Cantilever Retaining Wall EuroNorm EN-7, Design Approach 1 Page 1 of 6

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MATERIAL PROPERTIESSoil friction, φ = 27.6 degrees

Soil cohesion, c = zero17 kN/m³25 kN/m³

DIMENSIONS OF RETAINING STRUCTUREWall :- Height, H = 4 m

Thickness, w = 0.25 mHeight of bearing, Z = 5.5 m

Footing :- Width, B = 4.25 mLength, L = 1 m

Thickness, v = 0.4 mToe width, t = 1.5 m

Heel width, h = 2.5 mAre Rankine criteria satisfied on virtual back? Is h ≥ 2.422 m ? Yes

Unit weight of soil, γ b =Unit weight of reinforced concrete, γ c =

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CHARACTERISTIC DESIGN ACTIONS (not factored)

VARIABLE LOADS 10 kPa Moments2 kPa about Fulcrum

Forces Clockwise +ve27.5 kN/m 79.063 kN-m/m

Bridge Bearing :- 0 kN/m 0 kN-m/m0 kN/m 0 kN-m/m

PERMANENT LOADS Bridge Bearing :- 68 kN/m 110.5 kN-m/m Self Weight of :- Stem, S = WGs,k = 25.0 kN/m 40.6 kN-m/m

Footing, F = WGf,k = 42.5 kN/m 90.3 kN-m/mFill over heel, WGh,k = 170.0 kN/m 510.0 kN-m/m

ACTION EFFECTS ON VIRTUAL BACK Earth Press. Coeft. Ka = 0.3673.2 kN/m -7.1 kN-m/m

60.4 kN/m -88.5 kN-m/m

305.5 kN/m 751.4 kN-m/m-0.045 m

333.0 kN/m 830.5 kN-m/m63.6 kN/m -95.6 kN-m/m

Eccentricity (all Action Effects) from centre of footing; -0.082 m

Surcharge :- over heel, qQstb,k =beyond heel, qQdst,k =

QQstb,k = qQstb,k (B-t) =Horizontal, HQb,k =

Vertical, VQb,k =

Vertical, VGb,k =

Variable :- surcharge, QQdst,k = Ka∙qQdst,k∙(H+v) =Permanent :- soil weight, PG,k = Ka∙γb∙(H+v)²/2 =

Total Permanent Vertical Load, ΣVG,k=ΣWG,k+VGb,k=Eccentricity from centre of footing of ΣVG,k with PG,k ; emin =

Sum of all Vertical Loads, ΣVG,k+VQb,k+QQstb,k =Sum of all Horizontal Forces, PG,k+QQdst,k+HQb,k =

emax =

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PARTIAL FACTORS DESIGN VALUESComb.1 Comb.2 M1 M2

Materials M1 M2 27.6 22.7 degrees1 1.25 Ka = 0.367 0.443 ratio

Actions A1 A2 A1 A21.35 1 453.7 341.3 kN/m

1 1 237.5 237.5 kN/m1.5 1.3 81.5 72.9 kN/m

4.8 5.1 kN/m

-119.5 -107.0 kN-m/m-10.7 -11.2 kN-m/m

φd =On Tan φ, γφ =

Permanent, γG = Vd=γGΣVG,k+γQ(QQstb,k+VQb,k)=γG,fav = Vd,fav = γG,favΣWGk =

Variable, γQ = Pd = γGPG,k =Qd = γQKaqdst,k(H+v) =

Moment of Pd = ‒Pd(H+v)/3 =Moment of Qd = ‒Qd(H+v)/2 =

N.B. these moments are anti-clockwise about fulcrum and therefore negative.

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Coefficient of friction (concrete on soil), μ =Tan δA1 A2

Sum of Horizontal Action Effects, Pd+Qd = HEd,dst = 86.3 78.0 kN/m124.2 99.3 kN/m70% 79%

-130.2 -118.1 kN-m/m640.9 640.9 kN-m/m20% 18%

Eccentricity from centre of footing:- e = B/2 R ΣM/ΣV-0.026 -0.076 m

Equivalent footing width, Ref. Meyerhof G (1963), B' = B-2e = 4.301 4.403 m55.2 53.9 kPa

CHECK STABILITY WITH MINIMUM LOAD OVER FOOTING :– Vb = Hb = qstb = zero

Assume δd = φd for cast insitu footing (no plastic sheet)

Frictional resistance against sliding, HRd = Vd,fav ∗ Tan δd =For sliding:- Degree of utilisation ΛGEO,1 = HEd,dst/HRd =

Overturning action effect, MPd + MQd = MEd,dst =Restoring action effect, MRd,fav = γG,fav(ΣMoments of WG,k) =

For overturning:- Degree of utilisation ΛGEO,1 = MEd,dst/MRd,fav =

B W2 R (MRd + MEd,dst) WVd,fav =

Equivalent bearing pressure = Vd,fav/B' =

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CHECK STABILITY WITH MAXIMUM LOAD OVER FOOTING A1 A286.3 78.0 kN/m

237.2 142.7 kN/m36% 55%

-130.2 -118.11133.0 854.2 kN-m/m

11% 14%-0.086 -0.032 m

B' = B-2e = 4.421 4.314 m102.6 79.1 kPa

CHECK BEARING CAPACITY Bowles (1997) Tables 4-1 & 4-3 after Meyerhof (1963)where c=0 & qY= 6.8 kPa

14.089 8.3957

Shape Factors:- A1 A2 13.685 6.18622.6606 2.4156 10.773 12.875 degrees

2.7875 2.6071 2.7264 2.2565-0.0625 -0.0625 1.0155 1.0141

0.7749 0.73440.3717 0.1873269.52 85.601 kPa

38% 92%

Total horizontal action effect, Pd+Qd+γQHQb,k = HEd,dst =Frictional resistance against sliding, HRd = Vd * Tan δ =

For sliding:- Degree of utilisation ΛGEO,1 = HEd,dst/HRd =

Overturning action effect, MPd + MQd + γQHQb,k(Z+v) = MEd,dst =Restoring action effect, MRd =

Degree of utilization ΛGEO,1 = MEd,dst / MEd,stb =Eccentricity from centre of footing, e= B/2 R (MRd + MEd,dst) WVd =

Equivalent bearing pressure = Vd,fav/B' =

qult = cNcscdcic+qYNqsqdqiq+ 0.5γbB'Nγsγdγiγ

Nq=e(πTanφd)Tan²(45+φd/2) =

Nγ=2(Nq -1)Tan(φd) =sq=1+(B'/L) Sin(φd) θ = Tan-1(HEd,dst WVd,fav) =

sc=(sq*Nq-1)/(Nq-1) Kp =sγ=1–0.3*B/L dq = dγ = 1+0.1√Kp(D/B) =

iq = (1-θ/90) =iγ = (1-θ/φ)² =

qult =Degree of utilization ΛGEO,1 = q'/qult =