**1. A sample of dry sand was subjected to a triaxial test, with a confining pressure of 250 kN/m ^{2}. The angle of shearing resistance was found to be 36^{o}C. At what value of the major principal stress is the sample likely to fail?**

__Solution__

__Given__

Minor principal stress, σ_{3} = 250kN/m^{2},

Angle of shearing resistance, ϕ = 36^{o}

__To find,__

Major principal stress, σ_{1} =?

Use, σ_{1} = σ_{3} tan^{2} (45 + ϕ/2) + 2c tan (45 + ϕ/2)

**Note:** the soil sample is a dry sand, hence, cohesion, c = 0

Angle of inclination of the failure plane, θ = 45 + ϕ/2 = 45 + (36/2) = 45 + 18 = 63^{o}

σ_{1} = 250 tan^{2} (63) + 2 x 0 x tan (63) = 250 tan^{2} (63) = 962.959999 ≈ 963 kN/m^{2}

**2. A direct shear test was performed on a 6 cm x 6 cm sample of dry sand. The normal load was 360 N. The failure occurred at a shear load of 180 N. Plot the Mohr strength envelope and determine angle of shearing resistance, ϕ.**

**Assume cohesion, c = 0. Also determine principal stress at failure.**

__Solution__

Normal load = 360 N = (360/1000) kN = 0.36 kN

Shear load = 180 N = (180/1000) kN = 0.18 kN

Area of sand = (0.06 x 0.06) m^{2} = 0.0036 m^{2}

Normal stress, σ = normal load/area = 0.36/0.0036 = 100 kN/m^{2}

Shear stress, τ = shear load/area = 0.18/0.0036 = 50 kN/m^{2}

A plot of shear stress against the normal stress will give a straight line that passes through the origin.

The angle, ϕ is obtained as follows using the formula below

τ = σtan ϕ

ϕ = tan^{-1} (τ/σ) = tan^{-1} (50/100) = 26.57^{o}

The principal stresses at failure are σ and τ.

**Note**, σ = (σ_{1} + σ_{3})/2 + ((σ_{1} – σ_{3})/2) cos 2θ

And, τ = ((σ_{1} – σ_{3})/2) sin 2θ

Again, θ = (45 + ϕ/2) = 45 + (26.57/2) ≈ 58.285^{o}

100 = (σ_{1} + σ_{3})/2 + ((σ_{1} – σ_{3})/2) cos 2 x 58.285 (1)

50 = ((σ_{1} – σ_{3})/2) sin 2 x 58.285 (2)

From Eqn (2), ((σ_{1} – σ_{3})/2) = 50/0.8944

((σ_{1} – σ_{3})/2) = 55.9034

100 = (σ_{1} + σ_{3})/2 + 55.9034 x cos (2 x 58.285)

(σ_{1} + σ_{3})/2 = 100 – (55.9034 cos 116.57)

(σ_{1} – σ_{3})/2 = 55.904 » σ_{1} – σ_{3 }= 55.904 x 2 = 111.8068 (3)

(σ_{1} + σ_{3})/2 = 125.0056 » σ_{1} + σ_{3 }= 125.0056 x 2 = 250.0112 (4)

Eqns (3) + (4) = 2 σ_{1} = 361.818

σ_{1} = 361.818/2 = 180.909 ≈ 181 kN/m^{2}

from Eqn (4), σ_{3} = 250.0012 – σ_{1} = 250.0012 – 180.909 = 69.1022 ≈ 69.1 kN/m^{2}

**3. An unconfined compression test was conducted on an undisturbed sample of clay. The clay had a diameter of 37.5 mm and was 80 mm long. The load at failure measured by the proving ring was 28 N and the axial deformation of the sample at failure was 13 mm. determine the unconfined compressive strength and the undrained shear strength of the clay.**

__Solution__

Original area of the specimen, A_{o} = (π/4) x 37.5^{2} = 1105 mm^{2}

Axial strain, ε = 13/80 = 0.162

The corrected area of the specimen at failure is given by,

A_{f} = A_{o}/(1-ε) = 1105/ (1 – 0.162) = 1315 mm^{2}

Unconfined compressive strength, q_{u }= 28 N/1315 mm^{2} = (28/1000) kN / (1315/1000^{2}) m^{2}

= 21.292776 ≈ 21.3 kN/m^{2}

The undrained shear strength of the clay is one-half its unconfined compressive strength. Hence, C_{u} = q_{u}/2 = 21.3/2 = 10.05 kN/m^{2}

**4. A cohesive soil has an angle of shearing resistance of 15 ^{o} and a cohesion of 31 kN/m^{2}. If the specimen of this soil is subjected to a triaxial compression test, find the value of the lateral pressure in the cell for failure to occur at a total axial stress of 206.8 kN/m^{2}**.

__Solution__

Given: ϕ = 15^{o}; c = 31 kN/m^{2}; σ_{1} = 206.8 kN/m^{2}

Use the equation below:

σ_{1} = σ_{3 }tan^{2 }(45 + ϕ/2) + 2c tan (45 + ϕ/2)

**Note**, ϕ/2 = 15/2 = 7.5 » 45 + ϕ/2 = 45 + 7.5 = 52.5^{o}

206.8 = σ_{3 }tan^{2 }(52.5) + 2c tan (52.5) » σ_{3} = (206.8 – 2c tan 52.5)/ tan^{2} 52.5

σ^{3} = (206.8 – 62 tan 52.5) / tan^{2} 52.5 = 74.1876448 ≈ 74.19 kN/m^{2}

**5. A consolidated-drained triaxial test was conducted on a normally consolidated clay. The results are as follows: σ _{3} = 276 kN/m^{2}; (Δσ_{d})_{f} = 276 kN/m^{2}**

**Determine,**

**(a) Angle of internal friction, ϕ ^{1 }**

**(b) Angle θ that the failure plane makes with the major principal plane**

**(c) Normal stress, σ _{1} and shear stress, τ_{f}, on the failure plane**

__Solution__

For normally consolidated soil, the failure envelope equation is

τ_{f }= σ^{1} tan ϕ (because c^{1} = 0) – see Figure 1

**Figure 1; Ex.5: Mohr’s circle failure envelope for a normally consolidated clay**

(a) For the triaxial test, the effective major and minor principal stresses at failure are as follows:

σ_{1}^{1} = σ_{1} = σ_{3} + (Δσ_{d})_{f } = 276 + 276 = 552 kN/m^{2}

and σ_{3}^{1} = σ_{3} = 276 kN/m^{2}(Δσ_{d})_{f}

sin ϕ^{1} = (0.5 (σ_{1}^{1} – σ_{3}^{1}))/ c^{1} cot ϕ^{1} + 0.5 (σ_{1}^{1} + σ_{3}^{1}) (but c = 0)

sin ϕ^{1} = (0.5 (σ_{1}^{1} – σ_{3}^{1}))/ 0.5 (σ_{1}^{1} + σ_{3}^{1}) = AB/OA

sin ϕ^{1} = (σ_{1}^{1} – σ_{3}^{1}))/ (σ_{1}^{1} + σ_{3}^{1}) = (552 – 276)/ (552 – 276) = 0.333

ϕ^{1} = sin^{-1} 0.333 = 19.45^{o}

(b) θ = 45 + ϕ^{1}/2 = 45 + 19.45/2 = 54.725^{o}

(c) σ_{1} (on the failure plane) = ((σ_{1}^{1} + σ_{3}^{1})/2) + ((σ_{1}^{1} – σ_{3}^{1})/2) cos2θ

τ_{f} = ((σ_{1}^{1} – σ_{3}^{1})/2) sin2θ

σ_{1} = ((552 + 276)/2) + ((552 – 276)/2) cos (2 x 54.725^{o}) = 368.048 kN/m^{2}

τ_{f} = ((552 – 276)/2) sin (2 x 54.725^{o}) = 130.125 kN/m^{2}

**6. For the triaxial test described in Example 5 above,**

**(a) Determine the effective normal stress on the planes of maximum shear stress.**

**(b) Explain why the shear failure occurred along a plane with θ = 54.73 ^{o} and not along the plane of maximum shear stress**

__Solution__

(a) The maximum shear stress will occur on the plane with θ = 45^{o}

σ^{1} = ((σ_{1}^{1} + σ_{3}^{1})/2) + ((σ_{1}^{1} – σ_{3}^{1})/2) cos2θ = ((552 + 276)/2) + ((552 – 276)/2) cos (2 x 45^{o}) = 414 kN/m^{2}

(b) The shear stress that will cause failure along the plane with θ = 45^{o }is,

τ_{f }= σ^{1} tan ϕ^{1} = 414 tan 19.45 = 146.2 kN/m^{2}

However, the shear stress induced on that plane is

τ = ((σ_{1}^{1} – σ_{3}^{1})/2) sin2θ = ((552 – 276)/2) sin (2 x 45) = 138 kN/m^{2}

Because τ = 138 kN/m^{2} ˂ 146.2 kN/m^{2} = τ_{f}, the specimen did not fail along the plane of maximum shear stress.

**7. A consolidated-undrained triaxial test is performed on a specimen of saturated clay. The value of σ _{3} is 196.1 kN/m^{2}. At failure, (σ_{1} – σ_{3}) = 274.6 kN/m^{2} and u = 176.5 kN/m^{2}. If the failure plane in this test makes an angle of 57^{o} with the horizontal, calculate the normal and shear stresses on the failure surface and the maximum shear stress in the specimen.**

__Solution__

σ_{3} = 196.1 kN/m^{2}; σ_{1} – σ_{3} = 274.6 kN/m^{2}; u = 176.5 kN/m^{2}; θ = 57^{o}

σ_{1} = σ_{3} + (σ_{1} – σ_{3}) = 196.1 + 274.6 = 470.7 kN/m^{2}

» σ_{1}^{1} = σ_{1} – u = 470.7 – 176.5 = 294.2 kN/m^{2}

σ_{3}^{1} = 196.1 – 176.5 = 19.6 kN/m^{2}

σ_{1}^{1} – σ_{3}^{1} = σ_{1} – σ_{3} = 274.6 kN/m^{2}

σ_{1}^{1} + σ_{3}^{1} = 294.2 + 19.6 = 313.8 kN/m^{2}

σ^{1} = ((σ_{1}^{1} + σ_{3}^{1})/2) + ((σ_{1}^{1} + σ_{3}^{1})/2) cos 2θ = (313.8/2) + (274.6/2) cos (2 x 57) = 101.055 kN/m^{2}

τ = ((σ_{1}^{1} – σ_{3}^{1})/2) sin 2θ = (274.6/2) sin (2 x 57) = 125.43 kN/m^{2}

Maximum shear stress (this would occur at θ = 45^{o}) = τ = ((σ_{1}^{1} – σ_{3}^{1})/2) sin 2θ = 137.3 sin (2 x 45) = 137.3 kN/m^{2}

**8. The stresses on a failure plane in a drained test on a cohesionless soil are as shown below (Figure 2):**

**Normal stress, σ = 100 kN/m ^{2}**

**Shear stress, τ = 40 kN/m ^{2}**

**(a) Determine the angle of shearing resistance and the angle which the failure plane makes with the major principal plane.**

**(b) Find the major and minor principal stresses**

**(c) Determine the resultant stress on the plane of failure**

__Solution__

(a) Refer to Figure 2: Example 8

The failure (Coulomb’s) line passes through the origin and the point A with coordinates (100, 40).

Angle of shearing resistance, ϕ^{1}, tan ϕ^{1} = 40/100, ϕ^{1} = tan^{-1} (40/100) = 21.80^{o}

Figure 2: Example 8

The angle which the failure plane makes with the major principal plane

θ= 45 + ϕ^{1}/2 = 45 + 10.9 = 55.9^{o}

(b) The centre, C of the Mohr circle is located by drawing a normal AC to line OA at A. Mohr circle is drawn through point A.

Graphical approach:

Point B is 73.5 kN/m^{2} = σ_{3}

Line BC ≈ 43 (from 116.5 – 73.5)

» 116.5 + 43 = 159.5 kN/m^{2}

(c) Analytical approach

Length OA = √ 100^{2} + 40^{2} = 107.7

AC = OA tan ϕ^{1} = 107.7 x tan 21.80^{o} = 43.08 kN/m^{2}

OC = OA sec ϕ^{1} = 107.7 sec 21.8 = 115.993 » σ^{1} ≈ 116 kN/m^{2}

OD = OC + AC = 116 + 43.08 = 159.08 kN/m^{2}

OB = OC – AC = 116 – 43.08 = 72.92 » σ_{3} ≈ 73 kN/m^{2}

**(9) **

**(a) In-situ vane tests were made below the bottom of a bore hole in saturated sandy clay with a 160 mm high and 80 mm diameter shear vane. The maximum torque applied was 35 Nm when failure occurred. Assuming the sandy clay to be isotropic, derive an expression relating the torque, T to the undrained shear strength C _{u} and calculate the value of C_{u} (kN/m^{2}) for the sandy soil.**

**(b) If the angle of shearing resistance (ϕ) of the above soil is 15 ^{o}, find the value of the lateral pressure in the cell during a triaxial compression test which will cause failure to occur under a total axial stress of 207 kN/m^{2}.**

__Solution__

(a) Derivation of expression relating the torque, T to the undrained shear strength, C_{u}

Figure 3: Example 9

If T is the maximum torque applied at the head of the torque rod to cause failure, it should be equal to the sum of the resisting moment of the shear force along the side surface of the soil cylinder (M_{s}) and the resisting moment of the shear force at each end (M_{e}).

T = M_{s} + (M_{e} + M_{e})-*two ends*

The resisting moment M_{s} can be given as

M_{s} = ((πdh) C_{u})-** surface area **x (d/2)-

*moment arm*The resisting moments, M_{e} can be given as

M_{e} = [(C_{u} x (πd^{2}/4)-** surface area** x (2/3 x d/2)-

**] x 2 -(**

*moment arm*

*for the two ends)*T = ((πdh) C_{u}) x (d/2) + [(C_{u} x (πd^{2}/4) x (2/3 x d/2)] x 2

T = (C_{u} x (πd^{2}h)/2 + (C_{u} x (πd^{3 }x 4))/ 24

T = πC_{u} [((πd^{2}h)/2) + ((πd^{3 }x 4)/ 24)]

C_{u} = T/ [π [((d^{2}h)/2) + (d^{3 }/ 24)]]

For T = 35 Nm = 35 x 10^{-3} kNm; h = 160 mm = 0.16 m and d = 80 mm = 0.08 mm

C_{u} = 35 x 10^{-3} / [π [((0.08^{2} x 0.16)/2) + (0.08^{3 }/ 24)]] = 35 x 10^{-3}/ 0.00187657 = 18.65107397 ≈ 18.65 kN/m^{2}

(b) ϕ = 15^{o}; Total axial stress, σ_{1} = 207 kN/m^{2}

σ_{1} = σ_{3}tan^{2} (45 + ϕ/2) + 2 C_{u} tan (45 + ϕ/2)

ϕ/2 = 15/2 = 7.5^{o}; θ = 45 + 7.5 = 52.5^{o}

σ_{3} = (σ_{1} – 2 C_{u} tan (45 + ϕ/2))/ tan^{2} (45 + ϕ/2) = (207 – 2 x 18.651 tan 52.5)/ tan^{2} 52.5 = 93.256845 ≈ 93.3 kN/m^{2}.

**10. An embankment is to be constructed on reclaimed area with a soil whose properties are c ^{1} = 51 kN/m^{2}; ϕ^{1} = 25^{o} (in effective stress terms) and γ = 17.0 kN/m^{2}. The pore pressure parameters as found from a triaxial test are A = 0.5 and B = 0.9.**

**Find the shear strength of the soil at the base of the embankment just after the height of the fill has been raised from 0 m to 6 m. Assume a negligible pore pressure dissipation during this stage of construction and that the lateral pressure is three-quarters (3/4) of the vertical pressure.**

__Solution__

**Given Data**: c^{1} = 51 kN/m^{2}, ϕ^{1} = 25^{o}; γ = 17.0 kN/m^{2}

Pore pressure parameters, A = 0.5; B = 0.9

Assumed: Δσ_{3} = (3/4) Δσ_{1}

σ_{1} = 0; Δσ_{1} = 6 x 17 = 102 kN/m^{2}

Δσ_{3} = (3/4) Δσ_{1} = ¾ x 102 = 76.5 kN/m^{2}

Δu = B [Δσ_{3} + A (Δσ_{1} – Δσ_{3})] = 0.9 [76.5 + 0.5 (102 – 76.5)] = 80.325 kN/m^{2}

Total stress, σ = σ_{1} + Δσ_{1} = 0 + (6 x 17) = 102 kN/m^{2}

Effective stress, σ^{1} = σ – u = 102 – 80.325 = 21.675 kN/m^{2}

Shear strength, τ = c^{1} + σ^{1} tan ϕ^{1} = 51 + 21.675 tan 25 = 61.107 kN/m^{2}

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