Foundation Engineering

What to remember

• Bearing capacity (Terzaghi, strip): qu = c Nc + γ Df Nq + 0.5 γ B Nγ. For φ = 0, Nc = 5.7, Nq = 1, Nγ = 0. Square footing uses 1.3 and 0.4; circular uses 1.3 and 0.3.
• Safe load = lower of two criteria: shear failure (ultimate capacity divided by a factor of safety, normally 3) and permissible settlement.
• Deep foundations: pile capacity Qu = end bearing + skin friction; pile group efficiency, negative skin friction, well foundations and caissons are standard exam topics.

1. Types of foundations

• Shallow foundations (Df ≤ about B): isolated footing, strip (wall) footing, combined footing, strap footing and raft (mat).
• Deep foundations (Df much greater than B): piles, piers, well foundations (wells) and caissons.

A good foundation must be safe against shear failure and its settlement, particularly differential settlement, must be within permissible limits. The depth must be below the zone of seasonal moisture change, scour, frost and filled ground.

Rankine's minimum depth of foundation: Df = (q/γ) [(1 − sinφ)/(1 + sinφ)]². Example: q = 100 kPa, γ = 20 kN/m³, φ = 30°: Df = 5 × (1/9) = 0.56 m.

2. Bearing capacity

Terminology:

• Ultimate bearing capacity qu: the pressure at which the soil fails in shear.
• Net ultimate qnu = qu − γ Df (the overburden removed).
• Net safe bearing capacity qns = qnu / F.
• Safe bearing capacity qs = qns + γ Df.
• Allowable bearing pressure: the pressure allowed after considering both shear and settlement.

Usual factor of safety F = 3 for shear failure.

Terzaghi's equations (general shear failure):

Footing	qu
Strip	c Nc + γ Df Nq + 0.5 γ B Nγ
Square	1.3 c Nc + γ Df Nq + 0.4 γ B Nγ
Circular	1.3 c Nc + γ Df Nq + 0.3 γ B Nγ

Nc, Nq, Nγ depend only on φ and increase with φ. For purely cohesive soil, Terzaghi gives Nc = 5.7 (Prandtl's value is 5.14).

Example (clay, φ = 0): strip footing, c = 40 kPa, Df = 1 m, γ = 18 kN/m³: qu = 40 × 5.7 + 18 = 246 kPa. qnu = 228 kPa. With F = 3, qns = 76 kPa and qs = 94 kPa. Another example: qnu = 450, F = 3, γDf = 20: qs = 150 + 20 = 170 kPa.

Local shear failure (loose sand, soft clay): use reduced strengths c' = 2c/3 and tanφ' = (2/3) tanφ.

Modes of failure: general shear (dense sand, stiff clay; sudden with bulging), local shear and punching shear (loose soils).

Water table: a high water table reduces capacity. The correction factor Rw1 applies to the surcharge term: it is 0.5 when the water table is at ground level and 1.0 when it is at depth Df or more. The factor Rw2 applies to the width term: it is 0.5 when the water table is at the footing base and 1.0 when it is at depth B or more below the base. Use γ' (submerged) below the water table.

IS 6403 form: qnu = c Nc sc dc ic + q (Nq − 1) sq dq iq + 0.5 B γ Nγ sγ dγ iγ Rw, with shape (s), depth (d) and inclination (i) factors (Meyerhof/Hansen). Bearing capacity of sand increases with footing width B; for clay (φ = 0) the ultimate bearing capacity is independent of B.

Eccentric loading: qmax = (P/BL)(1 + 6e/B) and qmin = (P/BL)(1 − 6e/B); e ≤ B/6 for no tension. Example: P = 1200 kN, B = L = 3 m, e = 0.3 m: qmax = 133.3 × 1.6 = 213.3 kPa.

3. Settlement and field tests

Settlement types: immediate (elastic), primary consolidation and secondary compression. Differential settlement and tilt are more harmful than uniform settlement.

Standard Penetration Test (SPT): a split spoon sampler (51 mm outer diameter) is driven by a 63.5 kg hammer falling freely 750 mm. The blows for the first 150 mm (seating) are ignored; the number of blows for the next 300 mm is N. Corrections: overburden pressure and dilatancy. For N above 15 in saturated fine or silty sand: N' = 15 + 0.5 (N − 15). Example: N = 25 gives N' = 20. SPT is mostly used in sands; results correlate with φ and bearing capacity.

Plate Load Test: a square plate (usually 300 to 750 mm) is loaded in increments on a prepared pit; load-settlement curve gives bearing capacity.

• Clay: ultimate bearing capacity is independent of size; settlement is proportional to width: Sf = Sp (Bf/Bp). Example: Bp = 0.3 m, Bf = 1.5 m, Sp = 12 mm: Sf = 60 mm.
• Sand: bearing capacity increases with width; settlement: Sf = Sp [Bf (Bp + 0.3) / (Bp (Bf + 0.3))]². Example: Bp = 0.3, Bf = 1.5, Sp = 10 mm: ratio = 0.9/0.54 = 1.667, Sf = 10 × 2.78 = 27.8 mm.
• Limitation: the plate stresses only shallow depth, so it cannot capture deep compressible layers.

Soil exploration: boreholes by auger, wash boring, percussion or rotary methods. Depth of exploration is generally about 1.5 times the footing width below the foundation level (more for large layouts). Undisturbed samples use thin-walled tubes. Area ratio Ar = (Do² − Di²)/Di² × 100 and it should not exceed about 10% for good undisturbed samples. Example: Do = 110 mm, Di = 100 mm: Ar = 21%, too high.

4. Combined footings, rafts and retaining walls

• Combined footing: supports two or more columns; used when columns are close or one is near the property line. Its centroid should coincide with the resultant of column loads. Example: loads 600 kN and 400 kN, spacing 4 m: resultant 1.6 m from the heavier load.
• Strap (cantilever) footing: two footings joined by a strap beam; used when an edge column cannot be centred on its footing.
• Raft: used for weak soil or high loads (when isolated footings would cover much of the plan area), and it reduces differential settlement. A floating raft is one in which the weight of excavated soil equals the building weight so the net pressure is almost zero.

Retaining wall stability:

Check	Formula
Overturning	FOS = ΣMresisting / ΣMoverturning
Sliding	FOS = μ ΣV / ΣH
Bearing	maximum base pressure ≤ allowable pressure
Tension at base	eccentricity e ≤ B/6 (middle third)

Example: resisting moment 600, overturning 300: FOS = 2. Example: μ = 0.5, ΣV = 400, ΣH = 100: FOS = 2. Check the current code for required factors.

5. Pile foundations

Piles transfer load through weak soil to a firmer layer (end-bearing) or by skin friction (friction piles). Types: driven precast, driven cast-in-situ (such as Franki), bored cast-in-situ, steel H-piles and timber piles. Bored piles cause little vibration and are useful near existing buildings.

Static capacity: Qu = Qp + Qs.

• End bearing in clay: Qp = Ap c Nc with Nc = 9. In sand: Qp = Ap q Nq.
• Skin friction in clay (adhesion method): Qs = α cu As, where α is the adhesion factor. Example: cu = 50 kPa, α = 0.5, As = 12 m²: Qs = 300 kN. Example: Ap = 0.1 m², qp = 1000 kPa, As = 10 m², fs = 50 kPa: Qu = 100 + 500 = 600 kN.
• Safe load = Qu / FOS, with a factor of safety of 2.5 for static formula in common practice.

Dynamic (driving) formulae: Engineering News formula: Qa = W H / (6 (S + C)), where W is hammer weight, H drop height, S set per blow and C = 2.5 cm for a drop hammer (0.25 cm for steam hammer) when lengths are in cm. Example: W = 20 kN, H = 150 cm, S = 0.5 cm: Qa = 20 × 150 / (6 × 3) = 166.7 kN. Hiley's formula is more rational as it includes energy losses.

Pile groups: group efficiency = group capacity / (n × single-pile capacity). Friction piles in clay may fail as a block with efficiency below 1. Spacing for friction piles is generally about 3 times the diameter; closer spacing overlaps stress zones. Example: 9 piles of 300 kN each at efficiency 0.8 give 2160 kN.

Negative skin friction (down drag): occurs when surrounding soil settles more than the pile (for example fill placed on soft clay). It adds load to the pile instead of resisting it. A pile load test on a test pile gives the real capacity.

6. Wells, caissons and cofferdams

Well foundations are used for bridge piers in rivers. Parts: cutting edge, well curb, steining (the wall), bottom plug, sand filling, top plug and well cap. The well is sunk by dredging inside. The grip length is the depth below the maximum scour level. Lacey's regime scour depth: R = 0.47 (Q/f)^(1/3), where silt factor f = 1.76 √d (d in mm). For d = 4 mm, f = 3.52. The design scour depth is taken as a multiple of R (for example 2R near a pier nose or at a bend).

Caissons: open (well), box and pneumatic. A pneumatic caisson uses compressed air to keep water out of the working chamber; workers face decompression sickness. Cofferdam: a temporary enclosure that allows excavation below water.

Sheet piles: cantilever sheet piles for small heights; anchored bulkheads for larger retained heights.

Exam traps

• Nc = 5.7 in Terzaghi for φ = 0, but Prandtl/Skempton use 5.14.
• Square footing coefficients are 1.3 and 0.4; circular are 1.3 and 0.3.
• Net safe capacity is qnu/F; safe capacity adds γ Df.
• In clay, size does not change bearing capacity; in sand it does.
• Settlement ratio in clay is Bf/Bp, not squared.
• SPT first 150 mm is seating and not counted.
• Engineering News formula uses a factor of 6 and a constant C.
• Negative skin friction increases load on the pile.

One-liners

• 1. Terzaghi strip: qu = c Nc + γ Df Nq + 0.5 γ B Nγ.
• 2. Terzaghi shape factors: square 1.3 and 0.4; circular 1.3 and 0.3.
• 3. Local shear uses c' = 2c/3.
• 4. Net safe bearing capacity = qnu / 3 in common practice.
• 5. SPT uses a 63.5 kg hammer with a 750 mm fall.
• 6. Plate load test: clay settlement scales with width.
• 7. Pile capacity Qu = Qp + Qs.
• 8. Nc = 9 is used for end bearing of piles in clay.
• 9. Group efficiency = group capacity / (n × single pile capacity).
• 10. Well foundation parts include cutting edge, curb and steining.
• 11. Pneumatic caissons work under compressed air.
• 12. Middle third rule: e ≤ B/6.

Practice questions

1. In Terzaghi's equation for a strip footing, the bearing capacity factor Nc for φ = 0 is:

   1. 1.0
   2. 5.7
   3. 9.0
   4. 0
   Answer

   B. 5.7

   Terzaghi: Nc = 5.7, Nq = 1, Nγ = 0 for φ = 0.

2. Terzaghi's ultimate bearing capacity for a square footing is:

   1. 1.3 c Nc + γ Df Nq + 0.4 γ B Nγ
   2. 1.3 c Nc + 0.4 γ Df Nq + γ B Nγ
   3. c Nc + γ Df Nq + 0.5 γ B Nγ
   4. 1.3 c Nc + γ Df Nq + 0.3 γ B Nγ
   Answer

   A. 1.3 c Nc + γ Df Nq + 0.4 γ B Nγ

   The square footing uses shape factors 1.3 and 0.4.

3. In Terzaghi's equation for a circular footing, the width term has the coefficient:

   1. 0.4 γ B Nγ
   2. 1.3 γ B Nγ
   3. 0.5 γ B Nγ
   4. 0.3 γ B Nγ
   Answer

   D. 0.3 γ B Nγ

   Circular footing: 1.3 c Nc + γ Df Nq + 0.3 γ B Nγ.

4. A strip footing on clay (φ = 0) has c = 40 kPa, Df = 1 m and γ = 18 kN/m³. With a factor of safety of 3 on net capacity, the net safe bearing capacity is:

   1. 246 kPa
   2. 228 kPa
   3. 76 kPa
   4. 94 kPa
   Answer

   C. 76 kPa

   qnu = 40 × 5.7 = 228; qns = 228/3 = 76 kPa.

5. For the footing above, the safe bearing capacity (qns + γ Df) is:

   1. 82 kPa
   2. 94 kPa
   3. 246 kPa
   4. 76 kPa
   Answer

   B. 94 kPa

   76 + 18 × 1 = 94 kPa.

6. For local shear failure Terzaghi suggests using the reduced cohesion:

   1. c' = 3c/2
   2. c' = c/3
   3. c' = c/2
   4. c' = 2c/3
   Answer

   D. c' = 2c/3

   Also tanφ' = (2/3) tanφ.

7. When the water table rises to ground level, the correction factor Rw1 on the surcharge term is:

   1. 2.0
   2. 0.5
   3. 0.25
   4. 1.0
   Answer

   B. 0.5

   Rw1 = 0.5 at ground level and 1.0 when the water table is at depth Df or more.

8. If the net ultimate bearing capacity is 450 kPa, factor of safety is 3 and γ Df = 20 kPa, the safe bearing capacity is:

   1. 170 kPa
   2. 130 kPa
   3. 470 kPa
   4. 150 kPa
   Answer

   A. 170 kPa

   qs = 450/3 + 20 = 170 kPa.

9. The allowable bearing pressure of a foundation soil is governed by:

   1. Shear failure only
   2. Both shear failure and permissible settlement
   3. Settlement only
   4. Water content only
   Answer

   B. Both shear failure and permissible settlement

   The smaller of the shear-based and settlement-based pressures governs.

10. In the Standard Penetration Test, the weight of the hammer and its free fall are:

    1. 65 kg and 500 mm
    2. 75 kg and 1000 mm
    3. 63.5 kg and 750 mm
    4. 2.6 kg and 310 mm
    Answer

    C. 63.5 kg and 750 mm

    The SPT hammer is 63.5 kg dropping 750 mm.

11. The SPT N value is the number of blows for the penetration of:

    1. Last 300 mm of a 450 mm drive
    2. The full 450 mm
    3. The first 300 mm
    4. First 150 mm
    Answer

    A. Last 300 mm of a 450 mm drive

    The first 150 mm is seating and is not counted.

12. For saturated fine sand with a measured SPT N = 25 (N > 15), the dilatancy correction N' = 15 + 0.5(N − 15) gives:

    1. 10
    2. 15
    3. 25
    4. 20
    Answer

    D. 20

    15 + 0.5 × 10 = 20.

13. For a plate load test on clay, the ultimate bearing capacity of the actual footing compared with the test plate is:

    1. Decreases in proportion to width
    2. Is doubled
    3. Nearly the same (independent of size)
    4. Increases in proportion to width
    Answer

    C. Nearly the same (independent of size)

    For φ = 0 soils bearing capacity does not depend on B.

14. A 0.3 m plate settles 10 mm on sand. For a 1.5 m footing at the same pressure, Sf = Sp[Bf(Bp + 0.3)/(Bp(Bf + 0.3))]² gives:

    1. 10 mm
    2. 50 mm
    3. 16.7 mm
    4. 27.8 mm
    Answer

    D. 27.8 mm

    [1.5 × 0.6/(0.3 × 1.8)]² = (1.667)² = 2.78; Sf = 27.8 mm.

15. A 0.3 m plate settles 12 mm on clay. For a 1.5 m footing at the same pressure the settlement is:

    1. 12 mm
    2. 60 mm
    3. 2.4 mm
    4. 300 mm
    Answer

    B. 60 mm

    Sf = Sp × Bf/Bp = 12 × 5.

16. Piles that transfer most load to a hard stratum at the tip are called:

    1. End-bearing piles
    2. Friction piles
    3. Tension piles
    4. Batter piles
    Answer

    A. End-bearing piles

    Tip resistance dominates in end-bearing piles.

17. A pile has end area 0.1 m² with end resistance 1000 kPa and shaft area 10 m² with skin friction 50 kPa. Its ultimate capacity is:

    1. 100 kN
    2. 500 kN
    3. 600 kN
    4. 1500 kN
    Answer

    C. 600 kN

    Qu = 0.1 × 1000 + 10 × 50 = 100 + 500.

18. For end bearing of a pile in clay, the bearing capacity factor commonly used is:

    1. Nc = 1
    2. Nc = 20
    3. Nc = 5.7
    4. Nc = 9
    Answer

    D. Nc = 9

    Qp = Ap c Nc with Nc = 9.

19. A pile with cu = 50 kPa, adhesion factor α = 0.5 and shaft area 12 m² has skin friction capacity:

    1. 600 kN
    2. 150 kN
    3. 300 kN
    4. 75 kN
    Answer

    C. 300 kN

    Qs = α cu As = 0.5 × 50 × 12.

20. A drop hammer of 20 kN falls 150 cm; the set is 0.5 cm per blow. By the Engineering News formula Qa = WH/[6(S + 2.5)] the safe load is nearly:

    1. 100 kN
    2. 166.7 kN
    3. 333 kN
    4. 1000 kN
    Answer

    B. 166.7 kN

    20 × 150/(6 × 3) = 3000/18 = 166.7 kN.

21. Nine piles each with a single-pile capacity of 300 kN act as a group with efficiency 0.8. The group capacity is:

    1. 2160 kN
    2. 240 kN
    3. 3375 kN
    4. 2700 kN
    Answer

    A. 2160 kN

    0.8 × 9 × 300 = 2160 kN.

22. Friction piles in a group are usually spaced centre to centre at about:

    1. 3 times the pile diameter
    2. 1 times the diameter
    3. 6 times the diameter
    4. 10 times the diameter
    Answer

    A. 3 times the pile diameter

    Closer spacing causes overlapping of stressed zones.

23. Negative skin friction on a pile arises when:

    1. Water table falls to a hard layer
    2. The pile settles more than the soil
    3. The surrounding soil settles more than the pile
    4. The pile is hollow
    Answer

    C. The surrounding soil settles more than the pile

    Downdrag adds load to the pile, for example fill on soft clay.

24. Which of the following is a driven cast-in-situ pile?

    1. Timber pile
    2. Franki pile
    3. Steel H-pile
    4. Precast concrete pile
    Answer

    B. Franki pile

    Franki piles are formed by driving a casing and ramming concrete.

25. The cutting edge of a well foundation is located:

    1. At the top of the well cap
    2. In the middle of steining
    3. Above water level
    4. At the bottom of the well curb
    Answer

    D. At the bottom of the well curb

    The cutting edge helps the well sink under its own weight.

26. The grip length of a well foundation is the depth:

    1. Of the top plug
    2. Below the maximum scour level
    3. Above low water level
    4. Of the well cap
    Answer

    B. Below the maximum scour level

    A grip below scour level ensures stability against lateral loads.

27. By Lacey's formula f = 1.76 √d (d in mm), the silt factor for d = 4 mm is:

    1. 1.76
    2. 0.88
    3. 3.52
    4. 7.04
    Answer

    C. 3.52

    f = 1.76 × 2 = 3.52.

28. Rankine's minimum depth of foundation for q = 100 kPa, γ = 20 kN/m³ and φ = 30° is nearly:

    1. 1.0 m
    2. 1.67 m
    3. 5.0 m
    4. 0.56 m
    Answer

    D. 0.56 m

    Df = (q/γ)((1 − sinφ)/(1 + sinφ))² = 5 × 1/9 = 0.56 m.

29. A raft foundation is generally preferred when:

    1. The soil is solid rock
    2. The soil is weak and isolated footings would cover a large part of the plan
    3. Only one light column is present
    4. The water table is very low and loads small
    Answer

    B. The soil is weak and isolated footings would cover a large part of the plan

    A raft spreads load widely and reduces differential settlement.

30. In a floating (compensated) raft, the net pressure on the soil is nearly zero because:

    1. The raft is made very thick
    2. Piles carry all loads
    3. The water table is very high only
    4. The weight of excavated soil equals the weight of the building
    Answer

    D. The weight of excavated soil equals the weight of the building

    Excavation removes soil weight comparable to the load applied.

31. A footing for two columns with one of them on the property line is best solved by:

    1. A strap (cantilever) footing
    2. Two isolated square footings
    3. A pedestal
    4. A grillage only
    Answer

    A. A strap (cantilever) footing

    The strap beam transfers the eccentric moment.

32. A 3 m × 3 m footing carries 1200 kN with eccentricity 0.3 m along one side. The maximum pressure is:

    1. 133.3 kPa
    2. 160 kPa
    3. 213.3 kPa
    4. 266.7 kPa
    Answer

    C. 213.3 kPa

    (1200/9)(1 + 6 × 0.3/3) = 133.3 × 1.6.

33. For a retaining wall with resisting moment 600 kN·m and overturning moment 300 kN·m, the factor of safety against overturning is:

    1. 2.0
    2. 1.5
    3. 3.0
    4. 0.5
    Answer

    A. 2.0

    FOS = 600/300.

34. A wall with μ = 0.5, ΣV = 400 kN and ΣH = 100 kN has a factor of safety against sliding of:

    1. 4.0
    2. 0.5
    3. 1.25
    4. 2.0
    Answer

    D. 2.0

    FOS = μΣV/ΣH = 200/100.

35. For no tension at the base of a wall of base width 3 m, the resultant must lie within:

    1. e ≤ 1.5 m
    2. The middle third, so e ≤ 0.5 m
    3. The first third, e ≤ 1 m
    4. The middle half
    Answer

    B. The middle third, so e ≤ 0.5 m

    e ≤ B/6 = 0.5 m.

36. A caisson that uses compressed air to keep water out of the working chamber is a:

    1. Pneumatic caisson
    2. Open caisson
    3. Floating caisson
    4. Box caisson
    Answer

    A. Pneumatic caisson

    Pneumatic caissons allow excavation in deep water below river beds.

37. The area ratio of a sampler with outer diameter 110 mm and inner cutting-edge diameter 100 mm is:

    1. 11%
    2. 10%
    3. 21%
    4. 1.1%
    Answer

    C. 21%

    Ar = (110² − 100²)/100² × 100 = 21%.

38. Rotary drilling methods are mostly used for exploring:

    1. Very soft clays only
    2. Surface soil only
    3. Peat only
    4. Rock and hard strata
    Answer

    D. Rock and hard strata

    Rotary core drilling recovers rock cores.

39. The depth of exploration below an isolated footing is generally taken as about:

    1. 0.5 m
    2. 1.5 times the footing width
    3. 10 times the width
    4. 0.1 times the width
    Answer

    B. 1.5 times the footing width

    Stress influence extends to about 1.5 B.

40. Which of the following statements are correct? 1. In sand, ultimate bearing capacity increases with footing width. 2. In clay with φ = 0, ultimate bearing capacity increases with footing width.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    For φ = 0 soils bearing capacity is independent of width.

41. Which of the following statements are correct? 1. Piles transfer loads by end bearing and skin friction. 2. The group efficiency of friction piles in clay can be less than one.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    C. Both 1 and 2

    Both are correct; block failure can control close groups.

42. Which of the following statements are correct? 1. Settlement of foundations includes immediate, consolidation and secondary components. 2. Immediate settlement in saturated clay is due to expulsion of pore water.

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Answer

    A. 1 only

    Immediate settlement is mostly elastic distortion at constant volume; pore water expulsion causes consolidation settlement.

43. Which of the following pairs is correctly matched?

    1. Pile – cutting edge
    2. Raft – grip length
    3. Well foundation – steining
    4. Strap footing – Franki
    Answer

    C. Well foundation – steining

    Steining is the wall of a well; cutting edge belongs to wells.

44. In the Terzaghi equation, increasing the angle of shearing resistance φ results in:

    1. Lower Nq only
    2. Higher values of Nc, Nq and Nγ
    3. No change in any factor
    4. Lower Nc and higher Nq
    Answer

    B. Higher values of Nc, Nq and Nγ

    All three factors increase with φ.

45. Which of the following gives the net ultimate bearing capacity?

    1. qu − 2γ Df
    2. qu + γ Df
    3. qu / 3
    4. qu − γ Df
    Answer

    D. qu − γ Df

    The overburden pressure at foundation level is subtracted.