Question 9 of 9: Schmertmann settlement of a continuous footing; Rankine pressure in a φ u = 0 clay

Question 9: Schmertmann settlement of a continuous footing; Rankine pressure in a φ_u = 0 clay (24 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.

Part (a) — Schmertmann settlement of the continuous footing

Given.

Find. The expected maximum elastic settlement, which is the settlement computed on the softest of the three modulus profiles.

Approach. The clue asks for engineering judgement, and the judgement is simply this: settlement is inversely proportional to modulus, so the maximum settlement comes from Data Set 2, whose moduli are the lowest at every depth. Because L/B exceeds 15 the plane-strain strain-influence diagram must be used, with I_z = 0.2 at the base, a peak at one full width below the base, and zero at four widths.

1. Select the data set, and say why. Settlement varies inversely with E_s, so the maximum settlement is produced by the profile with the smallest moduli at every depth. Data Set 2 (3000 / 6000 / 12000 / 8000 kN/m²) is uniformly the softest and is therefore the only set that needs to be evaluated for the answer the question asks for. This is the engineering judgement the clue points at.
2. Establish the net applied pressure and the depth correction. The overburden removed by the excavation is q̅ = γD_f = 20(2.0) = 40 kPa, so the net pressure increase felt by the soil beneath the base is $$\Delta q=100-40=60\ \text{kPa}$$ and the embedment correction is $$C_1=1-0.5\left(\frac{\bar q}{\Delta q}\right)=1-0.5\left(\frac{40}{60}\right)=0.667\ \ (\ge 0.5\ \checkmark)$$
3. Locate and evaluate the peak strain influence factor. For a continuous (plane-strain) footing the peak occurs one full width below the base, at z = B = 2.5 m, that is at a depth of 4.5 m, where the effective overburden is σ′_zp = 20(4.5) = 90 kPa. Hence $$I_{zp}=0.5+0.1\sqrt{\frac{\Delta q}{\sigma'_{zp}}}=0.5+0.1\sqrt{\frac{60}{90}}=0.5+0.1(0.8165)=\boxed{0.582}$$ The diagram therefore runs linearly from I_z = 0.2 at the base to 0.582 at z = 2.5 m, and back linearly to zero at z = 4B = 10 m, i.e. at a depth of 12 m.
4. Evaluate the creep factor. With a service life of 12 years, $$C_2=1+0.2\log_{10}\!\left(\frac{t}{0.1}\right)=1+0.2\log_{10}(120)=1+0.2(2.0792)=1.416$$
5. Sum I_zΔz/E_s over the influence zone. The influence zone runs from the base at 2 m depth to 12 m depth, so it straddles exactly two of the tabulated bands. Over the first, 2–8 m depth (z = 0 to 6 m), the diagram covers the whole rising limb and part of the falling limb: the rising portion contributes ½(0.200 + 0.582)(2.5) = 0.977 and the falling portion, where I_z drops from 0.582 to 0.310, contributes ½(0.582 + 0.310)(3.5) = 1.561, a total of 2.538 m. Over the second, 8–12 m depth (z = 6 to 10 m), the remaining triangle contributes ½(0.310)(4.0) = 0.620 m. Therefore $$\sum\frac{I_z}{E_s}\Delta z=\frac{2.538}{6000}+\frac{0.620}{12000}=4.230\times 10^{-4}+0.517\times 10^{-4}=4.747\times 10^{-4}\ \text{m}^{3}/\text{kN}$$
6. Compute the settlement. $$S_e=C_1C_2\,\Delta q\sum\frac{I_z}{E_s}\Delta z=0.667(1.416)(60)\left(4.747\times 10^{-4}\right)=0.0269\ \text{m}$$ $$\boxed{S_e\approx 26.9\ \text{mm}\ \text{(Data Set 2, the maximum)}}$$
7. Report the spread, because the spread is the finding. Repeating the identical calculation on the other two profiles — only the two divisors change — gives 16.7 mm for Data Set 1 and 13.7 mm for Data Set 3. The three answers differ by a factor of two on a deposit the question itself calls "erratic", and that spread is the engineering message: a strip footing on this ground could settle anywhere between about 14 and 27 mm depending on which part of the site it lands on, which is precisely the condition that produces damaging differential settlement between adjacent footings. The design should be based on the 26.9 mm figure, and the erratic modulus profile is a strong argument for a raft or for a continuous footing running across the variability rather than isolated pads sitting in it.

Part (b) — Rankine active pressure and tensile crack in a saturated clay backfill

Given. Wall height H = 5 m; saturated clay backfill with φ_u = 0, c_u = 50 kPa and γ = 20 kN/m³; horizontal ground surface; no surcharge.

Find. (i) the Rankine active pressure distribution behind the wall, and (ii) the depth of the tensile crack.

Approach. With φ_u = 0 the Rankine active coefficient is unity, so the active pressure is simply the total vertical stress reduced by twice the undrained strength; the tensile crack extends to the depth at which that expression changes sign.

1. Write the Rankine active pressure. For a horizontal surface and a smooth vertical wall, $$\sigma_a=\gamma z\,K_a-2c_u\sqrt{K_a},\qquad K_a=\tan^{2}\!\left(45^{\circ}-\frac{\phi_u}{2}\right)=\tan^{2}45^{\circ}=1$$ so the expression collapses to $$\sigma_a=\gamma z-2c_u=20z-100\ \ \text{(kPa, with } z \text{ in metres)}$$
2. Evaluate the ordinates. At the ground surface and at the base of the wall, $$\sigma_a(0)=-2c_u=\boxed{-100\ \text{kPa (tension)}},\qquad \sigma_a(5.0)=20(5.0)-100=\boxed{0\ \text{kPa}}$$ The distribution is a straight line running from 100 kPa of tension at the surface to exactly zero at the base of the 5 m wall. Soil cannot sustain tension against a wall, so no active pressure at all is transmitted over the full height in the short term.
3. Determine the depth of the tensile crack. The crack extends to the depth at which the computed active pressure passes through zero: $$z_c=\frac{2c_u}{\gamma\sqrt{K_a}}=\frac{2(50)}{20}=\boxed{5.0\ \text{m}}$$ which for this particular combination of numbers equals the full height of the wall.
4. Interpret the result. Two consequences follow and both are worth stating. First, the net short-term active thrust on the wall, taking the tensile block as ineffective, is $$P_a=\tfrac{1}{2}\left(\gamma H-2c_u\right)\left(H-z_c\right)=\tfrac{1}{2}(0)(0)=0\ \text{kN/m}$$ Undrained, this clay exerts no thrust on the wall at all; indeed the critical height of an unsupported vertical cut in it is H_c = 4c_u/γ = 10.0 m, twice the wall height, so the face would in principle stand unsupported. Second — and this is why no wall is ever designed on that basis — the result is temporary and dangerous. If the crack fills with surface water it applies a hydrostatic thrust of $$P_w=\tfrac{1}{2}\gamma_w z_c^{2}=\tfrac{1}{2}(9.81)(5.0)^{2}=122.6\ \text{kN/m}$$ acting 1.67 m above the base, on a wall that the undrained analysis says needs to resist nothing. Over the longer term the clay softens, the undrained strength is no longer the operative parameter, and the wall must be designed on drained parameters c′ ≈ 0 and φ′, which for a stiff clay near 25° would give K_a = 0.406 and a thrust of about 101 kN/m. The design case is therefore the greater of the water-filled crack and the long-term drained pressure, and the crack should be sealed at the surface, drained, and prevented from filling.