# Worked Example | Analysis and Design of Steel Sheet Pile Wall (EN 1997-1)

This article contains a solved example on the analysis and design of steel sheet pile wall in accordance with BS EN1997-1:2004 – Code of Practice for Geotechnical design and the UK National Annex.

Geometry
Total length of sheet pile provided Hpile = 14500 mm
Number of different types of soil Ns = 2
Retained height of soil dret = 3500 mm
Depth of unplanned excavation dex = 500 mm
Total retained height ds = 4000 mm
Angle of retained slope β = 0.0 deg
Depth from ground level to top of water table retained side dw = 1500 mm
Depth from ground level to top of water table retaining side;  dwp = 4000 mm

Variable surcharge po,Q = 10.0 kN/m2

Soil layer 1
Characteristic shearing resistance angle ϕ’k,s1 = 30.0 deg
Characteristic wall friction angle  δk,s1 = 20.0 deg
Moist density of soil γm,s1 = 15.0 kN/m3
Characteristic saturated density of retained soil γs,s1 = 17.0 kN/m3
Height of soil 1  h1 = 8500 mm

Soil layer 2
Characteristic shearing resistance angle f’k,s2 = 27.0 deg
Characteristic wall friction angle dk,s2 = 16.0 deg
Moist density of soil γm,s2 = 16.0 kN/m3
Characteristic saturated density of retained soil γs,s2 = 19.0 kN/m3
Height of soil 2 h2 = 7000 mm

Partial factors on actions – Section A.3.1 – Combination 1
Permanent unfavourable action γG = 1.35
Permanent favourable action γG,f = 1.00
Variable unfavourable action γQ = 1.50
Angle of shearing resistance γf’ = 1.00
Weight density γg = 1.00

Design soil properties – soil 1
Design effective shearing resistance angle ϕ’d = tan-1[tan(ϕ’k)/γf’] = 30.0 deg
Design wall friction angle  δd = tan-1[tan(ϕk)/γf’] = 20.0 deg
Design moist density of retained soil γm.d1 = γmγ = 15.0 kN/m3
Design saturated density of retained soil γs.d1 = γsγ = 17.0 kN/m3
Design buoyant density of retained soil γd.d1 = γs.d1 – γw = 7.2 kN/m3

Active pressure using Coulomb theory Ka1 = sin(α + ϕ’d)2 / (sin(α)2 × sin(α – δd) × (1 + √(sin(ϕ’d + δd) × sin(ϕ’d – β)/(sin(α – δd) ´ sin(α + β))))2) = 0.297

Passive pressure using Coulomb theory Kp1 = sin(90 – ϕ’d)2 / (sin(90 + δd) × [1 – √[sin(ϕ’d + δd) × sin(ϕ’d) / (sin(90 + δd))]]2) = 6.105

Design soil properties – soil 2
Design effective shearing resistance angle ϕ’d = tan-1(tan(ϕ’k) /γf’) = 27.0 deg
Design wall friction angle  δd = tan-1(tan(δk)/γf’) = 16.0 deg
Design moist density of retained soil γm.d2 = γmγ = 16.0 kN/m3
Design saturated density of retained soil γs.d2 = γsγ = 19.0 kN/m3
Design buoyant density of retained soil γd.d2 = γs.d2 – γw = 9.2 kN/m3

Active pressure using Coulomb theory Ka2 = sin(α + ϕ’d)2 / (sin(α)2 × sin(α – δd) × (1 + √(sin(ϕ’d + δd) × sin(ϕ’d – β) / (sin(α – δd) × sin(α + β))))2) = 0.336

Passive pressure using Coulomb theory Kp2 = sin(90 – ϕ’d)2 / (sin(90 + δd) × [1 – √[sin(ϕ’d + δd) × sin(f’d) / (sin(90 + δd))]]2) = 4.416

Overburden on active side
Overburden at 0 mm below GL in soil 1; OB’a11 = po,Q × γQ = 15.0 kN/m2
Overburden at 1500 mm below GL in soil 1; OB’a21 = γG × γm.d1 × ha1 + OB’a11 = 45.4 kN/m2
Overburden at 4000 mm below GL in soil 1; OB’a31 = γG × γd.d1 × ha2 + OB’a21 = 69.6 kN/m2
Overburden at 8500 mm below GL in soil 1; OB’a41 = γG × γd.d1 × ha3 + OB’a31 = 113.3 kN/m2
Overburden at 8500 mm below GL in soil 2; OB’a42 = γG × γd.d1 × ha3 + OB’a31 = 113.3 kN/m2
Overburden at 11544 mm below GL in soil 2; OB’a51 = γG × γd.d2 × ha4 + OB’a42 = 151.1 kN/m2

Overburden on passive side
Overburden at 4000 mm below GL in soil 1; OB’p31 = 0 kN/m2 = 0.0 kN/m2
Overburden at 8500 mm below GL in soil 1; OB’p41 = γG,f × γd.d1 × hp3 + OB’p31 = 32.4 kN/m2
Overburden at 8500 mm below GL in soil 2; OB’p42 = γG,f × γd.d1 × hp3 + OB’p31 = 32.4 kN/m2
Overburden at 11544 mm below GL in soil 2; OB’p51 = γG,f × γd.d2 × hp4 + OB’p42 = 60.3 kN/m2

Pressure on active side
Active at 0 mm below GL in soil 1; p’a11 = Ka1 × OB’a11 = 4.5 kN/m2
Active at 1500 mm below GL in soil 1; p’a21 = Ka1 × OB’a21 = 13.5 kN/m2
Active at 4000 mm below GL in soil 1; p’a31 = Ka1 × OB’a31 + γγ × γw × (dL3 – dw) = 53.8 kN/m2
Active at 8500 mm below GL in soil 1; p’a41 = Ka1 × OB’a41 + γγ × γw × (dL4 – dw) = 126.4 kN/m2
Active at 8500 mm below GL in soil 2; p’a42 = Ka2 × OB’a42 + γγ × γw × (dL4 – dw) = 130.8 kN/m2
Active at 11544 mm below GL in soil 2; p’a51 = Ka2 × OB’a51 + γγ × γw × (dL5 – dw) = 183.8 kN/m2

Pressure on passive side
Passive at 4000 mm below GL in soil 1; p’p31 = Kp1 × OB’p31 + γG,f × γw × (dL3 – max(ds, dw)) = 0.0 kN/m2
Passive at 8500 mm below GL in soil 1; p’p41 = Kp1 × OB’p41 + γG,f × γw × (dL4 – max(ds, dw)) = 241.7 kN/m2
Passive at 8500 mm below GL in soil 2; p’p42 = Kp2 × OB’p42 + γG,f × γw × (dL4 – max(ds, dw)) = 187.0 kN/m2
Passive at 11544 mm below GL in soil 2; p’p51 = Kp2 × OB’p51 + γG,f × γw × (dL5 – max(ds, dw)) = 340.4 kN/m2

By iteration the depth at which the active moments equal the passive moments has been determined as 11544 mm as follows:-

Moment level 1;Ma11 = 0.5 × p’a11 × ha1 × ((H – dL2) + 2/3 × ha1) = 36.9 kNm/m
Moment level 1; Ma12 = 0.5 × p’a21 × ha1 × ((H – dL2) + 1/3 × ha1) = 106.7 kNm/m
Moment level 2; Ma21 = 0.5 × p’a21 × ha2 × ((H – dL3) + 2/3 × ha2) = 155.3 kNm/m
Moment level 2; Ma22 = 0.5 × p’a31 × ha2 × ((H – dL3) + 1/3 × ha2) = 563.5 kNm/m
Moment level 3; Ma31 = 0.5 × p’a31 × ha3 × ((H – dL4) + 2/3 × ha3) = 731.8 kNm/m
Moment level 3; Ma32 = 0.5 × p’a41 × ha3 × ((H – dL4) + 1/3 × ha3) = 1292.3 kNm/m
Moment level 4; Ma41 = 0.5 × p’a42 × ha4 × ((H – dL5) + 2/3 × ha4) = 404.0 kNm/m
Moment level 4; Ma42 = 0.5 × p’a51 × ha4 × ((H – dL5) + 1/3 × ha4) = 283.8 kNm/m

Moment level 3; Mp31 = 0.5 × p’p31 × hp3 × ((H – dL4) + 2/3 × hp3) = 0.0 kNm/m
Moment level 3; Mp32 = 0.5 × p’p41 × hp3 × ((H – dL4) + 1/3 × hp3) = 2471.0 kNm/m
Moment level 4; Mp41 = 0.5 × p’p42 × hp4 × ((H – dL5) + 2/3 × hp4) = 577.6 kNm/m
Moment level 4; Mp42 = 0.5 × p’p51 × hp4 × ((H – dL5) + 1/3 × hp4) = 525.7 kNm/m

Total active moment; SMa = 3574.5 kNm/m
Total passive moment; SMp = 3574.5 kNm/m

Required pile length
Length of pile required to balance moments; H = 11544 mm

Depth of equal pressure; dcontra = 5432 mm
Add 20% below this point; de_add = 1.2 × (H – dcontra) = 7334 mm

Minimum required pile length; Htotal = dcontra + de_add = 12766 mm

PassProvided length of sheet pile greater than minimum required length of pile

Pile capacity (EN1993-5)
Maximum moment in pile (from analysis); Mpile = max(abs(Mmin), abs(Mmax)) / 1m = 547.0 kNm/m
Maximum shear force in pile (from analysis); Vpile = 364.7 kN/m
Nominal yield strength of pile; fy_pile = 355 N/mm2
Name of sheet pile;  Arcelor PU(18)
Classification of pile; 2
Plastic modulus of pile; Wpl.y = 2134 cm3/m

Shear buckling of web (cl.5.2.2(6))
Width of section; c = h / sin(αpile) = 510 mm
Thickness of web;  tw = s = 9.0 mm
ε = √(235/fy_pile) = 0.814
c/tw = 56.6 = 69.6ε < 72ε
PASS – Shear buckling of web within limits

Bending 2
Interlock reduction factor (cl.5.2.2); βB = 1
Design bending resistance (eqn.5.2);
Mc,Rd = Wpl.y × fy_pile × βB / γM0 = 757.6 kNm/m
PASSMoment capacity exceeds moment in pile

Shear
Projected shear area of web (eqn.5.6); Av = s × (h – t) = 3769 mm2
Design shear resistance (eqn.5.5); Vpl,Rd = Av × fy_pile / (√(3) × γM0) / b = 1287.6 kN/m
PASSShear capacity exceeds shear in pile

Partial factors on actions – Section A.3.1 – Combination 2

Permanent unfavourable action;  γG = 1.00
Permanent favourable action; γG,f = 1.00
Variable unfavourable action; γQ = 1.30
Angle of shearing resistance; γϕ’ = 1.25
Weight density; γγ = 1.00

Design soil properties – soil 1
Design effective shearing resistance angle; ϕ’d = tan-1(tan(ϕ’k)/γϕ’) = 24.8 deg
Design wall friction angle;  δd = tan-1(tan(δk)/γϕ’) = 16.2 deg
Design moist density of retained soil; γm.d1 = γmγ = 15.0 kN/m3
Design saturated density of retained soil; γs.d1 = γsγ = 17.0 kN/m3
Design buoyant density of retained soil; γd.d1 = γs.d1 – γw = 7.2 kN/m3

Active pressure using Coulomb theory; Ka1 = sin(α + ϕ’d)2 / (sin(α)2 × sin(α – δd) × (1 + √(sin(ϕ’d + δd) × sin(ϕ’d – β)/(sin(α – δd) ´ sin(α + β))))2) = 0.364

Passive pressure using Coulomb theory; Kp1 = sin(90 – ϕ’d)2 / (sin(90 + δd) × [1 – √[sin(ϕ’d + δd) × sin(ϕ’d) / (sin(90 + δd))]]2) = 3.977

Design soil properties – soil 2

Design effective shearing resistance angle; ϕ’d2 = tan-1(tan(ϕ’k)/γϕ’) = 22.2 deg
Design wall friction angle; δd2 = tan-1(tan(δk)/γϕ’) = 12.9 deg
Design moist density of retained soil; γm.d2 = γmγ = 16.0 kN/m3
Design saturated density of retained soil; γs.d2 = γsγ = 19.0 kN/m3
Design buoyant density of retained soil; γd.d2 = γs.d2 – γw = 9.2 kN/m3

Active pressure using Coulomb theory; Ka2 = sin(α + ϕ’d)2 / (sin(α)2 × sin(α – δd) × (1 + √(sin(ϕ’d + δd) × sin(ϕ’d – β) / (sin(α – δd) × sin(α + β))))2) = 0.406

Passive pressure using Coulomb theory; Kp2 = sin(90 – ϕ’d)2 / (sin(90 + δd) × [1 – √[sin(ϕ’d + δd) × sin(f’d) / (sin(90 + δd))]]2) = 3.154

Overburden on active side
Overburden at 0 mm below GL in soil 1; OB’a11 = po,Q × γQ = 13.0 kN/m2
Overburden at 1500 mm below GL in soil 1;  OB’a21 = γG × γm.d1 × ha1 + OB’a11 = 35.5 kN/m2
Overburden at 4000 mm below GL in soil 1; OB’a31 = γG × γd.d1 × ha2 + OB’a21 = 53.5 kN/m2
Overburden at 8500 mm below GL in soil 1; OB’a41 = γG × γd.d1 × ha3 + OB’a31 = 85.8 kN/m2
Overburden at 8500 mm below GL in soil 2; OB’a42 = γG × γd.d1 × ha3 + OB’a31 = 85.8 kN/m2
Overburden at 12532 mm below GL in soil 2;OB’a51 = γG × γd.d2 × ha4 + OB’a42 = 122.9 kN/m2

Overburden on passive side
Overburden at 4000 mm below GL in soil 1; OB’p31 = 0 kN/m2 = 0.0 kN/m2
Overburden at 8500 mm below GL in soil 1; OB’p41 = γG,f × γd.d1 × hp3 + OB’p31 = 32.4 kN/m2
Overburden at 8500 mm below GL in soil 2; OB’p42 = γG,f × γd.d1 × hp3 + OB’p31 = 32.4 kN/m2
Overburden at 12532 mm below GL in soil 2;OB’p51 = γG,f × γd.d2 × hp4 + OB’p42 = 69.4 kN/m2

Pressure on active side

Active at 0 mm below GL in soil 1; p’a11 = Ka1 × OB’a11 = 4.7 kN/m2
Active at 1500 mm below GL in soil 1; p’a21 = Ka1 × OB’a21 = 12.9 kN/m2
Active at 4000 mm below GL in soil 1; p’a31 = Ka1 × OB’a31 + γG × γw × (dL3 – dw) = 44.0 kN/m2
Active at 8500 mm below GL in soil 1; p’a41 = Ka1 × OB’a41 + γG × γw × (dL4 – dw) = 99.9 kN/m2
Active at 8500 mm below GL in soil 2; p’a42 = Ka2 × OB’a42 + γG × γw × (dL4 – dw) = 103.5 kN/m2
Active at 12532 mm below GL in soil 2; p’a51 = Ka2 × OB’a51 + γG × γw × (dL5 – dw) = 158.1 kN/m2

Pressure on passive side

Passive at 4000 mm below GL in soil 1; p’p31 = Kp1 × OB’p31 + γG,f × γw × (dL3 – max(ds, dw)) = 0.0 kN/m2
Passive at 8500 mm below GL in soil 1; p’p41 = Kp1 × OB’p41 + γG,f × γw × (dL4 – max(ds, dw)) = 172.8 kN/m2
Passive at 8500 mm below GL in soil 2;p’p42 = Kp2 × OB’p42 + γG,f × γw × (dL4 – max(ds, dw)) = 146.2 kN/m2
Passive at 12532 mm below GL in soil 2; p’p51 = Kp2 × OB’p51 + γG,f × γw × (dL5 – max(ds, dw)) = 302.7 kN/m2

By iteration the depth at which the active moments equal the passive moments has been determined as 12533 mm as follows:-

Moment level 1; Ma11 = 0.5 × p’a11 × ha1 × ((H – dL2) + 2/3 × ha1) = 42.7 kNm/m
Moment level 1; Ma12 = 0.5 × p’a21 × ha1 × ((H – dL2) + 1/3 × ha1) = 111.8 kNm/m
Moment level 2; Ma21 = 0.5 × p’a21 × ha2 × ((H – dL3) + 2/3 × ha2) = 164.8 kNm/m
Moment level 2; Ma22 = 0.5 × p’a31 × ha2 × ((H – dL3) + 1/3 × ha2) = 515.1 kNm/m
Moment level 3; Ma31 = 0.5 × p’a31 × ha3 × ((H – dL4) + 2/3 × ha3) = 696.2 kNm/m
Moment level 3; Ma32 = 0.5 × p’a41 × ha3 × ((H – dL4) + 1/3 × ha3) = 1244.0 kNm/m
Moment level 4; Ma41 = 0.5 × p’a42 × ha4 × ((H – dL5) + 2/3 × ha4) = 561.3 kNm/m
Moment level 4; Ma42 = 0.5 × p’a51 × ha4 × ((H – dL5) + 1/3 × ha4) = 428.7 kNm/m

Moment level 3; Mp31 = 0.5 × p’p31 × hp3 × ((H – dL4) + 2/3 × hp3) = 0.0 kNm/m
Moment level 3; Mp32 = 0.5 × p’p41 × hp3 × ((H – dL4) + 1/3 × hp3) = 2151.5 kNm/m
Moment level 4; Mp41 = 0.5 × p’p42 × hp4 × ((H – dL5) + 2/3 × hp4) = 792.7 kNm/m
Moment level 4; Mp42 = 0.5 × p’p51 × hp4 × ((H – dL5) + 1/3 × hp4) = 820.5 kNm/m

Total active moment; SMa = 3763.9 kNm/m
Total passive moment; SMp = 3763.7 kNm/m

Required pile length
Length of pile required to balance moments; H = 12533 mm

Depth of equal pressure; dcontra = 5694 mm
Add 20% below this point; de_add = 1.2 × (H – dcontra) = 8207 mm
Minimum required pile length; Htotal = dcontra + de_add = 13901 mm

PASS – Provided length of sheet pile greater than minimum required length of pile

Pile capacity (EN1993-5)

Maximum moment in pile (from analysis); Mpile = max(abs(Mmin), abs(Mmax)) / 1m = 549.1 kNm/m
Maximum shear force in pile (from analysis); Vpile = 358.1 kN/m
Nominal yield strength of pile; fy_pile = 355 N/mm2
Name of pile;  Arcelor PU(18)
Classification of pile; 2
Plastic modulus of pile; Wpl.y = 2134 cm3/m

Shear buckling of web (cl.5.2.2(6))

Width of section; c = h / sin(apile) = 510 mm
Thickness of web; tw = s = 9.0 mm
ε = √(235/fy_pile)= 0.814
c / tw = 56.6 = 69.6ε < 72ε

PASSShear buckling of web within limits

Bending
Interlock reduction factor (cl.5.2.2); βB = 1
Design bending resistance (eqn.5.2);Mc,Rd = Wpl.y × fy_pile × βB / γM0 = 757.6 kNm/m

PASSMoment capacity exceeds moment in pile

Shear
Projected shear area of web (eqn.5.6); Av = s × (h – t) = 3769 mm2
Design shear resistance (eqn.5.5); Vpl,Rd = Av × fy_pile / (√(3) × γM0) / b = 1287.6 kN/m

PASSShear capacity exceeds shear in pile