Retaining Structures (30 marks): 07-Str-B5, May 2015, Q4

Question 4 — Retaining Structures (30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Friction angle (above and below water)	φ	36°
Bulk unit weight above the water table	γ	17 kN/m³
Effective unit weight below the water table	γ′	10.2 kN/m³
Surface surcharge	q	10 kPa
Depth to water table below the top (printed H)	H	6.40 m
Submerged soil above the dredge line	—	2.40 m
Retained height (top to dredge line)	H_t	6.4 + 2.4 = 8.80 m
Tie-rod level below the top	—	1.50 m
Factor of safety on passive resistance	FS	1.4
Tie-rod spacing	s	2.0 m

Reading the figure. The printed H = 6.4 m dimension runs from the top of the backfill to the water line, and the separate 2.4 m dimension runs from the water line to the dredge line, so the groundwater table is 6.4 m down and the wall retains 6.4 + 2.4 = 8.8 m of soil.

Find. (a) the Rankine active pressure diagram, separating the soil and surcharge contributions; (b) the embedment depth d below the dredge line for a factor of safety of 1.4; (c) the force carried by each tie rod.

Approach. Use the free-earth-support method: the wall is assumed rigid and simply supported at the tie rod, with the toe free to rotate. Apply the factor of safety by reducing the passive coefficient, take moments of the whole pressure diagram about the tie to find d, and then obtain the tie force from horizontal equilibrium.

(a) Compute the Rankine coefficients. For a smooth vertical wall with a horizontal surface and φ = 36°, $$K_a = \frac{1 - \sin 36^{\circ}}{1 + \sin 36^{\circ}} = 0.2596, \qquad K_p = \frac{1}{K_a} = 3.852$$ The same coefficients apply above and below the water table because the friction angle is unchanged; only the unit weight changes.
Build the pressure due to the soil. Active pressure is proportional to the vertical effective stress, so the diagram breaks at the water table: $$\sigma_a(0) = 0, \qquad \sigma_a(6.4\ \text{m}) = \gamma z K_a = 17(6.4)(0.2596) = 28.25\ \text{kPa}$$ $$\sigma_a(8.8\ \text{m}) = 28.25 + \gamma^{\prime}(2.4)K_a = 28.25 + 10.2(2.4)(0.2596) = 28.25 + 6.36 = 34.60\ \text{kPa}$$ The kink at 6.4 m is the signature of the water table: the gradient falls from 4.41 kPa/m to 2.65 kPa/m because buoyancy removes 6.8 kN/m³ from the effective unit weight.
Add the pressure due to the surcharge. A uniform surcharge produces a uniform increment of vertical stress at every depth, so its lateral effect is a rectangle of constant intensity from the surface to the toe: $$\sigma_{a,q} = qK_a = 10 \times 0.2596 = 2.60\ \text{kPa (constant with depth)}$$ Combining, the total active pressure at the dredge line is $$\boxed{\sigma_a(\text{dredge}) = 34.60 + 2.60 = 37.20\ \text{kPa}}$$
Account for the water. The figure shows free water on the dredged side standing at the same elevation as the groundwater table behind the wall, so the hydrostatic pressures on the two faces are equal and opposite and cancel. No net water-pressure block is included, and the whole analysis is carried out in effective stresses below 6.4 m.

Figure 4.1 — Net lateral pressure on the wall. Above the dredge line the diagram is entirely active (surcharge rectangle plus the two-slope soil diagram); below it the factored passive resistance overtakes the active pressure 1.46 m down.

(b) Resolve the active diagram above the dredge line into components. Taking moments about the tie rod at 1.5 m depth (positive arms below the tie):

Question 4(b) — active thrust components above the dredge line, per metre run
Component	Force (kN/m)	Depth of centroid (m)	Arm about tie (m)	Moment (kN·m/m)
Surcharge rectangle, 2.60 kPa over 8.80 m	22.85	4.40	2.90	66.25
Soil triangle, 0 → 28.25 kPa over 0–6.4 m	90.39	4.27	2.77	250.07
Soil rectangle, 28.25 kPa over 6.4–8.8 m	67.79	7.60	6.10	413.52
Soil triangle, 0 → 6.36 kPa over 6.4–8.8 m	7.63	8.00	6.50	49.57
Totals	188.65	—	—	779.42

Write the net pressure below the dredge line. The factor of safety is applied by mobilising only $K_p/FS$ of the passive resistance. With x measured below the dredge line, $$\sigma_p(x) = \gamma^{\prime}\frac{K_p}{FS}x = 10.2\left(\frac{3.852}{1.4}\right)x = 28.06x, \qquad \sigma_a(x) = 37.20 + \gamma^{\prime}K_a x = 37.20 + 2.65x$$ $$p_{net}(x) = 25.41x - 37.20 \ \text{kPa}, \qquad p_{net} = 0 \ \text{at} \ x_0 = 1.46\ \text{m}$$
Take moments about the tie rod. The free-earth-support condition is that the net moment about the anchor vanishes. With the lever arm to a point x below the dredge line equal to $(8.8 - 1.5 + x) = (7.3 + x)$, $$\int_{0}^{d}\big(25.41x - 37.20\big)(7.3 + x)\,dx = 779.42$$ $$\Rightarrow\; 8.472\,d^{3} + 74.17\,d^{2} - 271.54\,d - 779.42 = 0$$
Solve for the embedment. The positive root is $$d = 4.180\ \text{m} \;\Rightarrow\; \boxed{d = 4.2\ \text{m (adopt)}}$$ giving a total sheet-pile length of $8.8 + 4.2 = 13.0$ m before any allowance for driving tolerance. Many designers would add a further 20 % to the theoretical embedment when the factor of safety is applied to K_p rather than to the depth; that would give d = 5.0 m and a 13.8 m pile, which is the length that should be ordered if over-dredging is credible.
(c) Obtain the tie force from horizontal equilibrium. The net resultant below the dredge line is $$R_{net} = \int_{0}^{d}\!\big(25.41x - 37.20\big)dx = \tfrac{25.41}{2}(4.180)^{2} - 37.20(4.180) = 222.02 - 155.48 = 66.54\ \text{kN/m}$$ so that, per metre run of wall, $$T = \sum P_a - R_{net} = 188.65 - 66.54 = 122.12\ \text{kN/m}$$
Convert to the force in each tie. With ties at 2.0 m centres, $$\boxed{T_{tie} = 122.12 \times 2.0 = 244\ \text{kN per tie}}$$ Allowing a factor of safety of about 1.5 on the tie steel and corrosion loss on a marine wall, this calls for roughly a 45 mm diameter Grade 350W round bar ($A = 1590$ mm², yield load 557 kN) or an equivalent proprietary tie, with a waling designed for the same load and an anchor block placed outside the active wedge behind the wall.

Question 4 — final results
Item	Result
Rankine coefficients	K_a = 0.2596, K_p = 3.852, K_p/FS = 2.751
(a) Active pressure due to soil	0 at surface; 28.25 kPa at 6.4 m (water table); 34.60 kPa at 8.8 m (dredge line)
(a) Active pressure due to surcharge	2.60 kPa, uniform over the full depth
(a) Total at dredge line	37.20 kPa
Active thrust above dredge line	188.65 kN/m at 779.42 kN·m/m about the tie
Depth to zero net pressure	1.46 m below the dredge line
(b) Required embedment	d = 4.18 m → adopt 4.2 m (total pile length 13.0 m)
(c) Tie force	122.1 kN/m of wall → 244 kN per tie at 2.0 m centres

Topic: Free-earth-support design of an anchored sheet-pile wall in sand with a surcharge and a groundwater table.

Key relations: $K_a = (1-\sin\phi)/(1+\sin\phi)$, $K_p = 1/K_a$; active pressure from vertical effective stress $\sigma_a = K_a\sigma^{\prime}_v + K_a q$; net pressure below dredge $p_{net} = \gamma^{\prime}(K_p/FS - K_a)x - \sigma_{a,dredge}$; moment equilibrium about the anchor gives d; horizontal equilibrium gives T.

Why this works: A relatively short, flexible sheet pile embedded in sand rotates about the anchor rather than bending into an S-shape, so the toe is effectively free and the wall behaves as a propped cantilever with a single support at the tie. That single kinematic assumption makes the problem statically determinate: two equilibrium equations for two unknowns, d and T. Taking moments about the anchor first is what decouples them, because T has zero lever arm there. Applying the factor of safety to K_p rather than to the computed depth is the conservative choice because the passive resistance grows with the square of depth, so a small deficit in embedment costs a disproportionate amount of resistance.

Common pitfall: Applying the full unit weight below the water table, or forgetting that the surcharge contributes a rectangle that continues to the toe on the active side. The first adds about 6.1 kPa (16 %) to the active pressure at the dredge line; the second removes roughly 23 kN/m of thrust and 66 kN·m/m of moment, under-estimating both the embedment and the tie force. A third trap — the one this figure sets — is reading the 6.4 m dimension as the full retained height: it stops at the water line, and the 2.4 m below it must be added. A fourth is including a net hydrostatic block when the water levels are equal on the two faces — here they are, so it cancels; if the wall were dewatered on the dredged side, an unbalanced water pressure would dominate the whole design.

Source: Das, Principles of Foundation Engineering, 9th ed., Ch. 9 (sheet-pile walls — anchored walls, free-earth-support method, Rowe's moment reduction); Canadian Foundation Engineering Manual, 4th ed., Ch. 26 (earth pressures and anchored walls).

Question 4 of 6: Retaining Structures (30 marks)

Question 4 — Retaining Structures (30 marks)