16-Civ-B19 Foundation Engineering · December 2018

Question 4 of 5: Mat Foundations — Types, Bearing Capacity and Settlement

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

Notes on this paper

Paper format. National Exams, December 2018 — 16-Civ-B19 Foundation Engineering. Three hours, OPEN BOOK (one textbook plus one hand-written 8.5″ × 11″ aid sheet, both sides). Five questions, all to be answered, all of equal weight (20 marks each, 100 marks total). Any non-communicating calculator is permitted. The paper mixes SI and US customary units question by question, so each answer below stays in the units the question is posed in.

Reference texts

Das, B.M., Principles of Foundation Engineering, 9th ed. — the standard text for this syllabus. Ch. 3 (ultimate bearing capacity), Ch. 5 (vertical stress increase in soil), Ch. 6 (shallow-foundation settlement and plate load tests), Ch. 7 (mat foundations), Ch. 9 (pile foundations) and Ch. 10 (pile groups) supply every method used here.
Das, B.M., Principles of Geotechnical Engineering, 9th ed. — effective stress, consolidation and the Boussinesq stress-distribution theory the above builds on.
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed., 2006 — the Canadian practice document: limit-states terminology, geotechnical resistance factors, settlement tolerances and pile design guidance.
National Building Code of Canada (NBCC) 2020, Part 4 — load combinations and the limit-states framework within which the CFEM resistance factors sit.
Bowles, J.E., Foundation Analysis and Design, 5th ed. — alternative bearing-capacity and settlement formulations, useful as a cross-check.

Check: three conflicts in the printed paper, resolved as follows.

Q.2(a) embankment dimensions. The top of the figure reads 14 m + 5 m + 14 m = 33 m and the bottom chain reads 5 m + 11.5 m + 16.5 m = 33 m, so the two are consistent: A₂ lies 5 m inside the left toe and A₁ lies 16.5 m from each toe, i.e. on the embankment centreline.
Q.4(b) undrained shear strength. The paper prints c_u = 18 250 lb/ft². That is about 874 kPa — an order of magnitude stronger than any normal saturated clay and at the clay/weak-rock boundary. The answer below uses the printed value, and also states the result for c_u = 1 825 lb/ft² (87 kPa, a very stiff clay) in case the extra digit is a typographical slip.
Q.4(c) mat dimensions. The text says 75 ft × 100 ft while the figure it refers to is labelled 90 ft × 120 ft. The figure is solved as the primary case (the question says “as shown in figure below”) and the 75 ft × 100 ft case is carried through in full alongside it, so either reading is answered.

Question 4: Mat Foundations — Types, Bearing Capacity and Settlement (20 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) — types of mat foundation and when one is required

A mat (raft) is a single continuous slab supporting all, or a large group, of a structure's columns and walls. Six configurations are in common use, in ascending order of stiffness and cost. The flat plate is a slab of uniform thickness, simplest to build and used where column loads and spacings are light and regular. The flat plate thickened under columns adds a local drop panel to resist punching shear where the slab is otherwise adequate. The beam-and-slab mat (two-way grid of beams running between columns, above or below the slab) gives a much stiffer mat for heavier or irregular loads. The slab with basement walls as deep beams uses the perimeter and interior basement walls as a rigid cellular grillage. The cellular or box (compensated) raft encloses voids between two slabs so that the soil excavated nearly balances the structural weight. Finally, a piled raft combines a mat with piles that take part of the load and control differential settlement.

A mat is called for when a spread-footing solution stops making sense. The classic trigger is that the required footing area exceeds roughly half the building footprint, which happens on weak soil or under very heavy loads — at that point merging the footings into one slab costs little more and behaves far better. A mat is also chosen when the soil is erratic, containing pockets and lenses that would settle individual footings differently: the rigidity of the mat bridges them and converts differential settlement into a modest overall tilt. Other circumstances are a high water table where the basement slab must in any case be a continuous watertight element resisting uplift; structures highly sensitive to differential settlement; sites where the net pressure must be reduced by excavation, using a fully or partially compensated raft (net pressure $\Delta q = Q/A - \gamma D_f$ driven towards zero); and cases where uplift or overturning from wind or seismic action must be resisted by the weight and plan extent of a single element.

Part (b) — net ultimate bearing capacity of the 40 ft × 35 ft mat

Given. A mat 40 ft × 35 ft (so $B = 35$ ft is the shorter side and $L = 40$ ft), founded at $D_f = 6.0$ ft on saturated clay of undrained shear strength $c_u = 18\,250\ \text{lb/ft}^2$, loaded undrained ($\phi = 0$).

Find. The net ultimate bearing capacity $q_{net(u)}$.

Approach. Use the $\phi = 0$ form of the bearing-capacity equation with Meyerhof's shape and depth factors, which for a saturated clay reduces to a single compact expression.

State the $\phi = 0$ net bearing-capacity equation. For a rectangular mat on saturated clay, $$q_{net(u)} = 5.14\,c_u\left(1 + \frac{0.195\,B}{L}\right)\left(1 + 0.4\frac{D_f}{B}\right)$$ The 5.14 is $(\pi + 2)$, the Prandtl bearing-capacity factor for an undrained strip; the two brackets are the shape and depth corrections. The word net means the surcharge $\gamma D_f$ has already been removed, which is why no $qN_q$ term appears.
Evaluate the shape and depth factors. With $B = 35$ ft and $L = 40$ ft, $$1 + \frac{0.195(35)}{40} = 1 + 0.1706 = 1.1706, \qquad 1 + 0.4\frac{6}{35} = 1 + 0.0686 = 1.0686$$ The mat is nearly square, so the shape factor is close to the 1.195 of a perfect square; the depth factor is small because the mat is shallow relative to its width.
Substitute and evaluate. $$q_{net(u)} = 5.14(18\,250)(1.1706)(1.0686)$$ $$q_{net(u)} = 93\,805 \times 1.1706 \times 1.0686 = \boxed{117\,339\ \text{lb/ft}^2}$$ that is about 117.3 kip/ft², or 5 618 kN/m². At a customary mat factor of safety of 3 the allowable net pressure would be some 39 100 lb/ft².

Check: the printed $c_u$ is extraordinarily high. $c_u = 18\,250\ \text{lb/ft}^2$ is about 874 kPa. Real saturated clays run from a few kPa (soft) to roughly 200 kPa (hard), so this value sits at the clay/weak-rock boundary and no mat would ever be settlement-critical on it. The answer above uses the value as printed, which is what an examinee must do. If the intended value were $c_u = 1\,825\ \text{lb/ft}^2$ (87 kPa — a very stiff clay, and a one-digit misprint away), the identical arithmetic gives $q_{net(u)} = 11\,734\ \text{lb/ft}^2$ (562 kN/m²). Both are stated so that either reading of the paper is answered.

Part (c) — consolidation settlement at the centre of the mat

Given.

Quantity	Symbol	Value
Mat plan (figure / text)	$B \times L$	90 ft × 120 ft (text: 75 ft × 100 ft)
Total dead + live load	$Q$	$40 \times 10^3$ kip
Depth of mat	$D_f$	6 ft
Sand, 0–6 ft	$\gamma$	100 lb/ft³
Saturated sand, 6–30 ft	$\gamma_{sat}$	121.5 lb/ft³
NC clay, 30–48 ft	$\gamma_{sat}$, $C_c$, $e_0$	118 lb/ft³, 0.28, 0.90
Water table	—	6 ft below ground

Find. The primary consolidation settlement of the 18 ft clay layer beneath the centre of the mat.

Soil profile for Q.4(c). Settlement is computed for the 18 ft normally consolidated clay, using conditions at its mid-depth of 39 ft (33 ft below the base of the mat).

Approach. Compute the net pressure the mat applies, propagate it to the mid-depth of the clay with the four-quadrant Boussinesq influence factor, add it to the existing effective overburden there, and enter the normally-consolidated compression equation.

Find the gross and net applied pressures. Over the plan area shown on the figure, $$q = \frac{Q}{BL} = \frac{40 \times 10^6\ \text{lb}}{(90)(120)} = 3\,703.7\ \text{lb/ft}^2$$ The soil removed in excavating to 6 ft is credited back, so the net increase at founding level is $$\Delta q = q - \gamma D_f = 3\,703.7 - (100)(6) = 3\,103.7\ \text{lb/ft}^2$$ Using the net pressure is essential: the ground has already consolidated under the 600 lb/ft² that was excavated.
Locate the point at which stresses are evaluated. The clay runs from 30 ft to 48 ft, so its mid-depth is 39 ft below ground and $z = 39 - 6 = 33\ \text{ft}$ below the base of the mat. Mid-depth conditions represent an 18 ft layer adequately; a thicker layer would be sublayered.
Compute the effective overburden at mid-clay. Above the water table the total unit weight applies; below it the buoyant weight $\gamma' = \gamma_{sat} - \gamma_w$ applies with $\gamma_w = 62.4\ \text{lb/ft}^3$: $$\sigma'_0 = (100)(6) + (121.5 - 62.4)(24) + (118 - 62.4)(9)$$ $$\sigma'_0 = 600 + 1\,418.4 + 500.4 = 2\,518.8\ \text{lb/ft}^2$$
Propagate the net pressure to mid-clay. Divide the loaded rectangle into four equal quadrants meeting at the centre, each $45\ \text{ft} \times 60\ \text{ft}$, and use the corner influence factor $I(m,n)$ with $$m = \frac{45}{33} = 1.364, \qquad n = \frac{60}{33} = 1.818$$ The Boussinesq corner chart (Das Table 5.x) gives $I = 0.2174$, so $$\Delta\sigma = 4\,\Delta q\,I = 4(3\,103.7)(0.2174) = 2\,699\ \text{lb/ft}^2$$ Even 33 ft below the base, a mat this wide has shed only about 13 % of its net pressure — wide loads reach deep.
Enter the normally-consolidated settlement equation. The clay is normally consolidated, so the whole stress increase acts on the virgin compression line: $$S_c = \frac{C_c H_c}{1 + e_0}\log_{10}\!\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0}$$ $$S_c = \frac{(0.28)(18)}{1 + 0.90}\log_{10}\!\frac{2\,518.8 + 2\,699}{2\,518.8} = 2.653 \times \log_{10}(2.0716)$$ $$S_c = 2.653 \times 0.31629 = \boxed{0.839\ \text{ft} \approx 10.1\ \text{in} \;(256\ \text{mm})}$$
Carry the alternative mat size through. If the mat is the 75 ft × 100 ft of the question text rather than the 90 ft × 120 ft of the figure, the same load acts on a smaller area: $q = 5\,333.3$, $\Delta q = 4\,733.3\ \text{lb/ft}^2$, and with $m = 37.5/33 = 1.136$, $n = 50/33 = 1.515$ giving $I = 0.2024$, $$\Delta\sigma = 4(4\,733.3)(0.2024) = 3\,833\ \text{lb/ft}^2, \qquad S_c = 2.653\log_{10}\!\frac{6\,351.8}{2\,518.8} = 1.066\ \text{ft} = 12.8\ \text{in}$$ The smaller mat settles more, because the same load spread over less area both raises the net pressure and concentrates it.
Comment on the result. Ten to thirteen inches of consolidation is far beyond any serviceability tolerance — CFEM guidance and normal practice would look for 25–50 mm. The calculation is therefore not merely an arithmetic exercise: it demonstrates that a normally consolidated clay under this load needs either a compensated (deeper) raft to reduce $\Delta q$ towards zero, a piled raft carrying load past the clay, or preloading of the clay before construction.

Result	90 ft × 120 ft (figure)	75 ft × 100 ft (text)
Gross applied pressure, $q$	3 703.7 lb/ft²	5 333.3 lb/ft²
Net pressure, $\Delta q$	3 103.7 lb/ft²	4 733.3 lb/ft²
Effective overburden at mid-clay, $\sigma'_0$	2 518.8 lb/ft²
Influence factor $I(m,n)$ per quadrant	0.2174	0.2024
Stress increase at mid-clay, $\Delta\sigma$	2 699 lb/ft²	3 833 lb/ft²
Consolidation settlement, $S_c$	0.839 ft = 10.1 in (256 mm)	1.066 ft = 12.8 in (325 mm)
(b) Net ultimate bearing capacity, 40 × 35 ft mat	117 339 lb/ft² (5 618 kN/m²)

Topic. Mat foundations: undrained bearing capacity and one-dimensional consolidation settlement of an underlying normally consolidated clay.

Key relations. $q_{net(u)} = 5.14c_u(1 + 0.195B/L)(1 + 0.4D_f/B)$; net pressure $\Delta q = Q/A - \gamma D_f$; centre stress by four quadrants $\Delta\sigma = 4\Delta q\,I(m,n)$ with $m = (B/2)/z$, $n = (L/2)/z$; $S_c = \frac{C_cH_c}{1+e_0}\log_{10}\frac{\sigma'_0+\Delta\sigma}{\sigma'_0}$ for normally consolidated clay.

Why this works. A mat is wide enough that its bearing capacity is almost never the binding constraint — the shape and depth factors add only about 25 % to a Prandtl mechanism whose resistance is already large for any competent clay. What binds is settlement, because a wide load transmits its pressure to great depth: the influence factor at $z/B$ of about one-third is still above 0.2 per quadrant. The consolidation equation is logarithmic because the virgin compression line is straight in $e$–$\log\sigma'$ space, so it is the ratio of final to initial effective stress that sets the strain, which is why the buoyant unit weights below the water table matter so much: they roughly halve $\sigma'_0$ and thereby roughly double the log term.

Common pitfall. Using the gross pressure instead of the net in the settlement calculation (this alone inflates the answer by about 20 % here), and using total rather than buoyant unit weights below the water table when building $\sigma'_0$. A third is applying $I(m,n)$ with the full mat dimensions instead of the half-dimensions of a quadrant, and then forgetting the factor of four.

Source. Das, Principles of Foundation Engineering 9th ed., §7.2–7.5 (types of mat, bearing capacity, compensated foundations) and §5.4 / Ch. 6 (rectangular-area stress increase, consolidation settlement); CFEM 4th ed. §11 for settlement tolerances.

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