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Stability of gravity wall, two layers

Short answer

FS against overturning = 4.87, sliding = 1.79; eccentricity 0.53 m (< B/6), maximum base pressure ≈ 88 kN/m²; bearing capacity is adequate; the wall is stable.

A concrete gravity retaining wall of height 6 m retains two layers of horizontal backfill soil. The upper layer is 3 m deep cohesionless soil of unit weight 20 kN/m320\ \text{kN/m}^3 and angle of internal friction 32∘32^\circ and the bottom layer is c-ϕ\phi soil of unit weight 22 kN/m322\ \text{kN/m}^3, cohesion 30 kN/m230\ \text{kN/m}^2 and angle of internal friction 27∘27^\circ. There is water table at the depth of 2 m from the ground surface. The back of the retaining wall is smooth and vertical with the top width 1.2 m and bottom width 5 m. Check the stability of the retaining wall. Take unit weight of concrete as 25 kN/m325\ \text{kN/m}^3.

Assumptions: smooth vertical back; the sloping face is on the front (toe) side, so the wall is a rectangle 1.2×61.2\times6 m plus a triangle 3.8×63.8\times6 m; no embedment or passive resistance; water in the backfill acts at the wall back and uplift acts under the base. γw=9.81\gamma_w = 9.81 kN/m3^3, and the unit weights given are taken as saturated below the water table.

1. Active pressure (Rankine)

Ka1=tan⁡2(45∘−16∘)=0.3073,Ka2=tan⁡2(45∘−13.5∘)=0.3755,2cKa2=36.77K_{a1} = \tan^2(45^\circ-16^\circ) = 0.3073,\quad K_{a2} = \tan^2(45^\circ-13.5^\circ) = 0.3755,\quad 2c\sqrt{K_{a2}} = 36.77
  • Upper layer, 0-2 m (dry/moist): pp at 2 m =0.3073×20×2=12.29= 0.3073\times20\times2 = 12.29 kN/m2^2.
  • Upper layer, 2-3 m (submerged, γ′=10.19\gamma'=10.19): σv′=40+10.19=50.19\sigma_v'=40+10.19 = 50.19, p=15.42p = 15.42 kN/m2^2.
  • Lower layer (c-ϕ\phi): top p=0.3755×50.19−36.77=−17.92p = 0.3755\times50.19-36.77 = -17.92; base σv′=86.76\sigma_v' = 86.76, p=−4.19p = -4.19. Both are negative, so the c-ϕ\phi soil gives no effective active pressure (tension zone neglected).
  • Water pressure: 00 at 2 m depth, 9.81×4=39.249.81\times4 = 39.24 kN/m2^2 at the base.
PartForce (kN/m)Arm above base (m)Moment
Triangle 0-2 m12.294.66757.35
Rectangle 2-3 m12.293.50043.02
Triangle 2-3 m1.573.3335.22
Water 0.5 x 39.24 x 478.481.333104.64
Total ∑H\sum H104.63210.23

2. Wall weight and uplift

PartForce (kN/m)Arm from toe (m)
Rectangle 1.2 x 6 x 25180.04.400
Triangle 0.5 x 3.8 x 6 x 25285.02.533
Uplift (taken uniform, 39.24×539.24\times5)-196.22.500
∑V\sum V (net)268.8

Mr=180×4.4+285×2.533−196.2×2.5=1023.5M_r = 180\times4.4+285\times2.533 - 196.2\times2.5 = 1023.5 kN m/m.

3. Checks

  • Overturning: FSot=1023.5/210.2=4.87FS_{ot} = 1023.5/210.2 = 4.87 (>2). Safe. (Without uplift it would be 7.20.)
  • Sliding: base friction δ=23×27∘=18∘\delta = \tfrac23\times27^\circ = 18^\circ, adhesion ca=23×30=20c_a = \tfrac23\times30 = 20 kN/m2^2:
FSs=∑Vtan⁡18∘+caB∑H=268.8×0.3249+20×5104.6=1.79 (>1.5)FS_s = \frac{\sum V\tan18^\circ + c_aB}{\sum H} = \frac{268.8\times0.3249 + 20\times5}{104.6} = 1.79\ (>1.5)

Safe.

  • Eccentricity and base pressure: xˉ=(1023.5−210.2)/268.8=3.026\bar x = (1023.5-210.2)/268.8 = 3.026 m, e=∣2.5−3.026∣=0.526e = |2.5-3.026| = 0.526 m <B/6=0.833< B/6 = 0.833 m, so no tension. The resultant lies toward the heel:
qmax=87.7, qmin=19.9 kN/m2 (effective, after uplift)q_{max} = 87.7,\ q_{min} = 19.9\ \text{kN/m}^2\ (\text{effective, after uplift})
  • Bearing capacity (Vesic factors for ϕ=27∘\phi=27^\circ: Nc=23.94N_c=23.94, Nγ=14.47N_\gamma=14.47; B′=5−2e=3.95B' = 5-2e = 3.95 m, surface footing, γ′=12.19\gamma'=12.19):
qu=cNc+12γ′B′Nγ=30×23.94+0.5×12.19×3.95×14.47=1067 kN/m2q_u = cN_c + \tfrac12\gamma' B' N_\gamma = 30\times23.94 + 0.5\times12.19\times3.95\times14.47 = 1067\ \text{kN/m}^2

FSb=1067/qmaxFS_b = 1067/q_{max} is well above 3. Safe.

Answer: FSot=4.87FS_{ot} = 4.87, FSsliding=1.79FS_{sliding} = 1.79, e=0.53e = 0.53 m <B/6< B/6, qmax≈88q_{max} \approx 88 kN/m2^2 with a large bearing margin. The wall is stable (assuming the stated adhesion and uplift conditions).

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