07-Str-B1 · May 2013

Question 8 of 10: Stress Increase by the 2:1 Method and Consolidation Settlement

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

Notes on this paper

National Examinations — May 2013 — 07-Str-B1 Geotechnical Design. Three-hour, OPEN-BOOK exam; any non-communicating calculator permitted (the candidate must record its make and model). Format: Section A carries five short-answer questions of 7 marks each, of which any FOUR are to be answered; Section B carries the long design questions at 24 marks each, of which any THREE are to be answered. The paper instructs candidates to state any interpretive assumptions and to identify the source of every design chart or assumed value used. Every question in both sections is worked below, because the set is intended as a study resource.

Reference texts: Das, B.M., Principles of Foundation Engineering (9th ed., Cengage) — shallow foundations, consolidation settlement, sheet-pile walls, retaining walls and drilled shafts; Das, B.M., Principles of Geotechnical Engineering (9th ed., Cengage) — method of slices, lateral earth pressure, consolidation theory; Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM, 4th ed., 2006) — Canadian practice for SPT interpretation, pile design, limit-states design and tolerable settlement; Craig, R.F. / Knappett, J.A., Craig's Soil Mechanics (8th ed., CRC Press) — effective stress, shear strength and slope stability; Duncan, J.M., Wright, S.G. & Brandon, T.L., Soil Strength and Slope Stability (2nd ed., Wiley) — choice of strength parameters and factors of safety for short- and long-term analyses.

NOTE 1 — question numbering in the source. The printed paper labels the retaining-wall problem (Figure 4) and the drilled-pier problem (Figure 5) both as "Question 9", while the Section B heading reads "answer any THREE of the following FOUR questions". Section B therefore contains five printed problems under four numbers. They are set out below as Question 9 (retaining wall) and Question 10 (drilled pier) in printed order, so that each can be referred to unambiguously; the marks shown are those printed against each problem.

NOTE 2 — dimensions scaled from Figure 1. Figure 1 is a hand-drawn slope on a 1 m × 1 m grid with no written dimensions other than $R=10$ m. The geometry used in Question 6 was scaled from that grid: slope height 7 m over a 9 m horizontal run, a 3 m thick lower layer, and the centre of the trial circle 1.4 m horizontally beyond the toe and 8.0 m above it. Every one of these values reproduces the drawing to within about 0.2 m (one fifth of a grid square). Check against the original if the paper is used for marking rather than study.

Question 8: Stress Increase by the 2:1 Method and Consolidation Settlement (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 square pad footing on sand, with a compressible clay stratum beginning 1.5 m below the base of the footing.

Given data
Quantity	Symbol	Value
Column load	$Q$	200 kN
Footing dimensions (square)	$B\times B$	2.0 m × 2.0 m
Founding depth	$D_f$	1.5 m
Sand above the water table, 0 to 1.5 m	$\gamma$	19 kN/m³
Sand below the water table, 1.5 to 3.0 m	$\gamma_{sat}$	21 kN/m³
Clay, 3.0 to 6.0 m	$\gamma_{sat}$, $H_c$	20 kN/m³, 3.0 m
Initial void ratio of the clay	$e_0$	0.9
Compression and swelling indices	$C_c$, $C_s$	0.31, 0.09
Water table	—	At the base of the footing, 1.5 m below ground

Find. The average increase in vertical stress within the clay layer beneath the centre of the footing by the 2:1 dispersion method, and the resulting average consolidation settlement of that layer.

[Figure not reproduced: Figure 3 (redrawn) — soil profile with the 2.0 m square footing founded 1.5 m below ground and the 2:1 stress dispersion spreading the 200 kN column load into the clay layer. See the official exam paper.]

Approach. Spread the column load over an area that grows at one horizontal to two vertical from the edges of the footing, evaluate the resulting stress at the top, middle and bottom of the clay and average them by Simpson's rule; compute the initial effective overburden stress at mid-depth of the clay; then apply the one-dimensional consolidation equation over the full thickness of the layer.

Set up the 2:1 dispersion. The 2:1 method assumes the load spreads uniformly over an area whose plan dimensions increase by $z$ in each direction, $z$ being measured downward from the base of the footing. For a square footing, $$\Delta\sigma(z)=\frac{Q}{(B+z)^2}$$ This is an equilibrium statement, not an elasticity solution: the total force on every horizontal plane is the same $Q$, redistributed over a larger area.
Depths to the top, middle and bottom of the clay. The clay runs from 3.0 m to 6.0 m below ground, and the footing base is at 1.5 m, so measured from the base of the footing $$z_t=3.0-1.5=1.50\ \text{m},\qquad z_m=1.50+1.50=3.00\ \text{m},\qquad z_b=1.50+3.00=4.50\ \text{m}$$
Stress increase at the three levels. $$\Delta\sigma_t=\frac{200}{(2.0+1.5)^2}=\frac{200}{12.25}=16.33\ \text{kPa}$$ $$\Delta\sigma_m=\frac{200}{(2.0+3.0)^2}=\frac{200}{25.00}=8.00\ \text{kPa}$$ $$\Delta\sigma_b=\frac{200}{(2.0+4.5)^2}=\frac{200}{42.25}=4.73\ \text{kPa}$$
Average stress increase over the layer. The stress varies as $1/(B+z)^2$, so a straight mean of the end values would be poor. Simpson's rule over the three ordinates gives the weighted average that is standard for this calculation: $$\Delta\sigma_{av}=\frac{\Delta\sigma_t+4\Delta\sigma_m+\Delta\sigma_b}{6}=\frac{16.33+32.00+4.73}{6}=\frac{53.06}{6}=\boxed{\Delta\sigma_{av}=8.84\ \text{kPa}}$$
Initial effective overburden stress at mid-depth of the clay. Mid-depth is 4.5 m below ground surface. Above it lie 1.5 m of moist sand, 1.5 m of submerged sand and 1.5 m of submerged clay: $$\sigma'_0=19(1.5)+(21-9.81)(1.5)+(20-9.81)(1.5)$$ $$\sigma'_0=28.50+16.79+15.29=60.57\ \text{kPa}$$
Consolidation settlement of the layer. No preconsolidation pressure is given, so the clay is taken as normally consolidated and the virgin compression index governs the whole of the stress increase: $$S_c=\frac{C_c H_c}{1+e_0}\log\frac{\sigma'_0+\Delta\sigma_{av}}{\sigma'_0}$$ Substituting, $$S_c=\frac{0.31(3.0)}{1+0.9}\log\frac{60.57+8.84}{60.57}=\frac{0.930}{1.900}\log\frac{69.41}{60.57}$$ $$S_c=0.4895\times\log(1.1460)=0.4895\times0.05918=0.0290\ \text{m}$$ $$\boxed{S_c\approx29\ \text{mm}}$$
Check the result against the serviceability limit. A total settlement of 29 mm from consolidation alone is at the upper end of what is normally tolerated for an isolated column footing (25 mm total, with differential settlement of about 20 mm and an angular distortion of 1/500 for a framed structure). Immediate settlement of the two sand layers must be added to it, and the settlement will develop over years rather than at construction. The design is workable but the settlement criterion, not bearing capacity, is what limits it — which is precisely the point argued in Question 5.

Question 8 — results
Quantity	Value
Depth to top / middle / bottom of clay below the footing base	1.50 / 3.00 / 4.50 m
Stress increase at the top of the clay	16.33 kPa
Stress increase at mid-depth	8.00 kPa
Stress increase at the bottom of the clay	4.73 kPa
Average stress increase (Simpson)	$\boxed{\Delta\sigma_{av}=8.84\ \text{kPa}}$
Effective overburden at mid-depth of clay	60.57 kPa
Stress ratio $(\sigma'_0+\Delta\sigma_{av})/\sigma'_0$	69.41 / 60.57 = 1.146
Average consolidation settlement	$\boxed{S_c=29\ \text{mm}}$

Check — the assumption about stress history. The question supplies both $C_c$ and $C_s$ but no preconsolidation pressure $\sigma'_c$, so the clay has been taken as normally consolidated and $C_c$ applied to the whole stress increase. If the clay were overconsolidated with $\sigma'_c$ above 69.4 kPa the entire increase would lie on the recompression line and the settlement would fall in the ratio $C_s/C_c=0.09/0.31$, to about 8 mm; if $\sigma'_c$ lay between 60.6 and 69.4 kPa the settlement would be found from the two-part expression $S_c=\frac{C_sH}{1+e_0}\log\frac{\sigma'_c}{\sigma'_0}+\frac{C_cH}{1+e_0}\log\frac{\sigma'_0+\Delta\sigma}{\sigma'_c}$. The normally consolidated assumption is the conservative one and should be stated explicitly on the answer paper, as the exam's own Note 1 invites.

Topic. Stress distribution beneath a footing by the 2:1 approximation; one-dimensional consolidation settlement.

Key relations. 2:1 dispersion for a rectangular footing: $\Delta\sigma=\dfrac{Q}{(B+z)(L+z)}$. Simpson average over a layer: $\Delta\sigma_{av}=(\Delta\sigma_t+4\Delta\sigma_m+\Delta\sigma_b)/6$. Normally consolidated settlement: $S_c=\dfrac{C_cH}{1+e_0}\log\dfrac{\sigma'_0+\Delta\sigma}{\sigma'_0}$; overconsolidated with $\sigma'_c>\sigma'_0+\Delta\sigma$: replace $C_c$ by $C_s$.

Why this works. Consolidation settlement is the integral of vertical strain, and vertical strain in a normally consolidated clay is proportional to the logarithm of the effective stress ratio. Using the stress increase at mid-depth (or its Simpson average) with the full layer thickness replaces that integral with a single equivalent-strain calculation, which is accurate to a few percent for layers thinner than about twice the footing width.

Common pitfall. Measuring $z$ from the ground surface rather than from the base of the footing. Using $z=3.0$, 4.5 and 6.0 m would give an average increase of 5.1 kPa instead of 8.84 kPa and understate the settlement by about 40%. The second pitfall is using $\gamma_{sat}$ rather than $\gamma'$ below the water table when computing $\sigma'_0$, which would raise $\sigma'_0$ to 90 kPa and understate the settlement again.

Source. Das, Principles of Foundation Engineering, Ch. 5 (settlement of shallow foundations; 2:1 stress increase and average stress increase in a layer) and Ch. 1 (consolidation review); Das, Principles of Geotechnical Engineering, Ch. 11 (compressibility of soil, $C_c$ and $C_s$).

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