Soil Mechanics

What to remember

• Phase relations: e = Vv/Vs, n = e/(1+e), S e = w G, γd = G γw/(1+e). Learn the four unit weights: bulk, dry, saturated and submerged.
• Effective stress principle: σ' = σ − u. Consolidation, shear strength and settlement all depend on effective stress.
• Tests and values: Atterberg limits (LL, PL, SL), Darcy permeability, Proctor compaction, Terzaghi consolidation (Tv = cv t / H²), Mohr-Coulomb shear strength (τ = c + σ tanφ) and Rankine earth pressure (Ka = (1 − sinφ)/(1 + sinφ)).

1. Soil composition and phase relations

Soil has three phases: solids, water and air. Volumes: Vs (solids), Vw (water), Va (air), Vv = Vw + Va (voids), V = total.

Term	Definition	Relation
Void ratio e	Vv / Vs	e = n/(1 − n)
Porosity n	Vv / V	n = e/(1 + e)
Water content w	Ww / Ws	S e = w G
Degree of saturation S	Vw / Vv	0 (dry) to 1 (saturated)
Specific gravity G	solids density / water density	usually 2.60 to 2.80

Unit weights (γw = 9.81 kN/m³):

• Bulk: γ = (G + S e) γw / (1 + e)
• Dry: γd = G γw / (1 + e) = γ / (1 + w)
• Saturated: γsat = (G + e) γw / (1 + e)
• Submerged: γ' = γsat − γw = (G − 1) γw / (1 + e)

Example: G = 2.65, e = 0.65: γd = 2.65 × 9.81/1.65 = 15.76 kN/m³; γsat = 3.30 × 9.81/1.65 = 19.62 kN/m³; γ' = 9.81 kN/m³. Example: fully saturated soil, G = 2.7, w = 20%: e = wG/S = 0.54.

Relative density (granular soil): Dr = (emax − e)/(emax − emin).

2. Index properties and classification

Particle size (IS): gravel above 4.75 mm; sand 4.75 to 0.075 mm; silt 0.075 to 0.002 mm; clay below 0.002 mm. Sieve analysis is used for coarse particles and hydrometer analysis for fines.

• Uniformity coefficient Cu = D60/D10. Coefficient of curvature Cc = D30² / (D10 D60). Example: D10 = 0.1, D30 = 0.3, D60 = 0.6 mm gives Cu = 6 and Cc = 1.5.
• Well-graded gravel: Cu ≥ 4 and Cc between 1 and 3. Well-graded sand: Cu ≥ 6 and Cc between 1 and 3.

Atterberg limits (fine soil consistency): shrinkage limit (SL), plastic limit (PL), liquid limit (LL). Liquid limit is found by the Casagrande apparatus (25 blows) or the cone penetrometer; plastic limit by rolling a 3 mm thread.

• Plasticity index PI = LL − PL.
• Liquidity index LI = (w − PL)/PI. Consistency index CI = (LL − w)/PI.
• Activity A = PI / (percentage of clay-size particles). Inactive below 0.75, normal 0.75 to 1.25, active above 1.25. Montmorillonite is the most active (highly swelling); kaolinite the least; illite is in between.
• Sensitivity St = unconfined strength undisturbed / unconfined strength remoulded.
• Example: LL = 45, PL = 25, w = 35: PI = 20 and LI = 0.5.

IS soil classification: symbols G (gravel), S (sand), M (silt), C (clay), O (organic), Pt (peat), W (well graded), P (poorly graded). Plasticity: L (LL < 35), I (35 to 50), H (> 50). Examples: SW, SP, SM, SC, ML, CI, CH. Coarse-grained soil has more than 50% retained on the 75 micron sieve. The plasticity chart A-line is PI = 0.73 (LL − 20); soils above it are clays, below it silts or organic soils. At LL = 40 the A-line value is 14.6.

3. Permeability and seepage

Darcy's law: v = k i (v is discharge velocity), q = k i A. Hydraulic gradient i = head loss / length. Seepage velocity vs = v/n (always greater than v).

Typical k: clean gravel above 1 cm/s; sand 10⁻¹ to 10⁻³ cm/s; silt 10⁻³ to 10⁻⁶; clay below 10⁻⁷ cm/s.

Test	Soil	Formula
Constant head	Coarse (sand, gravel)	k = Q L/(A h t)
Falling head	Fine (silt, clay)	k = 2.303 (a L/(A t)) log10(h1/h2)

Hazen's formula k = C D10² (D10 in cm, C about 100, k in cm/s). Example: D10 = 0.02 cm gives k = 100 × 0.0004 = 0.04 cm/s.

Stratified deposits: equivalent horizontal permeability kh = Σ(k z)/Σz; vertical kv = Σz/Σ(z/k). Always kh ≥ kv. Example: two layers 1 m thick, k1 = 4 × 10⁻³ and k2 = 1 × 10⁻³ cm/s: kh = 2.5 × 10⁻³, kv = 2/(0.25 + 1) × 10⁻³ = 1.6 × 10⁻³ cm/s.

Flow net: flow lines and equipotential lines intersect at right angles. Discharge q = k H (Nf/Nd). Example: k = 10⁻⁵ m/s, H = 5 m, Nf = 4, Nd = 10: q = 2 × 10⁻⁵ m³/s per metre.

Quick sand condition: when upward gradient reaches critical gradient ic = (G − 1)/(1 + e) ≈ 1, effective stress becomes zero and a cohesionless soil loses strength.

Effective stress: σ' = σ − u. Capillary rise raises effective stress in the zone above the water table. Example: saturated clay with γsat = 19.62 kN/m³, water table at ground level: at 5 m depth σ = 98.1, u = 49.05 and σ' = 49.05 kN/m².

4. Compaction

Compaction removes air voids and increases dry density. The Proctor test relates dry density to water content; the peak gives maximum dry density (MDD) at optimum moisture content (OMC).

Test	Rammer	Drop	Layers	Blows/layer
Standard (light)	2.6 kg	310 mm	3	25
Modified (heavy)	4.9 kg	450 mm	5	25

Mould volume is 1000 cc. Modified compaction gives higher MDD and lower OMC. The zero-air-void line γd = G γw / (1 + w G) is a theoretical upper limit; the compaction curve always lies below it. On the dry side of OMC clay has a flocculated structure; on the wet side a dispersed one. Field rollers: sheepsfoot for clay, vibratory for sand and gravel, pneumatic-tyred for mixed soils. Field control uses the sand replacement or core cutter test.

5. Consolidation

Consolidation is the slow squeezing of water from a saturated clay under load, with a gradual transfer of load from pore water to the soil skeleton.

• Terzaghi's time factor Tv = cv t / H², where H is the drainage path (half the layer thickness for double drainage). U = 50% at Tv = 0.197; U = 90% at Tv = 0.848.
• Settlement: Sc = Cc H/(1 + e0) log10((σ0' + Δσ)/σ0'). Example: Cc = 0.3, H = 2 m, e0 = 1.0, σ0' = 100, Δσ = 100 kPa: Sc = 0.3 × 2/2 × 0.301 = 0.0903 m ≈ 90 mm.
• cv = k / (mv γw). Coefficient of compressibility mv = av/(1 + e0).
• Overconsolidation ratio OCR = σp'/σ0'. OCR = 1 is normally consolidated; above 1 is overconsolidated.
• Time to reach a given degree of consolidation is proportional to H². If drainage changes from one-way to two-way, time becomes one-fourth. Types of settlement: immediate, primary consolidation, secondary compression.

6. Shear strength

Mohr-Coulomb: τf = c + σ' tanφ. Granular soil has c = 0; saturated clay in undrained conditions has φu = 0.

• Failure plane makes angle 45° + φ/2 with the major principal plane.
• Principal stress relation: σ1 = σ3 tan²(45° + φ/2) + 2c tan(45° + φ/2).
• Tests: direct shear (simple, drainage poorly controlled), triaxial (UU, CU, CD), unconfined compression (UCS) and vane shear (soft clay in field). UCS gives cu = qu/2. Vane shear: cu = T / (π (D² H/2 + D³/6)).
• Skempton pore pressure: Δu = B [Δσ3 + A (Δσ1 − Δσ3)]; B = 1 for saturated soil.
• Dense sand dilates; loose sand contracts.

7. Lateral earth pressure and slope stability

Rankine theory (smooth vertical wall, horizontal backfill): Ka = (1 − sinφ)/(1 + sinφ) = tan²(45° − φ/2); Kp = 1/Ka; at rest K0 ≈ 1 − sinφ. For φ = 30°, Ka = 1/3, Kp = 3 and K0 = 0.5.

• Active pressure at depth z (dry cohesionless): Ka γ z. Total active thrust Pa = ½ Ka γ H², acting at H/3 above the base. Example: H = 6 m, γ = 18, Ka = 1/3: Pa = 0.5 × (1/3) × 18 × 36 = 108 kN/m.
• Cohesive soil: tension crack depth z0 = 2c / (γ √Ka). Example: c = 12 kPa, γ = 18, φ = 0: z0 = 1.33 m.
• Coulomb's theory allows wall friction and inclined backface.

Slopes: infinite dry cohesionless slope: FOS = tanφ / tanβ. Example: φ = 30°, β = 15°: FOS = 0.5774/0.2679 ≈ 2.16. Finite slopes: Swedish slip circle, method of slices, Taylor stability number Sn = c / (F γ H).

8. Stress distribution in soil

Boussinesq (point load Q): σz = 0.4775 Q / z² directly below the load; Westergaard gives 0.3183 Q/z². Example: Q = 1000 kN, z = 5 m: σz = 0.4775 × 1000/25 = 19.1 kPa. The 2:1 method spreads the load at a slope of 2 vertical to 1 horizontal. A pressure bulb is a contour of equal vertical stress, usually the 20 percent line (0.2 q) is taken as the limit of significant stress.

Exam traps

• Void ratio can exceed 1 but porosity cannot.
• Seepage velocity is more than discharge velocity (divide by n).
• Horizontal permeability of stratified soil is more than vertical.
• Liquid limit, not shrinkage limit, is the water content at which soil flows; PI = LL − PL.
• Standard Proctor is 2.6 kg/310 mm/3 layers; modified is 4.9 kg/450 mm/5 layers.
• Failure plane angle is 45° + φ/2 with the major principal plane (not with the minor).
• Ka is less than 1 and Kp is more than 1; Kp = 1/Ka.
• Consolidation time varies with H², not H.

One-liners

• 1. S e = w G.
• 2. γd = G γw/(1 + e) and γ' = γsat − γw.
• 3. PI = LL − PL; LI = (w − PL)/PI.
• 4. Activity = PI/ (clay fraction %).
• 5. Montmorillonite is the most active clay mineral.
• 6. Cu = D60/D10; Cc = D30²/(D10 D60).
• 7. Constant head test is for sand; falling head test is for silt and clay.
• 8. Critical hydraulic gradient ic = (G − 1)/(1 + e).
• 9. Effective stress is total stress minus pore water pressure.
• 10. Tv at 50% consolidation = 0.197.
• 11. Mohr-Coulomb: τ = c + σ tanφ.
• 12. Rankine Ka = (1 − sinφ)/(1 + sinφ).

Practice questions

1. The void ratio of a soil is defined as:

   1. Volume of solids / total volume
   2. Volume of voids / total volume
   3. Volume of voids / volume of solids
   4. Volume of water / volume of voids
   Answer

   C. Volume of voids / volume of solids

   e = Vv/Vs; the ratio Vv/V is porosity.

2. A soil has a void ratio of 0.5. Its porosity is:

   1. 0.250
   2. 0.500
   3. 0.667
   4. 0.333
   Answer

   D. 0.333

   n = e/(1 + e) = 0.5/1.5 = 0.333.

3. A fully saturated soil with G = 2.7 has water content 20%. Its void ratio is:

   1. 1.35
   2. 0.54
   3. 0.74
   4. 0.27
   Answer

   B. 0.54

   e = wG/S = 0.20 × 2.7/1 = 0.54.

4. Taking γw = 9.81 kN/m³, a soil with G = 2.65 and e = 0.65 has dry unit weight nearly:

   1. 16.30 kN/m³
   2. 18.90 kN/m³
   3. 19.62 kN/m³
   4. 15.76 kN/m³
   Answer

   D. 15.76 kN/m³

   γd = Gγw/(1 + e) = 25.997/1.65 = 15.76 kN/m³.

5. For G = 2.65 and e = 0.65, the saturated unit weight (γw = 9.81 kN/m³) is:

   1. 9.81 kN/m³
   2. 19.62 kN/m³
   3. 15.76 kN/m³
   4. 22.3 kN/m³
   Answer

   B. 19.62 kN/m³

   γsat = (G + e)γw/(1 + e) = 3.30 × 9.81/1.65 = 19.62 kN/m³.

6. The submerged unit weight of a saturated soil with γsat = 19.62 kN/m³ is:

   1. 10.81 kN/m³
   2. 19.62 kN/m³
   3. 9.81 kN/m³
   4. 29.43 kN/m³
   Answer

   C. 9.81 kN/m³

   γ' = γsat − γw = 19.62 − 9.81.

7. A soil with bulk unit weight 18.9 kN/m³ and water content 5% has a dry unit weight of:

   1. 17.0 kN/m³
   2. 19.8 kN/m³
   3. 16.2 kN/m³
   4. 18.0 kN/m³
   Answer

   D. 18.0 kN/m³

   γd = γ/(1 + w) = 18.9/1.05 = 18.0.

8. The plasticity index of a soil with LL = 45% and PL = 25% is:

   1. 20
   2. 70
   3. 35
   4. 10
   Answer

   A. 20

   PI = LL − PL = 20.

9. A soil has LL = 45, PL = 25 and natural water content 35%. The liquidity index is:

   1. 1.5
   2. 0.5
   3. 0.25
   4. 2.0
   Answer

   B. 0.5

   LI = (w − PL)/PI = 10/20 = 0.5.

10. A clay with PI = 30 and 40% clay-size particles has an activity of:

    1. 1.20
    2. 0.30
    3. 0.75
    4. 1.33
    Answer

    C. 0.75

    Activity = PI/clay % = 30/40 = 0.75.

11. Which of the following statements are correct? 1. Montmorillonite is the most active (swelling) clay mineral. 2. Kaolinite is the least active of the common clay minerals.

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

    C. Both 1 and 2

    Activity increases in the order kaolinite, illite, montmorillonite.

12. A clay has unconfined compressive strength 60 kPa undisturbed and 20 kPa remoulded. Its sensitivity is:

    1. 3
    2. 0.33
    3. 40
    4. 80
    Answer

    A. 3

    St = 60/20 = 3.

13. As per the Indian Standard, sand-size particles are those with diameter between:

    1. 0.0002 mm and 0.002 mm
    2. 4.75 mm and 80 mm
    3. 0.002 mm and 0.075 mm
    4. 0.075 mm and 4.75 mm
    Answer

    D. 0.075 mm and 4.75 mm

    IS classification: gravel above 4.75 mm, silt 0.002 to 0.075 mm.

14. A soil has D10 = 0.1 mm and D60 = 0.6 mm. The uniformity coefficient is:

    1. 0.17
    2. 6
    3. 60
    4. 0.6
    Answer

    B. 6

    Cu = D60/D10 = 6.

15. For D10 = 0.1 mm, D30 = 0.3 mm and D60 = 0.6 mm the coefficient of curvature is:

    1. 0.5
    2. 1.5
    3. 3.0
    4. 6.0
    Answer

    B. 1.5

    Cc = D30²/(D10 D60) = 0.09/0.06 = 1.5.

16. A sand is classed as well graded (SW) when:

    1. Cu ≥ 6 and 1 ≤ Cc ≤ 3
    2. Cu = 1 and Cc = 0
    3. Cu < 4 and Cc > 3
    4. Cu ≥ 4 and Cc > 6
    Answer

    A. Cu ≥ 6 and 1 ≤ Cc ≤ 3

    Well graded sand requires Cu of at least 6 and Cc between 1 and 3.

17. In the IS soil classification, the symbol CH denotes:

    1. Clean sand, hard
    2. Cohesive silt, hard
    3. Clay of high plasticity
    4. Coarse gravel of high density
    Answer

    C. Clay of high plasticity

    C = clay, H = high plasticity (LL > 50).

18. The A-line on the plasticity chart is PI = 0.73 (LL − 20). At LL = 40 the A-line value of PI is:

    1. 29.2
    2. 20.0
    3. 7.3
    4. 14.6
    Answer

    D. 14.6

    0.73 × 20 = 14.6.

19. Using Darcy's law, a soil with k = 10⁻³ cm/s under hydraulic gradient 0.5 has discharge velocity:

    1. 2 × 10⁻³ cm/s
    2. 5 × 10⁻³ cm/s
    3. 5 × 10⁻⁴ cm/s
    4. 10⁻³ cm/s
    Answer

    C. 5 × 10⁻⁴ cm/s

    v = k i = 10⁻³ × 0.5.

20. If discharge velocity is 0.002 cm/s and porosity is 0.4, the seepage velocity is:

    1. 0.005 cm/s
    2. 0.002 cm/s
    3. 0.02 cm/s
    4. 0.0008 cm/s
    Answer

    A. 0.005 cm/s

    vs = v/n = 0.002/0.4 = 0.005 cm/s.

21. Which of the following statements are correct? 1. The constant head permeability test is suited to coarse-grained soils. 2. The falling head test is suited to fine-grained soils.

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

    C. Both 1 and 2

    High-permeability soils give measurable flow under constant head; low-permeability soils need falling head.

22. By Hazen's formula k = 100 D10² (cm/s, D10 in cm), the permeability for D10 = 0.02 cm is:

    1. 0.4 cm/s
    2. 0.2 cm/s
    3. 0.002 cm/s
    4. 0.04 cm/s
    Answer

    D. 0.04 cm/s

    k = 100 × 0.0004 = 0.04 cm/s.

23. Two layers of 1 m each have k1 = 4 × 10⁻³ and k2 = 1 × 10⁻³ cm/s. The equivalent vertical permeability is:

    1. 2.5 × 10⁻³ cm/s
    2. 1.6 × 10⁻³ cm/s
    3. 4.0 × 10⁻³ cm/s
    4. 1.0 × 10⁻³ cm/s
    Answer

    B. 1.6 × 10⁻³ cm/s

    kv = 2/(1/4 + 1/1) × 10⁻³ = 1.6 × 10⁻³.

24. The critical hydraulic gradient for a sand with G = 2.7 and e = 0.7 is:

    1. 1.7
    2. 2.7
    3. 0.5
    4. 1.0
    Answer

    D. 1.0

    ic = (G − 1)/(1 + e) = 1.7/1.7 = 1.

25. A saturated clay has γsat = 19.62 kN/m³ and the water table is at ground level. The effective vertical stress at 5 m depth (γw = 9.81) is:

    1. 49.05 kN/m²
    2. 147.2 kN/m²
    3. 98.1 kN/m²
    4. 19.62 kN/m²
    Answer

    A. 49.05 kN/m²

    σ = 98.1, u = 49.05, σ' = 49.05 kN/m².

26. In a flow net with k = 10⁻⁵ m/s, H = 5 m, Nf = 4 and Nd = 10, the discharge per metre is:

    1. 1.25 × 10⁻⁴ m³/s
    2. 2 × 10⁻⁵ m³/s
    3. 2 × 10⁻⁴ m³/s
    4. 5 × 10⁻⁶ m³/s
    Answer

    B. 2 × 10⁻⁵ m³/s

    q = kH Nf/Nd = 10⁻⁵ × 5 × 0.4.

27. The rammer used in the standard Proctor compaction test weighs:

    1. 4.9 kg
    2. 2.6 kg
    3. 1.0 kg
    4. 6.5 kg
    Answer

    B. 2.6 kg

    Standard: 2.6 kg, 310 mm drop, 3 layers; modified: 4.9 kg.

28. The modified Proctor test uses soil placed in the mould in:

    1. 2 layers of 56 blows each
    2. 5 layers of 10 blows each
    3. 3 layers of 25 blows each
    4. 5 layers of 25 blows each
    Answer

    D. 5 layers of 25 blows each

    Modified: 4.9 kg rammer, 450 mm drop, 5 layers.

29. Which of the following statements are correct? 1. Modified Proctor compaction gives a higher maximum dry density than standard Proctor. 2. Modified Proctor compaction gives a higher optimum moisture content than standard Proctor.

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

    A. 1 only

    Greater energy raises MDD and lowers OMC.

30. The zero air void line represents:

    1. The optimum moisture line
    2. Maximum dry density at low water
    3. Theoretical full saturation (no air) at each water content
    4. The line of 50% saturation
    Answer

    C. Theoretical full saturation (no air) at each water content

    No compaction curve can lie above it.

31. The most suitable field roller for compacting cohesive clays is:

    1. Sheepsfoot roller
    2. Plain steel drum without ballast
    3. Smooth wheel roller
    4. Vibratory roller
    Answer

    A. Sheepsfoot roller

    Sheepsfoot rollers knead clays; vibratory rollers suit sands and gravel.

32. A clay layer takes 2 years to consolidate with drainage on one face. With drainage on both faces, the time becomes:

    1. 4 years
    2. 1 year
    3. 0.5 year
    4. 2 years
    Answer

    C. 0.5 year

    t ∝ H²; drainage path halves so time reduces to one-fourth.

33. The time factor Tv at 50% average consolidation is nearly:

    1. 0.5
    2. 0.197
    3. 0.848
    4. 1.0
    Answer

    B. 0.197

    Tv = 0.197 for U = 50% and 0.848 for U = 90%.

34. A 2 m clay layer has Cc = 0.3, e0 = 1.0, σ0' = 100 kPa and Δσ = 100 kPa. The consolidation settlement is nearly:

    1. 180 mm
    2. 300 mm
    3. 30 mm
    4. 90 mm
    Answer

    D. 90 mm

    Sc = 0.3 × 2/2 × log10 2 = 0.0903 m.

35. A clay with preconsolidation pressure 200 kPa under present overburden 100 kPa has an OCR of:

    1. 0.5
    2. 1
    3. 2
    4. 100
    Answer

    C. 2

    OCR = σp'/σ0' = 2, so it is overconsolidated.

36. The shear strength of soil by the Mohr-Coulomb law is:

    1. τ = c + σ tanφ
    2. τ = c − σ tanφ
    3. τ = c tanφ
    4. τ = σ/ tanφ
    Answer

    A. τ = c + σ tanφ

    Cohesion plus frictional resistance.

37. For a soil with φ = 30°, the failure plane makes what angle with the major principal plane?

    1. 45°
    2. 60°
    3. 75°
    4. 30°
    Answer

    B. 60°

    Angle = 45° + φ/2 = 60°.

38. An unconfined compression test gives qu = 100 kPa for saturated clay. The undrained cohesion cu is:

    1. 25 kPa
    2. 200 kPa
    3. 100 kPa
    4. 50 kPa
    Answer

    D. 50 kPa

    cu = qu/2.

39. For a saturated clay tested in an unconsolidated undrained triaxial test, the angle of shearing resistance φu is:

    1. Zero
    2. 45°
    3. 90°
    4. 30°
    Answer

    A. Zero

    Saturated undrained clay shows φu = 0.

40. For φ = 30°, the Rankine active earth pressure coefficient Ka is:

    1. 3
    2. 1/3
    3. 1/2
    4. 1/9
    Answer

    B. 1/3

    Ka = (1 − sin30°)/(1 + sin30°) = 0.5/1.5.

41. The total Rankine active thrust on a 6 m high wall with γ = 18 kN/m³ and Ka = 1/3 is:

    1. 216 kN/m
    2. 324 kN/m
    3. 108 kN/m
    4. 54 kN/m
    Answer

    C. 108 kN/m

    Pa = 0.5 × (1/3) × 18 × 6² = 108 kN/m.

42. A purely cohesive soil with c = 12 kPa and γ = 18 kN/m³ has a depth of tension crack of:

    1. 0.67 m
    2. 2.67 m
    3. 0.50 m
    4. 1.33 m
    Answer

    D. 1.33 m

    z0 = 2c/(γ√Ka) = 24/18 = 1.33 m (Ka = 1).

43. The factor of safety of an infinite dry cohesionless slope with φ = 30° and slope angle 15° is nearly:

    1. 2.16
    2. 3.46
    3. 0.50
    4. 1.00
    Answer

    A. 2.16

    FOS = tanφ/tanβ = 0.5774/0.2679.

44. For a point load Q = 1000 kN, the Boussinesq vertical stress directly below at depth 5 m is nearly:

    1. 40 kPa
    2. 8.0 kPa
    3. 95.5 kPa
    4. 19.1 kPa
    Answer

    D. 19.1 kPa

    σz = 0.4775 Q/z² = 477.5/25.

45. For a normally consolidated sand with φ = 30°, the coefficient of earth pressure at rest K0 (Jaky, 1 − sinφ) is:

    1. 0.33
    2. 1.0
    3. 0.5
    4. 3.0
    Answer

    C. 0.5

    K0 = 1 − sin 30° = 0.5, between Ka = 1/3 and Kp = 3.