Question 9 of 9: Schmertmann Settlement of a Continuous Footing on an Erratic Sand
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. EGBC / Engineers
Canada National Examination 16-Civ-B3 Geotechnical Design,
May 2018. Three hours, open book, any non-communicating
calculator. Section A — five discussion questions of 7
marks each, answer any four. Section B — four design
questions of 24 marks each, answer any three. Examinable total
$4\times 7 + 3\times 24 = 100$ marks.
Page 1 Note 6 requires the candidate to identify clearly the source
of every design chart and assumed value used, so the provenance of each
correlation is named where it is used, not only in the concept notes.
All nine questions are solved below, because the set is a study
resource rather than a three-hour sitting.
Reference texts for this subject. B. M. Das,
Principles of Foundation Engineering, 9th ed. (Cengage) — Ch. 3
(subsurface exploration and SPT corrections), Ch. 4 (bearing capacity), Ch. 5
(settlement, Schmertmann), Ch. 8 (retaining walls), Ch. 11–12 (pile
foundations and drilled shafts); B. M. Das, Principles of Geotechnical
Engineering, 9th ed. — Ch. 8 (shear strength), Ch. 15 (slope
stability); R. D. Holtz, W. D. Kovacs & T. C. Sheahan, An Introduction to
Geotechnical Engineering, 2nd ed.; Canadian Geotechnical Society,
Canadian Foundation Engineering Manual (CFEM), 4th ed. — the
Canadian design authority for the factors of safety and serviceability limits
quoted here; L. C. Reese & M. W. O'Neill, Drilled Shafts: Construction
Procedures and Design Methods (FHWA-HI-88-042).
Check — figure readings. Two dimensions are read from the drawings, as follows. (1) In Figure 2 the
“1 m” dimension is the height of the bell: its arrows point
inward at the flare, and scaling against the 4 m dimension on the same figure
puts the bell base exactly on the 12 m line. The pile is therefore
$L = 12$ m long with the bell top at 11 m, not
11 m long with its base floating 1 m clear of the layer base. (2) In
Figure 3 the “0.35 m” label carries extension lines
from the top and bottom corners of the base slab, so it is the base
thickness; the “0.5 m” at the left is measured to the base
underside, so only 0.15 m of soil covers the toe. The toe projection is
not dimensioned and follows from the printed values as
$4.0 - 0.3 - 2.0 = 1.7$ m. Figure 3 is not
drawn to scale — its toe is drawn about half its dimensioned length —
so the printed numbers govern, as page 1 Note 1 anticipates.
Question 9: Schmertmann Settlement of a Continuous Footing on an Erratic Sand (24 marks)
Find. The expected maximum elastic settlement
of the strip footing by Schmertmann's strain-influence-factor method.
Approach. Two judgements decide this problem before any
arithmetic. First, $L/B > 15$ makes the footing a plane-strain (strip) case, so
the strip influence diagram applies: $I_z = 0.2$ at the base, peaking at
$z = B$ and vanishing at $z = 4B$. Second, the question asks for the
maximum settlement and its clue invites us to choose one data set: the
maximum settlement comes from the softest profile, which is Data Set 2
throughout. The calculation is then $S_e = C_1C_2\Delta q\sum(I_z/E_s)\Delta z$
over the influence zone.
Figure 9-1. Plane-strain strain-influence diagram for the 2.5 m wide strip footing, superimposed on the Data Set 2 modulus profile. The influence zone runs from the base (2 m below ground) to 4B = 10 m below the base, i.e. to 12 m depth; the deposit's top layer lies entirely above the base and plays no part.
Select the data set, and say why. Data Set 2 is the softest
at every depth (3000, 6000, 12000, 8000 kPa against 5000, 10000, 15000, 12000 for
Set 1 and 8000, 12000, 20000, 15000 for Set 3). Since settlement is inversely
proportional to modulus and the question asks for the maximum expected
value in an admittedly erratic deposit, Data Set 2 is the correct choice; the
other two are computed at the end for comparison. This is the engineering
judgement the clue is asking for.
Separate the gross and net pressures. The overburden
removed by the excavation is
$$\bar{q} = \gamma D_f = 20(2.0) = 40\ \text{kPa},$$
so the net pressure increase that actually causes settlement is
$$\Delta q = q - \bar{q} = 150 - 40 = 110\ \text{kPa}.$$
Carrying the gross 150 kPa through as the pressure increase instead would give
94 mm rather than 63 mm — 48 per cent too high, because both
$\Delta q$ and $I_{zp}$ grow with it.
Choose the plane-strain influence diagram and locate its
peak. For $L/B \ge 10$ Schmertmann's plane-strain diagram applies:
$$I_z = 0.2\ \text{at}\ z = 0, \qquad I_z = I_{zp}\ \text{at}\ z = B,
\qquad I_z = 0\ \text{at}\ z = 4B .$$
The peak therefore sits a full width below the base, at
$z = 2.5$ m, that is 4.5 m below ground, and the influence zone extends to
$4B = 10$ m below the base, i.e. to 12 m depth. This is the single most
consequential choice in the question: the axisymmetric diagram (peak at
$B/2$, zero at $2B$) would give only 34.8 mm against the 63.2 mm obtained below,
45 per cent low.
Evaluate the peak influence factor. The effective
overburden at the depth of the peak is
$$\sigma'_{zp} = \gamma(D_f + B) = 20(2.0+2.5) = 90\ \text{kPa},$$
and therefore
$$I_{zp} = 0.5 + 0.1\sqrt{\frac{\Delta q}{\sigma'_{zp}}}
= 0.5 + 0.1\sqrt{\frac{110}{90}} = 0.5 + 0.111 = 0.611 .$$
Note that $\sigma'_{zp}$ belongs at the depth of the peak, not at the
base and not at mid-influence — a full $B$ below the base for a strip
footing.
Re-index the modulus table from the base of the footing.
The table is written from ground level while the influence diagram runs from the
footing base 2 m down. Mapping across, the 0–2 m layer is entirely above
the base and contributes nothing; the 2–8 m layer becomes 0–6 m below
the base at $E_s = 6000$ kPa; and the 8–12 m layer becomes 6–10 m
below the base at $E_s = 12000$ kPa. The influence zone stops exactly at 10 m
below the base, so the 12–16 m layer is never reached. Failing to re-index
is the most common arithmetic error in this problem.
Sum the influence over the three sub-layers. The 0–6 m
band must be split at $z = B = 2.5$ m where the diagram changes slope:
Sub-layer $z$ below base (m)
$\Delta z$ (m)
$E_s$ (kPa)
$I_z$ at top / bottom
$\bar{I_z}$
$\bar{I_z}\Delta z/E_s$ (m3/kN)
0.0 – 2.5
2.5
6000
0.200 / 0.611
0.4053
$1.689\times10^{-4}$
2.5 – 6.0
3.5
6000
0.611 / 0.326
0.4681
$2.731\times10^{-4}$
6.0 – 10.0
4.0
12000
0.326 / 0.000
0.1628
$0.543\times10^{-4}$
Sum
$4.962\times10^{-4}$
Eighty-nine per cent of the total comes from the upper 6 m, where the soft
6000 kPa layer coincides with the strongest part of the influence diagram.
Evaluate the two correction factors. The embedment
correction is
$$C_1 = 1 - 0.5\left(\frac{\bar{q}}{\Delta q}\right) = 1 - 0.5\frac{40}{110}
= 0.818 ,$$
and the creep correction for 12 years is
$$C_2 = 1 + 0.2\log_{10}\!\left(\frac{t}{0.1}\right)
= 1 + 0.2\log_{10}(120) = 1 + 0.2(2.079) = 1.416 .$$
The creep factor adds 42 per cent to the immediate settlement; had the question
asked for the settlement at the end of construction it would have been
$C_2 = 1.0$ and the answer 44.7 mm.
Combine. Schmertmann's expression gives
$$S_e = C_1C_2\,\Delta q\sum \frac{I_z}{E_s}\Delta z
= 0.818(1.416)(110)\left(4.962\times10^{-4}\right) = 0.0632\ \text{m},$$
that is
$$\boxed{\,S_e \approx 63\ \text{mm after 12 years}\,}$$
for the softest (Data Set 2) profile.
Bracket the answer with the other two data sets. Repeating
the same summation with Sets 1 and 3 gives
$\sum I_z\Delta z/E_s = 3.086\times10^{-4}$ and $2.535\times10^{-4}$, hence
$S_e = 39.3$ mm and 32.3 mm respectively. The maximum is therefore very nearly
double the minimum — which is exactly what the word “erratic”
in the question is warning about, and it is the reason a mat or a continuous
footing (rather than isolated pads) is the appropriate foundation type on this
site.
Judge the result against serviceability. Sixty-three
millimetres is far beyond the 25 mm normally tolerated for a strip footing, so
the proposal fails serviceability and the 150 kPa bearing pressure cannot be
used. Settlement is essentially proportional to the net pressure, and solving for
the pressure that returns 25 mm gives $\Delta q = 57.5$ kPa, a gross pressure of
about 98 kPa — a reduction of nearly half. Widening the footing helps
surprisingly little, because a wider strip drives its influence zone deeper into
the same soft 6000 kPa layer; the effective remedies are to reduce the pressure,
to found deeper so that the peak influence falls in the stiffer 12000 kPa
stratum, or to densify the upper 6 m by vibro-compaction or dynamic
compaction.
Quantity
Value
Overburden removed $\bar{q}$; net pressure $\Delta q$
40 kPa; 110 kPa
Influence-diagram type
plane strain (strip), $L/B > 15$
Depth of peak; $\sigma'_{zp}$
$z = B = 2.5$ m below base; 90 kPa
Peak influence factor $I_{zp}$
0.611
Influence zone
0 to $4B = 10$ m below base (2 to 12 m depth)
$\sum (I_z/E_s)\Delta z$ (Data Set 2)
$4.962\times10^{-4}$ m3/kN
$C_1$ (embedment) / $C_2$ (12-year creep)
0.818 / 1.416
Maximum settlement, Data Set 2
63 mm
Comparison: Data Set 1 / Data Set 3
39.3 mm / 32.3 mm
Immediate settlement (Set 2, $C_2 = 1$)
44.7 mm
Gross pressure for a 25 mm limit
about 98 kPa
Check — the two judgements, and
what turns on them. (1) Data Set 2 was chosen because the question
asks for the maximum. If the intent were a best estimate, Data Set 1 (the
middle profile) would be used and the answer would be 39 mm. All three are
reported. (2) Plane strain, not axisymmetric. $L/B > 15$ is stated in
the question and is decisive; the square-footing diagram on the same data gives
34.8 mm, 45 per cent low. (3) No water table is given, so the deposit is
treated as moist throughout at $\gamma = 20$ kN/m3; a table at founding level
would reduce $\sigma'_{zp}$ to about 65 kPa, raise $I_{zp}$ to 0.63 and increase
the settlement by roughly 3 per cent — not a sensitive assumption.
(4) The 150 kPa is treated as a gross pressure, consistent with
“the stress at the level of the foundation”; taking it as the net
increase (so that $\Delta q = 150$ kPa, $C_1 = 0.867$, $I_{zp} = 0.629$) would
give 94 mm. Chart source, as page 1 Note 6 requires:
Schmertmann, Hartman & Brown (1978) strain influence diagrams, as reproduced
in Das, Principles of Foundation Engineering, 9th ed.,
Fig. 5.9.
