In Terzaghi’s bearing capacity equation, the term that increases with the width of the footing is

all three, since all depend on the footing

Negative skin friction on a pile

acts downward and is added to the applied load

reduces the ultimate capacity and is subtracted from the resistance

acts downward and is added to the applied load

occurs only in cohesionless soils

is beneficial, since the soil grips the pile more tightly

Which of the following are correct about foundations on clay in the undrained condition (φ_u = 0)? (One or more)

The width term of the bearing capacity equation vanishes

Widening the footing does not increase the ultimate bearing pressure

Settlement, not bearing failure, often governs the design

The width term of the bearing capacity equation vanishes

Widening the footing does not increase the ultimate bearing pressure

Settlement, not bearing failure, often governs the design

The bearing capacity increases with the square of the width

Foundation Engineering: Stress Distribution, Bearing Capacity, Earth Pressure, Slopes and Piles

Foundation engineering is where the previous two chapters get spent, and it asks two questions of every structure: will the ground fail, and will it move too much? They are separate calculations with separate criteria, and in practice the second usually governs — a footing on a stiff clay is far more often limited by settlement than by a bearing failure it would never reach. Both start from how the load spreads, which is Boussinesq: a point load produces a vertical stress falling as 1/z², and superposing it under a loaded area gives the pressure bulb whose shape is the reason a footing twice as wide influences soil twice as deep. Then bearing capacity, Terzaghi's three-term expression q_u = cN_c + γD_f N_q + 0.5γBN_γ, in which each term names a different mechanism — cohesion, the surcharge alongside the footing, and the weight of the wedge beneath it. The water table enters through the unit weights and does more damage than candidates expect: bringing it up to the ground surface can halve the capacity, because submergence cuts the effective unit weight by nearly half in both the terms that depend on it. Earth pressure is the same soil pushed instead of loaded, and its three states — active, at rest, passive — differ by a factor of nine at φ = 30°, which is why the direction a wall is allowed to move matters more than its strength. Slopes are governed by whether water is flowing through them, and seepage parallel to the surface roughly halves the factor of safety of a cohesionless slope. And piles carry load along the shaft and at the base in a proportion that is usually the opposite of what the picture suggests: in a clay the shaft does most of it.

Sub-Surface Investigation: What the Ground Is, Before What It Can Carry

Every number in the rest of this chapter is only as good as the investigation that supplied it, and the syllabus names the tools: boreholes and sampling, the plate load test, and the standard and cone penetration tests. Boring establishes the sequence of strata and recovers samples — disturbed samples for classification and index properties, where the fabric does not matter, and undisturbed samples, taken in thin-walled tubes, for strength and compressibility, where it is the only thing that matters. A borehole should reach at least to the depth where the added stress falls to about a tenth of the existing overburden, which is roughly 1.5B below a footing, so the depth is set by the width of the foundation rather than by its load. An investigation that stops at 5 m has answered the question for a 1 m footing and not for the raft.

⚠️ A plate load test scales one way in sand and another in clay

The plate load test loads a small rigid plate, usually 300 mm square, and measures the settlement — then the result has to be scaled to the real footing, and the scaling law is not the same in the two materials. In clay, whose modulus is roughly constant with depth, settlement is proportional to width: S_f = S_p (B_f/B_p). In sand, whose stiffness increases with confining pressure and therefore with depth, the widely used relation is S_f = S_p [B_f(B_p + 0.3)/(B_p(B_f + 0.3))]². Put numbers on a 0.3 m plate that settles 5 mm under the design pressure, scaled to a 2 m footing. In sand: 5 × [2(0.6)/(0.3 × 2.3)]² = 5 × (1.739)² = 15.1 mm. In clay: 5 × (2/0.3) = 33.3 mm — more than twice as much from the same test result, because the clay relation is linear in B while the sand relation flattens out. That flattening is worth seeing: as B_f grows large the sand ratio tends to (0.6/0.3)² = 4, so no footing on sand settles more than four times its plate however wide it is, while on clay the settlement grows without limit. Two limitations follow directly and are the reason the test is used with care rather than trusted. The plate influences only a shallow bulb, about 1.5 × 0.3 = 0.45 m, so it cannot see a soft layer that a wide foundation will reach — the classic failure of the method. And on clay the test is quick enough to be undrained, so it measures immediate settlement and says nothing about the consolidation settlement that follows over years.

The standard penetration test drives a split-spoon sampler with a 63.5 kg hammer falling 750 mm and counts the blows for the second and third 150 mm of penetration — that count is N. It is crude, it is the most widely available soil datum in the world, and it is corrected before use: for overburden pressure, because the same sand gives a higher count at depth; for hammer energy, because a rope-and-pulley rig delivers about 60 per cent of the theoretical energy and a trip hammer more; and for dilatancy in fine saturated sands and silts, where the blow itself generates a pore pressure that stiffens the soil. From the corrected N come empirical estimates of relative density, friction angle and allowable bearing pressure — every one of them a correlation rather than a measurement, which is the honest description. The cone penetration test pushes an instrumented cone at a steady rate and logs tip resistance and sleeve friction continuously, giving a far better stratigraphic profile and no sample at all. The two are complements rather than rivals: CPT tells you where the layers change, boreholes and SPT tell you what they are.

Boussinesq, and the Pressure Bulb

Boussinesq solved the stress in an elastic half-space under a point load, and the vertical component is Δσ_z = (3Q/2πz²)[1 + (r/z)²]^(−5/2). Two features do all the work. Directly under the load r = 0, the bracket is 1, and the influence factor is 3/2π = 0.4775 — so Q = 1000 kN gives 119.4 kPa at 2 m depth and 29.8 kPa at 4 m, a quarter, because the stress falls as 1/z². Away from the axis it decays fast: at 2 m depth the factor falls from 0.4775 on the axis to 0.2733 one metre off and 0.0844 two metres off. Notice what the expression does not contain — no E, no Poisson’s ratio, no soil property at all. The vertical stress depends only on geometry, which is why the same chart is used for every soil and why the elastic assumption, plainly untrue for soil, is nevertheless serviceable for this one purpose.

Superpose the point solution over a loaded area and the contours of equal vertical stress form the pressure bulb. The contour usually taken as the practical limit is 0.2q, and for a square footing it reaches to about 1.5B below the base — which makes the depth of influence proportional to the width, not to the load. That single fact drives most of settlement practice. A narrow footing at high pressure disturbs only a shallow zone; a wide raft at low pressure reaches deep, so a soft layer at 8 m is irrelevant to a 1 m footing and decisive for a 10 m raft carrying the same building. It is also why a borehole depth is specified as a multiple of the foundation width, and why a site investigation that stops at 5 m has answered the wrong question for a raft.

Bearing Capacity, and What the Water Table Costs

Terzaghi’s expression for a strip footing is q_u = c N_c + γ D_f N_q + 0.5 γ B N_γ, and each term is a separate mechanism rather than a fitted coefficient. c N_c is the cohesion resisting along the failure surface. γ D_f N_q is the surcharge of the soil alongside the footing, which must be lifted for the failure wedge to escape — which is why simply founding deeper buys capacity even in a uniform soil. 0.5 γ B N_γ is the weight of the wedge beneath the footing, and it is the only term that grows with the width. The factors depend on φ alone, and steeply: at φ = 30°, N_q = 18.4, N_c = 30.1 and N_γ = 22.4, all of which roughly double for every 5° more. That steepness is worth respecting, because it means an error of a couple of degrees in φ moves the answer far more than an error in γ or B.

⚠️ Raise the water table to the surface and half the bearing capacity is gone

Take a strip footing 2 m wide at 1.5 m depth on a sand with γ = 18 kN/m³, c = 0 and φ = 30°. With the water table well below, the surcharge term is 18(1.5)(18.40) = 496.8 kPa and the width term 0.5(18)(2)(22.40) = 403.2 kPa, so q_u = 900.1 kPa. Net of the overburden already there, q_net = 873.1, and a factor of 3 on the net gives a safe bearing pressure of 318.0 kPa. Now move the water table and change nothing else. Bring it to foundation level and the soil beneath the footing is submerged, so the third term uses γ′ = 18 − 9.81 = 8.19 instead of 18: q_u falls to 680.3 kPa, a loss of 24 per cent. Bring it to the ground surface and both the surcharge and the wedge are submerged: q_u = 409.5 kPa, a loss of 54.5 per cent. More than half the bearing capacity has been removed by water, with no change to the soil, the footing or the load. The mechanism is simple once seen — submergence roughly halves the effective unit weight, and two of the three terms are directly proportional to it — but the size of the effect surprises people, and it is the reason a design must state the water table it assumed and the reason a seasonal high is used rather than the level found on the day of the investigation. Two related traps. If the water table sits at a depth greater than B below the footing it is usually taken as having no effect, since the failure wedge does not reach it, and intermediate depths are interpolated. And the loss is not a strength loss: φ has not changed, and the sand is exactly as strong as it was. What changed is the weight available to hold the failure wedge down.

The same 2 m strip footing at 1.5 m depth, φ = 30°, as the water table rises
Water table	Surcharge term	Width term	`q_u` (kPa)
Deep — no effect	496.8	403.2	900.1
At foundation level	496.8	183.5	680.3 (−24%)
At the ground surface	226.0	183.5	409.5 (−55%)

Earth Pressure and Slope Stability

The pressure a soil exerts on a wall depends on which way the wall is allowed to move, and by a startling margin. Let the wall move away and the soil expands laterally, mobilises its own shear strength to help hold itself up, and pushes with the active pressure K_a = (1 − sin φ)/(1 + sin φ). Push the wall into the soil and it must be forced to fail in the other direction, so the resistance is the passive pressure K_p = 1/K_a. Prevent any movement at all and you get the at-rest pressure, K₀ ≈ 1 − sin φ. At φ = 30° those are 1/3, 3 and 0.5 — so on a 5 m wall retaining soil at γ = 18, the thrusts are 75, 675 and 112.5 kN/m, all acting at H/3 above the base. The active and passive values differ by a factor of nine. That ratio is the practical content of the topic: the movement needed to reach the active state is tiny, a fraction of a per cent of the wall height, while reaching full passive resistance needs movement ten times larger — usually far more than a structure can tolerate — which is why passive resistance is heavily discounted in design while active pressure is taken at full value.

🎯 An infinite slope that stands dry and fails when water flows through it

For a long uniform slope in cohesionless soil the factor of safety has a form so simple it is worth memorising: F = tan φ/tan β, where β is the slope angle. Everything cancels — the depth of the failure plane, the unit weight, the length of the slope — leaving two angles. So a dry cohesionless slope stands if and only if it is flatter than the friction angle, whatever it is made of and however deep the slip. At φ = 30° and β = 20°, F = 0.5774/0.3640 = 1.586, comfortably stable. Now let water seep parallel to the slope surface, which is what happens in a hillside during prolonged rain. The pore pressure on the failure plane rises, the effective stress falls, and the factor of safety is multiplied by γ′/γ_sat: with γ_sat = 20 kN/m³, that ratio is 10.19/20 = 0.5095. The factor of safety drops from 1.586 to 0.808 — and the slope that was comfortably stable in the dry fails. Seepage has taken away almost exactly half the safety margin, and it has done so without any change in the geometry, the soil or the load. Three things follow. This is why slope failures cluster after rainfall rather than during it or in dry weather, and why the trigger is often the rise of the water table rather than the rain itself. It is why drainage — weep holes, counterfort drains, horizontal bores — is usually the cheapest slope stabilisation available, because it attacks the term that is doing the damage. And it is why a slope must be analysed for the worst credible groundwater condition rather than the one measured on a dry afternoon.

Finite Slopes, Combined Footings and Contact Pressure

The infinite-slope result applies to a long uniform slope whose failure plane is parallel to the surface, which is a good model for a shallow slip in a cohesionless soil and a poor one for a cutting in clay. A finite slope fails along a surface that is approximately circular, and the standard treatment is the method of slices: divide the mass above a trial circle into vertical slices, and for each slice compare the shear resistance available on its base with the shear the weight demands there. Fellenius’s ordinary method ignores the forces between slices altogether, which makes the arithmetic direct and the answer conservative, sometimes by 20 per cent. Bishop’s simplified method restores the horizontal inter-slice forces while still neglecting the vertical ones, and the resulting expression has the factor of safety on both sides, so it has to be solved by iteration — guess F, compute F, repeat until it settles, which it does in three or four passes. The extra work buys a materially less conservative and more realistic answer, which is why Bishop is the method in practical use and Fellenius the one that fits on an exam page.

🎯 Contact pressure is not uniform, and the shape depends on both the footing and the soil

Every bearing capacity calculation in this chapter assumed the pressure under the footing is uniform. It is not, and the way it departs from uniform depends on two things at once — how stiff the footing is, and what the soil is. Under a rigid footing the settlement must be the same everywhere, so the pressure distributes itself to make that happen: on clay, which can carry shear at its edges, the pressure concentrates at the edges and is theoretically infinite there before local yielding rounds it off; on sand, which has no strength at an unconfined edge, the pressure concentrates at the centre and falls to nearly zero at the perimeter. Under a flexible footing the reverse holds — the pressure is uniform and the settlement is what varies, dishing in the middle on clay and at the edges on sand. The uniform assumption is therefore a convenience that is wrong in all four cases and adequate in most, because bearing capacity depends on the total load rather than its distribution. It stops being adequate when the bending moment in the footing is what you are computing, which is exactly why a raft is analysed as a plate on an elastic foundation and not as a uniformly loaded slab. Combined footings and rafts exist for the cases a single pad cannot handle: two columns too close for separate footings, a column on a property line that cannot be centred under its own pad, or a total footing area so large a fraction of the plan that one continuous slab is cheaper. A raft has a further advantage worth naming — a floating or compensated raft is excavated deep enough that the weight of soil removed roughly equals the weight of the building, so the net change in effective stress at depth is near zero and so, therefore, is the consolidation settlement.

Piles: Where the Load Actually Goes

A pile carries load two ways, and the split is usually the opposite of what the drawing suggests. Shaft resistance acts along the whole embedded length; base resistance acts on a small area at the tip. Take a bored pile 0.4 m in diameter and 10 m long in a clay with c_u = 50 kPa and an adhesion factor α = 0.5. The shaft gives Q_s = α c_u (π d L) = 0.5(50)(π)(0.4)(10) = 314.2 kN. The base gives Q_p = 9 c_u A_p = 9(50)(0.1257) = 56.5 kN. The total is 370.7 kN, and the shaft carries 85 per cent of it. That is typical for a pile in clay, and it explains a good deal of practice: why pile capacity is roughly proportional to length rather than to diameter, why a slightly short pile is a serious defect while a slightly narrow one is not, and why the condition of the shaft — the smearing and softening left by the boring — matters more than the cleanliness of the base.

A group of piles is not simply the sum of its members, and two checks are needed. Individually, the group capacity is n times the single-pile value. As a block, the group and the soil between the piles fail together, with shaft resistance round the block perimeter and end bearing over the whole block area. The design capacity is the lesser. For nine of the piles above at a spacing of 3d, the block is 2.8 m square: the block capacity comes to 9128 kN against 3336 kN for the sum of the individuals, so individual failure governs at that spacing and the group efficiency is effectively 1. Bring the piles closer and the block value falls while the sum does not, until the block governs — which is the real meaning of group efficiency and why codes set minimum spacings of about 2.5d to 3d. One more effect belongs here and it has the opposite sign to everything above. If the ground around a pile settles more than the pile does — a fill placed recently, a clay still consolidating — the soil hangs on the shaft and drags it down. That is negative skin friction, and it is not a reduction in capacity but an additional load on the pile, which is why it is added to the applied load rather than subtracted from the resistance.

Ground improvement techniques

Every foundation decision so far has taken the soil as given and chosen a footing to suit it. Ground improvement — named explicitly in the GATE 2027 revision — inverts that: it changes the soil so that a cheaper foundation will do. The methods divide cleanly by what they remove, and choosing between them is almost entirely a question of the soil’s permeability, because that decides how fast water can leave.

The methods, by what they take out of the soil
Technique	What it removes or adds	Suits
Preloading with vertical (sand or wick) drains	Removes water — forces the consolidation settlement to happen before the structure is built	Soft saturated clay. The drains shorten the drainage path, and time goes as its square — the same `T_v` relation as in chapter 2
Vibro-compaction / dynamic compaction	Removes voids — rearranges grains into a denser packing	Loose sand and gravel. Useless in clay, where the grains cannot rearrange
Stone columns	Adds stiffness and a drainage path together	Soft clay and silt — the one method that both reinforces and drains
Grouting (permeation, compaction, jet)	Adds a binder into the voids	Coarse soils and rock fissures; permeation grout cannot enter a clay at all
Geosynthetic reinforcement	Adds tensile strength — which soil has none of	Embankments on soft ground, retaining walls, separation under pavements

⚠️ Permeability decides the method, and it is why sand and clay never share one

The commonest error in this topic is to treat the techniques as interchangeable strengtheners. They are not, and the dividing line is how fast water can leave. Vibro-compaction densifies a loose sand because water can escape from between the grains in seconds as they rearrange — in a clay the water cannot get out at that rate, so the vibration merely remoulds and weakens it. Conversely, preloading with vertical drains is the standard answer in soft clay precisely because the clay’s low permeability is the problem being solved, and it is pointless in sand, which consolidates almost instantly under its own construction sequence. The same logic runs through grouting: a permeation grout flows through the voids of a gravel and cannot enter the pores of a clay, so clay needs jet grouting or deep soil mixing, which cut and replace rather than permeate. Read that way, the table is one variable in five disguises, and the question to ask of any ground-improvement problem is not "which is strongest" but "can the water get out, and how fast".

Key takeaways

Boussinesq contains no soil property — vertical stress falls as 1/z² and depends only on geometry. The pressure bulb reaches about 1.5B, so the depth that matters scales with the width of the foundation, not the load.
q_u = cN_c + γD_f N_q + 0.5γBN_γ — three terms, three mechanisms. Only the last grows with width, which is why deepening a footing often buys more than widening it.
The water table costs 24% at foundation level and 55% at the surface on the worked footing — with no change to soil, size or load. It is not a loss of strength: φ is unchanged, only the weight holding the wedge down.
Active, at-rest and passive stand at 1 : 1.5 : 9 for φ = 30°. The active state needs a tiny movement and the passive a large one, which is why passive resistance is discounted in design and active pressure is not.
An infinite cohesionless slope has F = tan φ/tan β — two angles and nothing else. Seepage parallel to the surface multiplies it by γ′/γ_sat ≈ 0.5, taking the worked slope from 1.59 to 0.81.
In clay the shaft carries about 85% of a pile’s capacity, not the base. Group capacity is the lesser of the sum of individuals and the block; and negative skin friction is an added load, not a reduced resistance.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A vertical point load of 1000 kN acts at the surface of a soil. By Boussinesq theory, the vertical stress increase directly below the load at a depth of 2 m, in kPa, is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 119.37
119.37 kPa. Directly below the load r = 0, so the bracket [1 + (r/z)²]^(−5/2) equals 1 and the expression reduces to Δσ_z = 3Q/(2πz²) = 0.4775 Q/z² = 0.4775(1000)/4 = 119.37 kPa. The influence factor 3/2π = 0.4775 is worth knowing on sight because it is the peak of the whole Boussinesq chart. Two properties of the result matter more than the number. The stress falls as 1/z², so doubling the depth to 4 m quarters it to 29.84 kPa — depth attenuates load fast, which is the whole reason deep foundations work. And the expression contains no soil property at all — no E, no Poisson’s ratio, no φ, no unit weight. The vertical stress under a point load is a matter of geometry alone, which is why the same influence chart serves for every soil, and why the elastic half-space assumption, obviously false for a real soil, is nevertheless acceptable for this one purpose. The decay away from the axis is steep too: at the same 2 m depth the factor falls from 0.4775 on the axis to 0.2733 one metre off and 0.0844 two metres off. Superposing this solution under a loaded area gives the pressure bulb, whose 0.2q contour reaches about 1.5B below a square footing — which is the practical result to carry, since it means the depth of soil that matters scales with the width of the foundation and not with how heavily it is loaded.
In Terzaghi’s bearing capacity equation, the term that increases with the width of the footing is
1. c N_c
2. γ D_f N_q
3. 0.5 γ B N_γ
4. all three, since all depend on the footing
Show answer
Answer: C — `0.5 γ B N_γ`
0.5 γ B N_γ. Each of the three terms is a different mechanism, and reading them that way makes the answer obvious rather than memorised. c N_c is the cohesion mobilised along the failure surface, and it has no B in it. γ D_f N_q is the surcharge of soil beside the footing, which the failure wedge must lift to escape — it depends on the founding depth, which is why founding deeper buys capacity even in a uniform soil. 0.5 γ B N_γ is the self-weight of the wedge beneath the footing, and it is the only term proportional to the width. Option D is the trap: the factors N_c, N_q and N_γ depend on φ alone and not on the footing at all. The practical readings follow directly. For a clay in the undrained condition, φ_u = 0 makes N_γ = 0, so the width term vanishes entirely and the bearing capacity of a footing on clay is essentially independent of its width — widening a footing on clay reduces the pressure but does not raise the capacity per unit area. For a sand, c = 0 kills the first term and capacity comes entirely from depth and width. And since only the third term carries B, doubling a footing’s width on sand raises q_u by much less than double, so it is usually more effective to found deeper than to spread wider. One caution worth carrying into the exam: there is no single N_γ — Terzaghi, Meyerhof, Hansen and Vesic publish different expressions differing by up to 50 per cent, so a question must supply its factors or name the author, and an answer computed from unattributed factors cannot be checked.
For a footing on sand, raising the water table from well below the footing to the ground surface reduces the ultimate bearing capacity by roughly
1. 10%
2. 25%
3. 50%
4. nothing, since φ is unchanged
Show answer
Answer: C — 50%
About 50 per cent. Submergence roughly halves the effective unit weight — from 18 to 18 − 9.81 = 8.19 kN/m³ in the worked example — and for a sand two of the three terms in the bearing capacity equation are directly proportional to it. On the footing used in this chapter, B = 2 m at D_f = 1.5 m with φ = 30°: with the water table deep, q_u = 900.1 kPa; at foundation level, only the width term is submerged and q_u = 680.3 kPa, down 24%; at the ground surface, both the surcharge and the width terms are submerged and q_u = 409.5 kPa, down 54.5%. Option B is right for the intermediate case and is the answer that comes from stopping halfway. Option D is the trap, and it is worth answering carefully because its premise is true: φ really is unchanged, and the sand really is exactly as strong as it was. What has changed is not strength but weight — the surcharge that must be lifted for the failure wedge to escape, and the self-weight of the wedge itself, are both buoyed by the water. Bearing capacity is not a material property; it is a balance between a failure mechanism and the weight resisting it. Three practical consequences. A design must state the water table it assumed, since the same soil gives two answers a factor of two apart. The level to assume is the seasonal high, not the one found on the day of the investigation. And a water table more than about B below the footing is usually taken as having no effect at all, because the failure wedge does not extend that deep — with intermediate depths interpolated between the two extremes.
A smooth vertical wall 5 m high retains a cohesionless backfill with γ = 18 kN/m³ and φ = 30°, level at the top and dry. The total active thrust in kN per metre run is ______
Numerical answer — type the value.
Show answer
Answer: 75
75 kN/m. K_a = (1 − sin φ)/(1 + sin φ) = (1 − 0.5)/(1 + 0.5) = 1/3 exactly at φ = 30°, which is why that angle is so common in question papers. Then P_a = 0.5 K_a γ H² = 0.5(1/3)(18)(25) = 75 kN/m, acting at H/3 = 1.667 m above the base, since the pressure distribution is triangular. Compare the other two states of the same soil against the same wall. At rest, with K₀ = 1 − sin φ = 0.5, the thrust is 112.5 kN/m — half as much again. Passive, with K_p = 1/K_a = 3, it is 675 kN/m, nine times the active value. That factor of nine is the real content of the topic, and it is a factor of K_p/K_a = tan⁴(45 + φ/2), so it grows sharply with φ. What decides which state applies is not the soil but how much the wall is allowed to move: a movement of a fraction of a per cent of the wall height is enough to mobilise the full active state, while reaching full passive resistance needs a movement roughly ten times larger, usually far more than a structure can tolerate. That asymmetry is why design takes active pressure at its full value while heavily discounting passive resistance, and why a wall built rigidly enough not to move at all — a basement wall keyed into a floor slab — must be designed for the at-rest pressure and not the active, a 50 per cent larger load that is a common and expensive omission.
An infinite slope of dry cohesionless soil has φ = 30° and a slope angle of 20°. The factor of safety is ______ (round to three decimals)
Numerical answer — type the value.
Show answer
Answer: 1.586
1.586. For an infinite slope in dry cohesionless soil the factor of safety is F = tan φ/tan β = tan 30°/tan 20° = 0.5774/0.3640 = 1.5863. What makes the result striking is everything that cancelled out of it: the depth of the failure plane, the unit weight of the soil and the length of the slope all disappear, leaving two angles. So a dry cohesionless slope is stable if and only if it is flatter than the friction angle — regardless of how deep the potential slip surface is, which is why such a slope has no critical depth and why sand tips stand at their angle of repose whatever their size. Now add water and the picture changes completely. With seepage parallel to the slope surface, which is what a hillside experiences during prolonged rain, the pore pressure on the failure plane rises, the effective stress falls, and the factor of safety is multiplied by γ′/γ_sat. With γ_sat = 20 kN/m³ that ratio is 10.19/20 = 0.5095, so F drops from 1.586 to 0.808 — and the slope that was comfortably stable dry now fails. Seepage has removed almost exactly half the safety margin with no change in geometry, soil or load. This is the mechanism behind the familiar observation that slope failures cluster after heavy rainfall rather than during it: the trigger is the rise of the water table, which lags the rain. It is also why drainage is usually the cheapest stabilisation available — weep holes, counterfort drains, horizontal bores all attack the one term doing the damage — and why a slope must be analysed for the worst credible groundwater condition rather than the level measured on a dry day.
A bored pile 0.4 m in diameter and 10 m long is installed in clay with c_u = 50 kPa and adhesion factor α = 0.5. Taking N_c = 9 for the base, the ultimate capacity in kN is ______ (round to one decimal)
Numerical answer — type the value.
Show answer
Answer: 370.7
370.7 kN. Two contributions. Shaft: Q_s = α c_u (π d L) = 0.5(50)(π)(0.4)(10) = 314.2 kN. Base: Q_p = N_c c_u A_p = 9(50)(π × 0.4²/4) = 9(50)(0.1257) = 56.5 kN. Total 370.7 kN. The split is the part worth taking away: the shaft carries 84.7 per cent of the capacity and the base only 15 per cent, which is typical for a pile in clay and is the opposite of the picture most people carry, in which a pile is imagined as a column standing on something firm. Several pieces of practice follow from it. Capacity is roughly proportional to length, since the shaft term contains L, while the base term does not — so a pile cast 1 m short loses about 31 kN of shaft resistance and is a real defect, whereas a small under-run in diameter matters much less. The condition of the shaft matters more than the cleanliness of the base, which is why the smear and softening left by boring is captured in the adhesion factor α, and why α falls as the clay gets stiffer, from about 1.0 in a soft clay to 0.4 or less in a stiff one — the stiffer clay is stronger but the pile cannot mobilise as much of it. And a group needs a second calculation: the design capacity is the lesser of n times the single-pile value and the capacity of the whole group failing as a block. For nine of these piles at 3d spacing the block gives 9128 kN against 3336 kN for the sum, so individual failure governs and the group efficiency is effectively 1 at that spacing.
Negative skin friction on a pile
1. reduces the ultimate capacity and is subtracted from the resistance
2. acts downward and is added to the applied load
3. occurs only in cohesionless soils
4. is beneficial, since the soil grips the pile more tightly
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Answer: B — acts downward and is added to the applied load
It acts downward and is added to the applied load. Skin friction takes its sign from the relative movement between pile and soil. Normally the pile settles more than the surrounding ground, the soil resists that movement, and friction acts upward as useful resistance. But if the ground settles more than the pile — a fill placed recently over compressible soil, a clay layer still consolidating, a water table lowered by pumping — the soil moves down past the pile, hangs on the shaft, and drags it down. Option A is the trap and it is a matter of bookkeeping rather than semantics: the pile’s resistance has not changed at all, since the soil is just as strong as before. What has changed is the demand, so the drag force belongs on the load side of the check, and treating it as a capacity reduction gets a different and unconservative answer whenever a partial factor is applied to only one side. Option C has it backwards — negative skin friction is most severe in soft cohesive soils and recent fills, precisely because those are the deposits that go on settling for years after the pile is installed. Option D inverts the effect entirely. The standard remedies follow from the mechanism: coat the upper part of the shaft with bitumen so the soil slides past instead of gripping, sleeve it, or simply wait for the fill to consolidate before installing the piles. And the depth over which it acts is bounded by the neutral plane, the level at which pile and soil settle equally — above it friction drags down, below it friction resists as usual.
Which of the following are correct about foundations on clay in the undrained condition (φ_u = 0)? (One or more)
1. The width term of the bearing capacity equation vanishes
2. Widening the footing does not increase the ultimate bearing pressure
3. Settlement, not bearing failure, often governs the design
4. The bearing capacity increases with the square of the width
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Answer: A — The width term of the bearing capacity equation vanishes; B — Widening the footing does not increase the ultimate bearing pressure; C — Settlement, not bearing failure, often governs the design
A, B and C. With φ_u = 0 the bearing capacity factors collapse to N_c = 5.14, N_q = 1 and N_γ = 0. A follows immediately: N_γ = 0 kills the 0.5 γ B N_γ term entirely, and it is the only term containing B. B is the same statement read as engineering — since no term depends on the width, the ultimate bearing pressure of a footing on undrained clay is independent of how wide it is. Widening still reduces the applied pressure for a given column load, so it still helps, but it does not raise the capacity per unit area, and that is a genuinely counter-intuitive result worth having met once. C is the design reality and the reason this question exists: a footing on a stiff clay usually has a comfortable margin against bearing failure and is limited instead by settlement, both immediate and consolidation, so the calculation that governs is the one from the previous chapter rather than this one. It also connects back to the pressure bulb — a wider foundation reaches deeper, so widening it to reduce the pressure can increase the settlement by involving soft layers a narrow footing never reached. D is false and is the inversion the question is testing: the capacity does not grow with B at all under φ_u = 0, let alone with B²; even for a sand, where the width term survives, capacity is proportional to B and not to its square. The general habit worth forming is to look at which terms survive when a parameter goes to zero — c = 0 for a sand kills the cohesion term, φ = 0 for undrained clay kills the width term, and each simplification tells you what actually governs.