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16-Civ-B3 Geotechnical Design · May 2016

Question 6 of 9: Schmertmann settlement of a strip footing from a CPT profile (24 marks)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada National Examinations, May 2016 — 98-Civ-B3 Geotechnical Design. Three hours, OPEN BOOK, any non-communicating calculator. Section A carries five discussion questions of 7 marks each (answer any four); Section B carries four design questions of 24 marks each (answer any three); the examinable total is 4 × 7 + 3 × 24 = 100 marks. All nine questions are worked below, because the set is a study resource rather than a timed attempt.

Reference texts (98-Civ-B3 / 16-Civ-B3 Geotechnical Design).

Sources of design charts and assumed values (page-1 Note 6). Note 6 requires the candidate to identify the source of every design chart used and of every value assumed where the paper gives none. They are named at the point of use and collected here:

  • Strain-influence diagram and the C₁, C₂ correction factors (Q6) — Schmertmann, Hartman and Brown (1978), as tabulated in Das, Principles of Foundation Engineering, 9th ed., Section 5.6.
  • Rankine active coefficient for an inclined backfill (Q7) — Das, Principles of Geotechnical Engineering, 9th ed., Eq. (13.35).
  • Bearing-capacity factors N₢, Nᵤ, Nγ and the depth and load- inclination factors (Q7) — Vesic / Meyerhof as tabulated in Das, Principles of Foundation Engineering, 9th ed., Tables 3.3 and 3.4, applied to a retaining-wall base in Section 8.6.
  • Meyerhof bearing-capacity factor Nᵤ* for a driven pile point and the limiting point resistance (Q8) — Das, 9th ed., Section 11.9 and its interpolated Nᵤ* table; Nᵤ* = 143 at φ′ = 35°.
  • Adhesion factor α against cu/p₀ (Q8) — Das, 9th ed., Table 11.6 (after Terzaghi, Peck and Mesri); α = 0.68 at cu/p₀ = 0.5.
  • Earth-pressure coefficient K and interface friction angle δ′ for a driven high-displacement pile (Q8) — Das, 9th ed., Section 11.11: K ≈ 1.4K₀ and δ′ ≈ 0.8φ′ are assumed, and the critical-depth rule L′ = 15D is Das Eq. (11.42).
  • Unit weight of water γᵣ = 9.81 kN/m³ and g = 9.81 m/s² throughout; atmospheric pressure p₀ = 100 kPa.

Section A — discussion questions (7 marks each)

Section B — design questions (24 marks each)

Question 6 — Schmertmann settlement of a strip footing from a CPT profile (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.

Given.

QuantitySymbolValue
Footing plan dimensionsB × L2.5 m × 30 m
Founding depthDf3.0 m below ground surface
Line load carriedQ/L400 kN per metre run
Groundwater table—3.0 m depth (at founding level)
Unit weight above the water tableγ17 kN/m3
Unit weight below the water tableγsat20 kN/m3
Unit weight of waterγw9.81 kN/m3
Modulus profile (layers 2 to 7, Figure 1 table)Es 7353 / 10 049 / 16 667 / 22 059 / 14 216 / 26 471 kPa over 3–5, 5–6, 6–7, 7–8, 8–9 and 9–12 m depth

Find. The elastic (immediate) settlement of the strip footing shortly after construction, and the settlement including creep after 50 years, by Schmertmann's strain-influence-factor method.

Es = 4,902 kPa3 mEs = 7,353 kPa5 mEs = 10,049 kPa6 mEs = 16,667 kPa7 mEs = 22,059 kPa8 mEs = 14,216 kPa9 mEs = 26,471 kPa12 m2 mexisting ground surfaceGWTstrip footing, width B = 2.5 m00.20.40.6strain-influence factor Izpeak 0.6194 at z = B0.2 at the base1B2B3B4Bz below baseLeft: CPT-derived modulus profile. Right: Schmertmann strain-influence diagram for a strip footing (L/B ≥ 10).
Figure Q6. Modulus profile from the Robertson and Campanella CPT sounding of Figure 1, and the Schmertmann strain-influence diagram for a strip footing. Because L/B = 12, the plane-strain distribution applies: Iz = 0.2 at the base, a peak at one footing width below the base, and zero at four footing widths.

Approach. Establish that the footing is a strip ($L/B \ge 10$), reduce the applied pressure to a net pressure, build the plane-strain strain-influence diagram from the peak factor $I_{zp}$, integrate $I_z/E_s$ layer by layer over the four-width depth of influence, and apply the embedment factor $C_1$ and the creep factor $C_2$.

  1. Confirm the geometry and the applicable influence diagram. With $L/B = 30/2.5 = 12 \ge 10$ the footing is a strip and the plane-strain form of Schmertmann's diagram governs: $I_z = 0.2$ at the base, the peak $I_{zp}$ occurs at $z = B$ below the base, and $I_z$ falls linearly to zero at $z = 4B$. The depth of influence is therefore $4B = 10.0$ m below the base, that is from 3.0 m to 13.0 m below ground surface. (The axisymmetric form, with the peak at $z = B/2$ and zero at $z = 2B$, would apply to a square or circular footing and is not used here; the right-hand panel of Figure 1 shows the strip diagram, confirming the intent.)
  2. Reduce the line load to a contact pressure. The load is quoted per metre run, so the gross contact pressure is $$q = \frac{Q/L}{B} = \frac{400}{2.5} = 160.0\ \text{kPa}.$$
  3. Compute the effective overburden removed at founding level. The water table is at 3.0 m, exactly at the base, so the whole 3.0 m of soil excavated lies above the water table and acts at its bulk unit weight: $$\sigma'_{vD} = \gamma D_f = 17 \times 3.0 = 51.0\ \text{kPa}.$$ Schmertmann's method uses the net pressure, so $$\Delta q = q - \sigma'_{vD} = 160.0 - 51.0 = 109.0\ \text{kPa}.$$
  4. Find the effective stress at the depth of the peak influence factor. For a strip footing the peak sits at $z = B = 2.5$ m below the base, that is at 5.5 m below ground surface, which is below the water table, so the submerged unit weight $\gamma' = 20 - 9.81 = 10.19$ kN/m3 applies over that 2.5 m: $$\sigma'_{zp} = 51.0 + 10.19 \times 2.5 = 76.475\ \text{kPa}.$$
  5. Evaluate the peak strain-influence factor. Schmertmann's expression relates the peak to the ratio of net pressure to the effective stress at the peak depth: $$I_{zp} = 0.5 + 0.1\sqrt{\frac{\Delta q}{\sigma'_{zp}}} = 0.5 + 0.1\sqrt{\frac{109.0}{76.475}} = 0.5 + 0.1\left(1.1939\right) = 0.6194 .$$ The diagram is now fully defined: $I_z$ rises linearly from 0.2 at $z = 0$ to $\boxed{I_{zp} = 0.6194}$ at $z = 2.5$ m, then falls linearly to zero at $z = 10.0$ m.
  6. Divide the influence depth into sub-layers and sum $I_z\,\Delta z/E_s$. The modulus profile changes at 5, 6, 7, 8, 9 and 12 m depth and the influence diagram has its kink at 5.5 m depth, so those are the sub-layer boundaries. $I_z$ is taken at the mid-height of each sub-layer, which is exact for a linear diagram. The tabulated profile stops at 12 m while the influence zone runs to 13 m, so the modulus of layer 7 is carried down to 13 m; that sub-layer contributes only 0.6 per cent of the total, so the assumption is immaterial.
    z below base (m)Depth below ground (m)Δz (m) Es (kPa)Iz at mid-height IzΔz/Es (m3/kN)
    0 – 2.03.0 – 5.02.007 353 0.36781.0003 × 10−4
    2.0 – 2.55.0 – 5.50.5010 049 0.57742.873 × 10−5
    2.5 – 3.05.5 – 6.00.5010 049 0.59872.979 × 10−5
    3.0 – 4.06.0 – 7.01.0016 667 0.53683.221 × 10−5
    4.0 – 5.07.0 – 8.01.0022 059 0.45422.059 × 10−5
    5.0 – 6.08.0 – 9.01.0014 216 0.37162.614 × 10−5
    6.0 – 9.09.0 – 12.03.0026 471 0.20652.340 × 10−5
    9.0 – 10.012.0 – 13.01.0026 471 0.04131.560 × 10−6
    Sum 2.6245 × 10−4
  7. Apply the embedment correction $C_1$. Embedment relieves part of the strain because the soil at the base has already carried the overburden that was removed: $$C_1 = 1 - 0.5\left(\frac{\sigma'_{vD}}{\Delta q}\right) = 1 - 0.5\left(\frac{51.0}{109.0}\right) = 0.7661 .$$ The lower bound of 0.5 is not approached, so the value stands.
  8. Apply the creep correction $C_2$ for each required time. Schmertmann's creep factor is referred to a nominal "immediate" time of 0.1 year: $$C_2 = 1 + 0.2\log_{10}\!\left(\frac{t}{0.1}\right).$$ Soon after construction, $t = 0.1$ year and $C_2 = 1.000$. After 50 years, $C_2 = 1 + 0.2\log_{10}(500) = 1 + 0.2(2.6990) = 1.5398 .$
  9. Assemble the settlements. $$S_e = C_1 C_2\,\Delta q \sum \frac{I_z}{E_s}\Delta z .$$ Soon after construction, $$S_e = 0.7661 \times 1.000 \times 109.0 \times 2.6245\times10^{-4} = 0.02191\ \text{m},$$ that is $\boxed{S_e \approx 21.9\ \text{mm}}$. After 50 years the same product with $C_2 = 1.5398$ gives $S_e = 0.03374$ m, that is $\boxed{S_{e,50\ \text{yr}} \approx 33.7\ \text{mm}}$, of which 11.8 mm is creep.
  10. Comment on acceptability. An immediate settlement of about 22 mm is within the 25 mm that Canadian practice normally tolerates for an isolated footing, and the long-term value of about 34 mm is acceptable for a strip footing provided the differential movement along its 30 m length is controlled; since the modulus profile is reasonably uniform along the sounding, the strip should settle fairly evenly. The design is settlement-governed, exactly as argued in Question 1.
QuantityValue
Gross contact pressure, q160.0 kPa
Effective overburden at founding level, σ′vD51.0 kPa
Net applied pressure, Δq109.0 kPa
Effective stress at the peak depth (5.5 m), σ′zp76.48 kPa
Peak strain-influence factor, Izp0.6194 at 2.5 m below the base
Depth of influence4B = 10.0 m below the base (to 13.0 m depth)
Σ IzΔz/Es2.6245 × 10−4 m3/kN
Embedment factor, C10.7661
Creep factor, C2 (0.1 yr / 50 yr)1.000 / 1.5398
Settlement soon after construction21.9 mm
Settlement after 50 years33.7 mm

Check: assumptions and sensitivities recorded under page-1 Notes 1 and 6.

  • Modulus below 12 m. The tabulated profile ends at 12 m but the influence zone reaches 13 m. Layer 7 has been extended. That sub-layer contributes 1.56 × 10−6 of a total 2.6245 × 10−4, so even halving the assumed modulus changes the settlement by only about 0.1 mm immediately and 0.2 mm at 50 years.
  • The qc column. The table heads the cone resistance "kPa", but the tabulated moduli satisfy Es = 2.5 qc exactly (4902/200 = 7353/300 = … = 24.51) only if qc is read in tonnes per square metre, which is also what reconciles the values with the 0 to 160 kg/cm² axis of Figure 1 (for example layer 7 at 1080 t/m² = 108 kg/cm²). Nothing turns on this, because the moduli are given directly in kPa and have been used as tabulated; the observation is recorded because it is the internal check that the table and the chart in Figure 1 are the same data.
  • Strip versus axisymmetric influence diagram. Using the square-footing diagram by mistake (peak Izp = 0.6308 at z = B/2, zero at z = 2B) would give 15.1 mm immediately and 23.2 mm at 50 years — about 31 per cent low. The choice of diagram is the largest single judgement in this question, and L/B = 12 settles it unambiguously in favour of the plane-strain form.
  • Rigidity and shape refinements. Schmertmann's later work adds a shape factor C3 for very long footings; it is not part of the formulation in the reference text and has not been applied. Applying C3 = 0.73 would reduce both answers by 27 per cent, so a designer quoting the result should state which formulation was used.
  • Nature of the answer. Schmertmann's method returns the settlement of a footing on a granular deposit; it is not a consolidation calculation, and the "50 year" value is creep of the sand skeleton, not primary consolidation.