ₄ /B = 1.50, m ₄  = 0.575, A ₁ = 0.330, A ₂ = 0.130 and ΔH ₄  = 0.50

Substituting the above values in Eq. (43) we get, q _uR  = 258.05 kPa.

Appendix 2

BCR values based on empirical equation (Table 6).