Semester 2

Question 1

How do you find shear strength from a Mohr’s circle?

Answer 1

τ = σ' * tanφ' (+c')   => effective stress, Mohr-Coulomb (drained)
τ = Cu = (σ1-σ3)/2   => total stress, Tresca (undrained)

Question 2

How can u relate passive and earth pressure coefficient

Answer 2

Ka = 1/Kp

Question 3

What is the at rest pressure coefficient?

Answer 3

K0 = σ’h/σ’v (relates in situ h/v effective soil stress)
K0 ≈ 0.45 - 0.6 or 1-sinφ’ (loose sands, NC clays)
K0 ≈ 0.6 - 2 or 1(-sinφ’)*OCR^sinφ’(OC clays or recompressed soils)

Question 4

How do you determine the angle of friction of the soil from a Mohr’s circle?

Answer 4

sinφ’ = σ3/σ1

Question 5

How do you determine (τf,σf) from Mohr’s circle?

Answer 5

τf = σ3 * cosφ'
σf = (σ1-σ3) - x
x = σ3 * sinφ'

Question 6

How do you calculate the angle of shear failure in a compression test?

Answer 6

δ = 45’ + φ’/2

Question 7

What can you say about a soil density if it fails by shear or bulging?

Answer 7

shear -> high density (curve reaches peak then down to φ’crit)
bulging -> low density (curve rises and flattens at φ’crit)

Question 8

What are the advantages of using the plasticity method in soil mechanics?

Answer 8

1) dont need the full stress-strain relationship, only strength
2) using the critical state strength, we dont need to know initial soil state or stress history

Question 9

Define the lower bound “safe” theorem for perfectly plastic soil

Answer 9

If you can determine a set of stresses in the ground in equilibrium with the external loads, which does not exceed the soil strength, the the structure will not collapse.

Question 10

Define the upper bound “unsafe” theorem for perfectly plastic soil

Answer 10

If you take any compatible mechanism of the slip surfaces and consider an increment of movement, and you show that the work done by the soil resisting movement equals the work done by the external loads, the the structure must collapse.

Question 11

How do you calculate the lowest upper bound solution for bearing capacity factor Nq for Mohr-Coulomb (drained) analysis?

Answer 11

Nq = Kp * e^(πtanφ')
Kp = passive earth pressure coefficient 
φ' = friction angle

Question 12

What are the two equations for enhancement factors for undrained total stress analysis?

Answer 12

Skempton and Meyerhof: (σf-σ0) = (Nc * Sc * dc) * Cu
Brinch Hansen: (σf-σ0) = Nc * (1 + Sc + dc) * Cu
σ0 = ρgD = γ*D

Question 13

What is the equation for the enhancement factor for drained effective stress analysis?

Answer 13

σ'f = qf = [Nq*Sq*dq]*σ'0 + [Nγ*sγ*dγ*rγ*(γB/2 -Δu)]
(f)γ = analogous to (f)q applied to Nγ

Question 14

How do you calculate the external and internal work on a HODOGRAPH?

Answer 14

Ext. W = qb * Vq = qb * Vx * sinθ (+γAxVq if soil is not weightless)
Int. = sum (CuLV) for AB, BD, BC

Question 15

How would u calculate the long term safety factor of a footing according to Brinch Hanson?

Answer 15

F = Qf/Qa
Qf = qf*D*L = σ'f*D*L

Question 16

Why may short-term stability have a much larger factor of safety than long-term stability?

Answer 16

Due to the parameters 𝑐u versus 𝜙′. The 𝑐u is very high, suggesting a dense, stiff material. If 𝜙′ is very low, suggesting a critical-state value. This means that,
although 𝐹 is low for long-term stability, we can be happy that the soil strength has been determined conservatively.

Question 17

Why is ultimate bearing capacity higher for methods involving enhancement factors?

Answer 17

Enhancement factors take into account for:

i) foundation has finite length, additional contribution from edges.
ii) soil at the sides give strength, must shear before collapse.

Question 18

How do you calculate ultimate pile axial load using adhesion factors from Weltman & Healy (1978)?

Answer 18

Qp = Qs + Qb – W
Qb = Ab * 9 * Cu,b
Qs = As * τs = π*D*L*α*Cu,s

Question 19

How do you calculate the adhesion factor, α?

Answer 19

α = Cu / σv’
σv’ = L/2 * (γ-g)
or from the graph

Question 20

Discuss mechanisms which can reduce the internal friction angle of a soil below its
peak friction value during slope failure. What value would you use in design to
stabilise a slip that has already occurred?

Answer 20

Dense granular soils (sands, gravel + silts) and overconsolidated clays will exhibit a peak
friction value after which, continued shearing will result in a drop in frictional strength to
critical state. This process is called “strain softening”. If a clay which is sheared still further
(usually > 100 mm displacement is required on the shear plane), a “residual” friction angle
can result, which may be more than 10° less than the critical value. Typical mechanisms that
cause this are landsliding, glacial motions, solifluction with free-thaw cycles, etc… A residual
friction value is thought to arise due to the alignment of clay particles in the direction of
shearing, resulting in “slicken-sides”

Question 21

list different ways in which landslides can be triggered.

Answer 21

• Water (rainfall, glacial or snowmelt, groundwater rise….)
• Seismic action (earthquakes, blasting……)
• Increase of shear stress (e.g. loading at head)
• Loss of shear resistance (e.g. undercutting of toe)

Question 22

List different ways in which incipient instability may be identified during a walkover
survey.

Answer 22

Ponds, sinks, marshy ground;
• Tension cracks at head, toe slumping or bulging;
• Vegetation changes etc.

Question 23

Explain possible options by which slope instability may be mitigated for specific cases
such as: a slowly failing slope, with max depth to shear zone of 5 m in glacial till above a
railway

Answer 23

By installing soil nails or anchors; possibly piles (depending on extent of the instability)

Question 24

Explain possible options by which slope instability may be mitigated for specific cases
such as: a small 2.5 m high cliff of overconsolidated London clay which has suddenly softened
due to a burst water main 20 m behind the top of the shear zone.

Answer 24

Repair of the water main, provision of drainage layers at the shear zone to isolate from
water. For the slope itself, reinforced soil, soil nails or a retaining wall, depending on extent
of softening and location of proximate structures.

Question 25

What is soil “sensitivity” and how is it quantified?

Answer 25

Soil sensitivity St is formally defined as Cu(peak) / Cu(remoulded). It is the ratio of the undisturbed
strength to the remoulded strength of a fine grained (clay or silt) soil. It occurs as a result of
physiochemical influences on clay microstructure and describes how the undrained shear
strength of a clayey soil can reduce due to remoulding / disturbance / shearing.

Question 26

Give two advantages and two disadvantages of the SPT compared with the CPT test in
terms of how the test is conducted and the overall data obtained.

Answer 26

The SPT is able to undertaken on a range of soils from clay to gravel. Boreholes sunk during
SPT testing and split-samples taken can be used to examine and hence determine the soil
type directly. The CPT can have problems in coarse grained soils which it may find difficult to
push through without damaging the cone tip.
Conversely, the SPT only obtains data every 1-2 m, while the CPT provides a continuous
profile with depth. The CPT is generally more reliable (more sophisticated and quality
controlled) and can provide information on thin soil layers which cannot be obtained from
an SPT