A sheet pile wall is a special type of retaining wall that is generally made of steel, timber, and in the most limited case concrete structures. Steel piles are the commonest because they can be used on all kinds of terrain, they can be used at depths greater than 3 m, they are water-tight, and can be re-used. Timber sheet piles are generally cheaper than steel sheet piles but can only be used for temporary structures where the depth of driving does not exceed 3 m. Reinforced concrete sheet piles can only be used only when it is possible to jet them into fine sands or drive them into very soft soils. They are not suitable for tougher soils as they can generally break off under driving.

**Types of Sheet Pile Walls**

Sheet pile walls are generally classified based on their structural form and loading system. Under these, we have:

1. The cantilever sheet pile walls and

2. The anchored sheet pile walls.

The cantilever sheet pile walls are further classified into **free cantilever sheet pile walls** (cantilever sheet pile walls subjected to a concentrated load at the top) – these derive stability entirely from the lateral passive resistance of the soil below the dredge level into which they are driven, **cantilever sheet pile walls** (these retains backfill at a higher level on one side) – the stability is entirely from the lateral passive resistance of the soil into which the sheet pile is driven.

The anchored sheet pile walls which are held above the driven depth by anchors provided at a suitable level can be divided into two namely:

1. Fixed earth support piles, and

2. Free earth support piles

**Design of Sheet Piles**

The design of sheet pile walls lies in the determination of the DEPTH OF EMBEDMENT, d. There are unique equations for the design of cantilever sheet pile walls and anchored sheet pile walls. The equations also vary when the sheet pile walls are located in cohesionless soils against when they are located in cohesive soils (These can be assessed from specialized textbooks). The manual method of this design is tedious and I would show a manual approach to the design and a simple software application to verify the design and which can be used to achieve the same purpose with ease. When the cantilever sheet pile walls are located in cohesionless soil, the depth of embedment calculated should be increased by 20% to 40% while for cantilever sheet pile walls located in clay soils, the calculated depth of embedment should be increased by 40% to 60%.

**Design Parts**

In the article, I would design the cantilever sheet pile wall without backfill using manual and software applications. Figure 1 shows a cantilever sheet pile in a cohesionless soil deposit.

The pole rotates about the point P. The pressure above P is passive in the front and active on the back side. However, the pressures below point P are reversed, that is, there is active pressure in the front and passive in the backside. Figure 2 shows the actual pressure distribution. As the analysis using the actual pressure distribution is quite complicated, the pressure distribution is generally simplified as shown in Figure 3.

The depth, **a** of the point, P of the zero pressure is given by

p_{1} – γa (K_{p} – K_{a}) = 0 » a = p_{1}/ γ (K_{p} – K_{a})

Let the total active pressure above point P be P_{1} acting at a height, Z_{1} above P. The passive pressure is given by the diagram, PDE. The passive pressure intensity at the bottom tip A can be expressed as:

p_{2} = γ (K_{p} – K_{a}) (d – a) = γ (K_{p} – K_{a}) b

Where; **b **= d – a,

d is the depth of point A below the dredge level.

The passive pressure is indicated by the diagram EAF on the back side. The intensity of pressure at the tip A is given by:

p_{3} = γ (h + d) K_{p} – γdK_{a} » γ (h + d) K_{p} – γ (b + a) K_{a}

From the equation of equilibrium in the horizontal direction, P_{1} + P_{3} – P_{2} = 0

The total pressure P_{3} and P_{2} can be expressed in terms of p_{3} and p_{2} as follows:

P_{1} + ½ m (p_{2 }+ p_{3}) – ½ p_{2} b = 0———–(1)

Equivalence area diagrams are shown below:

**= **

**=**

From Eqn (1), m = (½ p_{2} b – P_{1})/ ½ (p_{2} + p_{3}) = (p_{2} b – 2P_{1})/ (p_{2} + p_{3})———(2)

Taking a moment of all forces about A,

P_{1} (b + z_{1}) – ½ p_{2} b (b/3) + ½ m (p_{2} + p_{3}) x (m/3) = 0———-(3)

Substitute equation (2) into (3)

P_{1} (b + Z_{1}) – (p_{2}b^{2}/6) + (p_{2} + p_{3})/6 [(p_{2}b – 2P_{1})/ (p_{2} + p_{3})]^{2} = 0———(4)

Eqn (4) can be re-written as

b^{4} + C_{1}b^{3} – C_{2}b^{2} – C_{3}b – C_{4} = 0———–(5)

C_{1} = p_{4}/ (γ (K_{p} – K_{a})); C_{2 }= 8P_{1}/ (γ (K_{p} – K_{a})); C_{3} = [6P_{1} [2γ (K_{p} – K_{a}) Z_{1} + p_{4}]]/ (γ (K_{p} – K_{a}))^{2}; C_{4} = [P_{1 }[6Z_{1}p_{4} + 4P_{1}]]/ (γ (K_{p} – K_{a}))^{2}

In which, p_{4} = γhK_{p} + γa (K_{p} – K_{a})

Eqn (5) is solved by trial and error method to determine b, then d = b + a.

The depth, d is for a factor of safety of unity. The required depth D is usually taken as D = 1.2d to 1.4d. This gives a factor of safety of about 1.5 to 2.0.

Alternatively, a factor of safety can be applied to the passive resistance. In that case, the value of K_{p} is usually taken as 1/2 to 2/3 of the normal value while computing b from Eqn (5), and the required depth D is taken as equal to d.

In the above discussions, the depth of the water table is not considered. If the water table on the front side is at the same level as on the rear side, the analysis remains unaltered except that the submerged unit weight (γ^{1}) should be used for the soil below the water table. However, if the difference in the two levels is greater than 1m, the pressure due to water on the sheet pile should be found from the flow net and properly accounted for in the analysis.

**Example**

Determine the required depth of embedment or penetration for the cantilever sheet pile wall shown in Figure 5 below. Take γ = 18 kN/m^{3}; ϕ = 38^{o}; h = 7 m

**Solution**

K_{a} = tan^{2} (45 – 38/2) = 0.238

K_{p} = tan^{2} (45 + 38/2) = 4.204

p_{1} = 0.238 x 18 x 7 =29.988 kN/m^{2}

a = p_{1}/ (γ (K_{p} – K_{a})) = 29.988/ (18 x (4.204 – 0.238)) = 29.988/ 71.388 = 0.42 m

P_{1} = ½ x 29.988 x 7 + ½ x 29.988 x 0.42 = 104.958 + 6.297 = 111.255 kN

Taking a moment about P and dividing by P_{1}

Z_{1} = [(104.958 x 2.753) + (6.297 x 0.28)]/ 111.255 = (288.949 + 1.7632)/111.255 = 2.613 m

p_{2} = γ (K_{p} – K_{a}) (b) = 18 (4.204 – 0.238) b = 71.388 b

p_{3} = γ (h + d) K_{p} – γdK_{a }= γ (h + b + a) K_{p} – γ (b + a) K_{a} = 18 (7 + b + 0.42) 4.204 – 18 (b + 0.42) 0.238

p_{3} = 75.672 (7.42 + b) – 4.284 (b + 0.42) = 561.486 + 75.672 b – 4.284 b – 1.799 = 559.687 + 71.388 b

From Eqn (2), m = (p_{2} b – 2P_{1})/ (P_{2} + P_{3}) = [71.388 b – (2 x 111.255)]/ (559.687 + 71.388b + 71.388b) = [71.388 b – (222.51)]/ (559.687 + 142.776 b)

From Eqn (4), P_{1} (b + Z_{1}) – (p_{2}b^{2}/6) + (p_{2} + p_{3})/6 [(p_{2}b – 2P_{1})/ (p_{2} + p_{3})]^{2} = 0 » [111.255 (b + 2.613)] – (71.388b^{3}/6) + [(559.687 + 142.776 b)/6] x [(71.388 b^{2} – (2 x 111.255))/ (559.687 + 71.388b + 71.388b)]^{2}

» 111.255 b + 290.7 – 11.898 b^{3 }+ (93.281 + 23.796 b) x [(71.388 b^{2} -222.51)/ (559.687 + 142.776 b)]^{2}

This equation can be solved by a suitable trial and error method. However, I would adopt the second approach to solve it because it is simple and conformable to the tool available for me to solve it.

**Alternatively,**

b^{4} + C_{1}b^{3} – C_{2}b^{2} – C_{3}b – C_{4} = 0

C_{1} = p_{4}/ (γ (K_{p} – K_{a}))?

C_{2 }= 8P_{1}/ (γ (K_{p} – K_{a}))?

C_{3} = [6P_{1} [2γ (K_{p} – K_{a}) Z_{1} + p_{4}]]/ (γ (K_{p} – K_{a}))^{2}?

C_{4} = [P_{1 }[6Z_{1}p_{4} + 4P_{1}]]/ (γ (K_{p} – K_{a}))^{2}?

p_{1} = γa (K_{p} – K_{a}) = 18 x 0.42 (4.204 – 0.238) = 29.983

p_{2 }= γb (K_{p} – K_{a}) = 18 x 7 (4.204 – 0.238) = 499.716

p_{3} = γ (h + b + a) K_{p} – γ (b + a) K_{a} = 18 (7 + b + 0.42) 4.204 – 18 (b + 0.42) 0.238 = 75.672 (7.42 + b) – 4.284 (b + 0.42) = 561.486 + 75.672b – 4.284 b – 1.799 = 559.687 + 71.388 b

p_{4} = γh K_{p} + γa (K_{p} – K_{a}) = 18 x 7 x 4.204 + 18 x 0.42 (4.204 – 0.238) = 529.704 + 29.983 = 559.687

C_{1} = p_{4}/ (γ (K_{p} – K_{a})) = 559.687/ (18 x 3.966) = 7.84

C_{2 }= 8P_{1}/ (γ (K_{p} – K_{a})) = (8 x 29.983)/ (18 x 3.966) = 3.36

C_{3} = [6P_{1} [2γ (K_{p} – K_{a}) Z_{1} + p_{4}]]/ (γ (K_{p} – K_{a}))^{2} = [6 x 29.983 [2 x 18 (4.204 – 0.238) 2.613 + 559.687]]/ (18 x 3.966)^{2 }= (179.898 (373.074 + 559.687))/ 5096.2465 = 167801.8384/5096.2465 = 32.927

C_{4} = [P_{1 }[6Z_{1}p_{4} + 4P_{1}]]/ (γ (K_{p} – K_{a}))^{2} = [29.983 [(6 x 2.613 x 559.687) + (4 x 29.983)]]/ (18 x 3.966)^{2} = (29.983 (8774.7728 + 119.932))/ 5096.2465 = 266689.934/5096.2465 = 53.331

C_{1} = **7.84**; C_{2} = **3.36**; C_{3} = **32.927**; C_{4} = **53.331**

Therefore, b^{4} + C_{1}b^{3} – C_{2}b^{2} – C_{3}b – C_{4} = 0 » b^{4} + 7.84 b^{3} – 3.36 b^{2} – 32.927 b – 53.331 = 0

Solving by trial and error method, b = 2.5

Therefore, d = b + a = 2.5 + 0.42 = 2.92

D = 1.5 d = 1.5 x 2.92 = 4.38, say 4.4 m

After determining the depth of embedment, the pile is checked to ensure that it passes all tests and to ensure that the depth of embedment determined is very satisfactory. This can be done with Tekla Tedds software. In a situation where the sheet pile has a surcharge load, it is also added and the depth of embedment is investigated. In Tekla Tedds, a water Table can be added, soil properties can be defined, cohesionless or cohesive soils can be defined and, the steel section to be used can also be defined.

**Steel sheet piling analysis & design** (using Tekla Tedds software)

In accordance with BS EN1997-1:2004 – Code of Practice for Geotechnical Design and the UK National Annex

Geometry

The total length of sheet pile provided; H_{pile} = **11400** mm

Number of different types of soil; N_{s} = **1**

Retained height; d_{ret} = **6500** mm

Depth of unplanned excavation; d_{ex} = **500** mm

Total retained height; d_{s} = **7000** mm

Angle of retained slope; b = **0.0** deg

Loading

Soil characteristic properties table

Soil | f’_{k }(deg) |
d_{k }(deg) |
g_{m} (kN/m^{3}) |
g_{s} (kN/m^{3}) |
h (mm) |

1 | 38.0 | 25.0 | 18.0 | 18.0 | 11400 |

Partial factors on actions – Section A.3.1 – Combination 1

Permanent unfavourable action; g_{G} = **1.35**

Permanent favourable action; g_{G,f} = **1.00**

Variable unfavourable action; g_{Q} = **1.50**

Angle of shearing resistance; g_{f}_{‘} = **1.00**

Weight density; g_{g} = **1.00**

Design properties table – combination 1

Soil | f’_{d} |
d_{d} |
g_{m.d} |
g_{s.d} |
K_{a} |
K_{p} |

1 | 38.0 | 25.0 | 18.0 | 18.0 | 0.217 | 13.901 |

Overburden on the active side

Overburden at 0 mm below GL in soil 1; OB’_{a11} = 0 kN/m^{2} = **0.0** kN/m^{2}

Overburden at 7000 mm below GL in soil 1; OB’_{a21} = g_{G} ´ g_{m.d1} ´ h_{a1} + OB’_{a11} = **170.1** kN/m^{2}

Overburden at 9670 mm below GL in soil 1; OB’_{a31} = g_{G} ´ g_{m.d1} ´ h_{a2} + OB’_{a21} = **235.0** kN/m^{2}

Overburden on the passive side

Overburden at 7000 mm below GL in soil 1; OB’_{p21} = 0 kN/m^{2} = **0.0** kN/m^{2}

Overburden at 9670 mm below GL in soil 1; OB’_{p31} = g_{G,f} ´ g_{m.d1} ´ h_{p2} + OB’_{p21} = **48.1** kN/m^{2}

Pressure on the active side

Active at 0 mm below GL in soil 1; p’_{a11} = K_{a1} ´ OB’_{a11} = **0.0** kN/m^{2}

Active at 7000 mm below GL in soil 1; p’_{a21} = K_{a1} ´ OB’_{a21} = **36.9** kN/m^{2}

Active at 9670 mm below GL in soil 1; p’_{a31} = K_{a1} ´ OB’_{a31} = **50.9** kN/m^{2}

Pressure on the passive side

Passive at 7000 mm below GL in soil 1; p’_{p21} = K_{p1} ´ OB’_{p21} = **0.0** kN/m^{2}

Passive at 9670 mm below GL in soil 1; p’_{p31} = K_{p1} ´ OB’_{p31} = **668.1** kN/m^{2}

By iteration the depth at which the active moments equal the passive moments has been determined as 9670 mm as follows:-

Active moment about 9670 mm

Moment level 1; M_{a11} = 0.5 ´ p’_{a11} ´ h_{a1} ´ ((H – d_{L2}) + 2/3 ´ h_{a1}) = **0.0** kNm/m

Moment level 1; M_{a12} = 0.5 ´ p’_{a21} ´ h_{a1} ´ ((H – d_{L2}) + 1/3 ´ h_{a1}) = **645.6** kNm/m

Moment level 2; M_{a21} = 0.5 ´ p’_{a21} ´ h_{a2} ´ ((H – d_{L3}) + 2/3 ´ h_{a2}) = **87.6** kNm/m

Moment level 2; M_{a22} = 0.5 ´ p’_{a31} ´ h_{a2} ´ ((H – d_{L3}) + 1/3 ´ h_{a2}) = **60.5** kNm/m

Passive moment about 9670 mm

Moment level 2; M_{p21} = 0.5 ´ p’_{p21} ´ h_{p2} ´ ((H – d_{L3}) + 2/3 ´ h_{p2}) = **0.0** kNm/m

Moment level 2; M_{p22} = 0.5 ´ p’_{p31} ´ h_{p2} ´ ((H – d_{L3}) + 1/3 ´ h_{p2}) = **793.8** kNm/m

Total moments about 9670 mm

Total active moment; SM_{a} = **793.7** kNm/m

Total passive moment; SM_{p} = **793.7** kNm/m

Required pile length

Length of pile required to balance moments; H = **9670** mm

Depth of equal pressure; d_{contra} = **7150** mm

Add 20% below this point; d_{e_add} = 1.2 ´ (H – d_{contra}) = **3023** mm

Minimum required pile length; H_{total} = d_{contra} + d_{e_add} = **10174** mm

Pass – Provided length of sheet pile greater than the minimum required length of the pile

Pile capacity (EN1993-5)

Maximum moment in pile (from analysis); M_{pile }= max(abs(M_{min}), abs(M_{max})) / 1m = **411.9** kNm/m

Maximum shear force in pile (from analysis); V_{pile }= **645.6** kN/m

Nominal yield strength of pile; f_{y_pile} = **270** N/mm^{2}

Name of pile; Arcelor AU25

Classification of pile; 2

Plastic modulus of pile; W_{pl.y} = **2866** cm^{3}/m

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

Width of section; c = h / sin(a_{pile}) = **551** mm

Thickness of web; t_{w} = s = **10.2** mm

e = Ö(235 N/mm^{2} / f_{y_pile}) = **0.933**

c / t_{w} = 54.1 = 57.9 ´ e <= 72 ´ e

PASS – Shear buckling of web within limits

**Bending**

Interlock reduction factor (cl.5.2.2); b_{B} = **0.75**

Design bending resistance (eqn.5.2); M_{c,Rd} = W_{pl.y} ´ f_{y_pile} ´ b_{B} / g_{M0} = **580.4** kNm/m

PASS – Moment capacity exceeds moment in pile

**Shear**

Projected shear area of web (eqn.5.6); A_{v} = s ´ (h – t) = **4442** mm^{2}

Design shear resistance (eqn.5.5); V_{pl,Rd} = A_{v} ´ f_{y_pile} / (Ö(3) ´ g_{M0}) / b = **923.3** kN/m

**PASS** – Shear capacity exceeds shear in a pile

**Combined bending and shear**

Shear presence mnt reduction factor (eqn.5.10); r = (2 ´ V_{pile} / V_{pl,Rd} – 1)^{2} = **0.159**

Reduced bending resistace (eqn.5.9); M_{V,Rd} = min((b_{B} ´ W_{pl.y} – r ´ A_{v}^{2} / (4 ´ s ´ b ´ sin(a_{pile}))) ´ (f_{y_pile} / g_{M0}), M_{c,Rd}) = **546.5** kNm/m

PASS – Reduced moment capacity exceeds moment in pile

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

Permanent unfavourable action; g_{G} = **1.00**

Permanent favourable action; g_{G,f} = **1.00**

Variable unfavourable action; g_{Q} = **1.30**

Angle of shearing resistance; g_{f}_{‘} = **1.25**

Weight density; g_{g} = **1.00**

Design properties table – combination 2

Soil | f’_{d} |
d_{d} |
g_{m.d} |
g_{s.d} |
K_{a} |
K_{p} |

1 | 32.0 | 20.5 | 18.0 | 18.0 | 0.275 | 7.038 |

Overburden on the active side

Overburden at 0 mm below GL in soil 1; OB’_{a11} = 0 kN/m^{2} = **0.0** kN/m^{2}

Overburden at 7000 mm below GL in soil 1; OB’_{a21} = g_{G} ´ g_{m.d1} ´ h_{a1} + OB’_{a11} = **126.0** kN/m^{2}

Overburden at 10597 mm below GL in soil 1; OB’_{a31} = g_{G} ´ g_{m.d1} ´ h_{a2} + OB’_{a21} = **190.7** kN/m^{2}

Overburden on the passive side

Overburden at 7000 mm below GL in soil 1; OB’_{p21} = 0 kN/m^{2} = **0.0** kN/m^{2}

Overburden at 10597 mm below GL in soil 1; OB’_{p31} = g_{G,f} ´ g_{m.d1} ´ h_{p2} + OB’_{p21} = **64.7** kN/m^{2}

Pressure on the active side

Active at 0 mm below GL in soil 1; p’_{a11} = K_{a1} ´ OB’_{a11} = **0.0** kN/m^{2}

Active at 7000 mm below GL in soil 1; p’_{a21} = K_{a1} ´ OB’_{a21} = **34.7** kN/m^{2}

Active at 10597 mm below GL in soil 1; p’_{a31} = K_{a1} ´ OB’_{a31} = **52.5** kN/m^{2}

Pressure on the passive side

Passive at 7000 mm below GL in soil 1; p’_{p21} = K_{p1} ´ OB’_{p21} = **0.0** kN/m^{2}

Passive at 10597 mm below GL in soil 1; p’_{p31} = K_{p1} ´ OB’_{p31} = **455.7** kN/m^{2}

By iteration the depth at which the active moments equal the passive moments has been determined as 10597 mm as follows:-

Active moment about 10597 mm

Moment level 1; M_{a11} = 0.5 ´ p’_{a11} ´ h_{a1} ´ ((H – d_{L2}) + 2/3 ´ h_{a1}) = **0.0** kNm/m

Moment level 1; M_{a12} = 0.5 ´ p’_{a21} ´ h_{a1} ´ ((H – d_{L2}) + 1/3 ´ h_{a1}) = **719.9** kNm/m

Moment level 2; M_{a21} = 0.5 ´ p’_{a21} ´ h_{a2} ´ ((H – d_{L3}) + 2/3 ´ h_{a2}) = **149.6** kNm/m

Moment level 2; M_{a22} = 0.5 ´ p’_{a31} ´ h_{a2} ´ ((H – d_{L3}) + 1/3 ´ h_{a2}) = **113.2** kNm/m

Passive moment about 10597 mm

Moment level 2; M_{p21} = 0.5 ´ p’_{p21} ´ h_{p2} ´ ((H – d_{L3}) + 2/3 ´ h_{p2}) = **0.0** kNm/m

Moment level 2; M_{p22} = 0.5 ´ p’_{p31} ´ h_{p2} ´ ((H – d_{L3}) + 1/3 ´ h_{p2}) = **982.6** kNm/m

Total moments about 10597 mm

Total active moment; SM_{a} = **982.8** kNm/m

Total passive moment; SM_{p} = **982.8** kNm/m

Required pile length

Length of pile required to balance moments; H = **10597** mm

Depth of equal pressure; d_{contra} = **7285** mm

Add 20% below this point; d_{e_add} = 1.2 ´ (H – d_{contra}) = **3975** mm

Minimum required pile length; H_{total} = d_{contra} + d_{e_add} = **11259** mm

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

Pile capacity (EN1993-5)

Maximum moment in pile (from analysis); M_{pile }= max(abs(M_{min}), abs(M_{max})) / 1m = **440.1** kNm/m

Maximum shear force in pile (from analysis); V_{pile }= **541.3** kN/m

Nominal yield strength of pile; f_{y_pile} = **270** N/mm^{2}

Name of pile; Arcelor AU25

Classification of the pile; 2

Plastic modulus of pile; W_{pl.y} = **2866** cm^{3}/m

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

Width of section; c = h / sin(a_{pile}) = **551** mm

Thickness of web; t_{w} = s = **10.2** mm

e = Ö(235 N/mm^{2} / f_{y_pile}) = **0.933**

c / t_{w} = 54.1 = 57.9 ´ e <= 72 ´ e

**PASS** – Shear buckling of web within limits

**Bending**

Interlock reduction factor (cl.5.2.2); b_{B} = **0.75**

Design bending resistance (eqn.5.2); M_{c,Rd} = W_{pl.y} ´ f_{y_pile} ´ b_{B} / g_{M0} = **580.4** kNm/m

PASS – Moment capacity exceeds moment in pile

**Shear**

Projected shear area of web (eqn.5.6); A_{v} = s ´ (h – t) = **4442** mm^{2}

Design shear resistance (eqn.5.5); V_{pl,Rd} = A_{v} ´ f_{y_pile} / (Ö(3) ´ g_{M0}) / b = **923.3** kN/m

**PASS** – Shear capacity exceeds shear in a pile

**Combined bending and shear**

Shear presence mnt reduction factor (eqn.5.10); r = (2 ´ V_{pile} / V_{pl,Rd} – 1)^{2} = **0.030**

Reduced bending resistace (eqn.5.9); M_{V,Rd} = min((b_{B} ´ W_{pl.y} – r ´ A_{v}^{2} / (4 ´ s ´ b ´ sin(a_{pile}))) ´ (f_{y_pile} / g_{M0}), M_{c,Rd}) = **574.0** kNm/m

**PASS** – Reduced moment capacity exceeds moment in pile

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