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16-Civ-B3 Geotechnical Design · December 2017

Question 8 of 9: Allowable bearing pressure on a strip footing from a 25 mm settlement limit

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

Notes on this paper

Paper format. National Examinations, December 2017 — 16-Civ-B3 Geotechnical Design; three hours, open book, any non-communicating calculator. Section A holds five discussion questions worth 7 marks each of which four are marked; Section B holds four design questions worth 24 marks each of which three are marked, so 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 marked script.

Reference texts. B. M. Das, Principles of Foundation Engineering, 8th–9th ed. (Cengage); B. M. Das, Principles of Geotechnical Engineering, 9th ed.; R. F. Craig / J. Knappett, Craig's Soil Mechanics, 9th ed.; Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed.; D. P. Coduto, Foundation Design: Principles and Practices; J. E. Bowles, Foundation Analysis and Design; ASTM D1586 (SPT), D5778 (CPTu), D2573 (field vane).

Source of design charts and assumed values (page 1, Note 6). Rankine active and passive coefficients from Das, Principles of Foundation Engineering, Ch. 7 (Eqs. 7.11 and 7.30); cantilever-wall stability procedure from Das Ch. 8 (Eqs. 8.11–8.14). Drilled-shaft adhesion factor alpha* = 0.55 and the 1.5 m surface exclusion from Reese & O'Neill (1989), tabulated in Das Ch. 12; bearing factor Nc* = 9 from Skempton (1951); block-failure check from Das Ch. 11 (Eq. 11.55). Compression index from Skempton's Cc = 0.009(LL − 10). Strain-influence factors and the C1, C2 corrections from Schmertmann, Hartman & Brown (1978) as presented in Das Ch. 5; Es = 500(N60 + 15) kPa from Das Table 5.7 (Bowles). Friction angle from the SPT via Wolff (1989), phi' = 27.1 + 0.3(N1)60 − 0.00054[(N1)60]2, with the Liao & Whitman (1986) overburden correction; Vesic bearing-capacity, shape and depth factors from Das Tables 4.2 and 4.3. Every assumed value (unit weights of concrete, specific gravity for the void ratio, pile spacing, factors of safety) is stated in a callout beside the step that uses it.

Question 8: Allowable bearing pressure on a strip footing from a 25 mm settlement limit (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. A continuous (strip) footing of width B = 2.5 m founded at Df = 1.5 m in sand with gamma = 20 kN/m3, c' = 0 and phi' = 38°. The modulus profile of Figure 2, measured downwards from the level of the foundation, is Es = 6000 kPa from 0 to 2 m, 12 000 kPa from 2 to 8 m and 10 000 kPa from 8 to 14 m. The peak strain influence factor is given as Izp = 0.5 and the creep time is t = 10 years. The permissible settlement is 25 mm.

Find. The maximum gross pressure q̄ that may be applied at founding level.

[Figure not reproduced: Figure 2 (redrawn). Strip footing, the plane-strain strain-influence diagram (I z = 0.2 at the base, peak 0.5 at z = B, zero at z = 4B) and the measured E s profile. Both depth axes are measured from the underside of the footing, as the exam figure draws them. See the official exam paper.]

Approach. Use Schmertmann's strain-influence method in its plane-strain form, integrate Iz/Es over the influence depth of 4B, apply the embedment correction C1 and the creep correction C2 for ten years, and solve the resulting settlement expression for the net pressure that produces exactly 25 mm.

  1. Choose the correct influence diagram. The footing is described as continuous, so $L/B \to \infty$ and the plane-strain (strip) diagram applies: $I_z = 0.2$ at the base, rising linearly to the peak at $z = B$ and falling linearly to zero at $z = 4B$. For this footing the peak sits at $z = 2.5$ m and the influence zone extends to $$z_{max} = 4B = 4(2.5) = 10.0\ \text{m}$$ below the base. Using the square-footing diagram instead (peak at $B/2$, zero at $2B$) would shorten the influence zone by more than half and under-predict the settlement badly.
  2. Surcharge at founding level. The pressure removed by the excavation, which is also the datum from which the net pressure is measured, is $$q=\gamma D_f = 20(1.5)=30.0\ \text{kPa}$$
  3. Set up the influence-factor ordinates. With $I_{zp}=0.5$ given, $$I_z(z)=0.2+\frac{(0.5-0.2)z}{B}\ \ \text{for } 0\le z\le B, \qquad I_z(z)=0.5\,\frac{4B-z}{3B}\ \ \text{for } B\le z\le 4B$$ so at the Es layer boundaries $I_z(0)=0.200$, $I_z(2.0)=0.440$, $I_z(2.5)=0.500$, $I_z(8.0)=0.1333$ and $I_z(10.0)=0$.
  4. Integrate Iz/Es layer by layer. Each sublayer is bounded by a change of modulus or by the kink in the diagram at $z = B$, and within a sublayer $I_z$ is linear so the mean ordinate may be used.
Sublayer z (m)Δz (m)Es (kPa)Iz at top / bottommean Iz(Iz/Es)Δz (m3/kN)
0 – 2.02.06 0000.2000 / 0.44000.32001.0667 × 10−4
2.0 – 2.50.512 0000.4400 / 0.50000.47000.1958 × 10−4
2.5 – 8.05.512 0000.5000 / 0.13330.31671.4514 × 10−4
8.0 – 10.02.010 0000.1333 / 0.00000.06670.1333 × 10−4
Σ(Iz/Es)Δz2.8472 × 10−4
  1. Creep correction. Schmertmann's time factor for ten years is $$C_2 = 1+0.2\log_{10}\!\left(\frac{t}{0.1}\right)=1+0.2\log_{10}(100)=1+0.2(2)=1.4$$ so forty per cent of the immediate settlement is added over the design life.
  2. Assemble the settlement expression. Writing the net pressure as $\Delta q = \bar{q}-q$, the embedment correction is $C_1 = 1-0.5\,q/\Delta q$ and $$S_e = C_1C_2\,\Delta q\sum\frac{I_z}{E_s}\Delta z = \left(1-\frac{0.5(30)}{\Delta q}\right)(1.4)\,\Delta q\,(2.8472\times10^{-4})$$ The bracket multiplies out neatly, because $C_1\Delta q = \Delta q - 15$: $$S_e = 1.4(2.8472\times10^{-4})(\Delta q - 15)=3.9861\times10^{-4}(\Delta q-15)$$
  3. Solve for the limiting net pressure. Setting $S_e = 0.025$ m, $$\Delta q - 15 = \frac{0.025}{3.9861\times10^{-4}}=62.72 \Rightarrow \Delta q = 77.7\ \text{kPa}$$ and the embedment correction that this implies, $C_1 = 1-15/77.7=0.807$, is above the lower limit of 0.5, so the expression used is valid.
  4. Convert to the pressure applied at foundation level. The question asks for the stress at the level of the foundation, which is the gross value: $$\bar{q}=\Delta q + q = 77.7+30.0=\boxed{107.7\ \text{kPa}\ \ (\text{say }105\ \text{kPa})}$$ The corresponding line load on the strip is $107.7 \times 2.5 = 269$ kN per metre run, of which $75$ kN/m is merely replacing the excavated soil.
  5. Confirm that settlement, not strength, is the governing criterion. For phi' = 38° the Reissner and Vesic factors are $N_q = e^{\pi\tan 38^{\circ}}\tan^{2}(45^{\circ}+19^{\circ})=48.93$ and $N_\gamma = 2(N_q+1)\tan 38^{\circ}=78.02$, and with the depth factor $F_{qd}=1+2\tan\phi'(1-\sin\phi')^{2}(D_f/B)=1.138$ (shape factors are unity for a strip), $$q_u = qN_qF_{qd}+\tfrac{1}{2}\gamma BN_\gamma = 30(48.93)(1.138)+\tfrac{1}{2}(20)(2.5)(78.02)=1671+1951=3622\ \text{kPa}$$ At a factor of safety of 3 this permits about 1207 kPa, more than eleven times the settlement-controlled value, so the 25 mm limit governs by a wide margin — the usual outcome for a footing on medium-dense sand.
QuantityValue
Influence depth 4B / peak depth B10.0 m / 2.5 m below the base
Σ(Iz/Es)Δz2.847 × 10−4 m3/kN
Surcharge q = gamma Df30.0 kPa
C2 (10 years) / C1 at the answer1.4 / 0.807
Net pressure for 25 mm, Δq77.7 kPa
Maximum stress at foundation level, q̄107.7 kPa, say 105 kPa
Corresponding line load269 kN per metre run
Ultimate bearing capacity / allowable at FS = 33622 kPa / 1207 kPa — not critical
Check — the two readings of the figure that decide this answer. (1) The Es axis in Figure 2 is drawn at the level of the underside of the footing and the depth axis descends from that point, so the moduli are indexed from the base, not from the ground surface. Had the profile been read from ground level, only the first 0.5 m of the influence zone would sit in the 6000 kPa layer instead of 2.0 m, the summation would fall to 2.456 × 10−4 and the answer would rise to about 118 kPa. (2) "Continuous foundation" fixes the plane-strain influence diagram; the axisymmetric diagram would give Σ(Iz/Es)Δz = 1.688 × 10−4 and an answer near 151 kPa. Both readings are stated on the drawing, so neither is a guess, but both are worth naming because each moves the result by tens of per cent. If the 25 mm were required as an immediate settlement with creep accepted on top, C2 would be 1.0 and the permissible pressure would rise to 133 kPa; the ten-year figure quoted in the question is the conservative and correct interpretation.