Ten (10) solved problems in geotechnical engineering

Reading Time: 9 minutes

1. A sample of dry sand was subjected to a triaxial test, with a confining pressure of 250 kN/m². The angle of shearing resistance was found to be 36^oC. At what value of the major principal stress is the sample likely to fail?

Solution

Given

Minor principal stress, σ₃ = 250kN/m²,

Angle of shearing resistance, ϕ = 36^o

To find,

Major principal stress, σ₁ =?

Use, σ₁ = σ₃ tan² (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

σ₁ = 250 tan² (63) + 2 x 0 x tan (63) = 250 tan² (63) = 962.959999 ≈ 963 kN/m²

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) m² = 0.0036 m²

Normal stress, σ = normal load/area = 0.36/0.0036 = 100 kN/m²

Shear stress, τ = shear load/area = 0.18/0.0036 = 50 kN/m²

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 = (σ₁ + σ₃)/2 + ((σ₁ – σ₃)/2) cos 2θ

And, τ = ((σ₁ – σ₃)/2) sin 2θ

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

100 = (σ₁ + σ₃)/2 + ((σ₁ – σ₃)/2) cos 2 x 58.285                               (1)

50 = ((σ₁ – σ₃)/2) sin 2 x 58.285                                                         (2)

From Eqn (2), ((σ₁ – σ₃)/2) = 50/0.8944

((σ₁ – σ₃)/2) = 55.9034

100 = (σ₁ + σ₃)/2 + 55.9034 x cos (2 x 58.285)

(σ₁ + σ₃)/2 = 100 – (55.9034 cos 116.57)

(σ₁ – σ₃)/2 = 55.904 » σ₁ – σ₃ = 55.904 x 2 = 111.8068                      (3)

(σ₁ + σ₃)/2 = 125.0056 » σ₁ + σ₃ = 125.0056 x 2 = 250.0112            (4)

Eqns (3) + (4) = 2 σ₁ = 361.818

σ₁ = 361.818/2 = 180.909 ≈ 181 kN/m²

from Eqn (4), σ₃ = 250.0012 – σ₁ = 250.0012 – 180.909 = 69.1022 ≈ 69.1 kN/m²

 

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 37.5² = 1105 mm²

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 mm²

Unconfined compressive strength, q_u = 28 N/1315 mm² = (28/1000) kN / (1315/1000²) m²

= 21.292776 ≈ 21.3 kN/m²

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/m²

 

4. A cohesive soil has an angle of shearing resistance of 15^o and a cohesion of 31 kN/m². 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/m².

Solution

Given: ϕ = 15^o; c = 31 kN/m²; σ₁ = 206.8 kN/m²

Use the equation below:

σ₁ = σ₃ tan² (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 = σ₃ tan² (52.5) + 2c tan (52.5) » σ₃ = (206.8 – 2c tan 52.5)/ tan² 52.5

σ³ = (206.8 – 62 tan 52.5) / tan² 52.5 = 74.1876448 ≈ 74.19 kN/m²

 

5. A consolidated-drained triaxial test was conducted on a normally consolidated clay. The results are as follows: σ₃ = 276 kN/m²; (Δσ_d)_f = 276 kN/m²

Determine,

(a) Angle of internal friction, ϕ¹

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

(c) Normal stress, σ₁ and shear stress, τ_f, on the failure plane

Solution

For normally consolidated soil, the failure envelope equation is

τ_f = σ¹ tan ϕ (because c¹ = 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:

σ₁¹ = σ₁ = σ₃ + (Δσ_d)_f  = 276 + 276 = 552 kN/m²

and σ₃¹ = σ₃ = 276 kN/m²(Δσ_d)_f

sin ϕ¹ = (0.5 (σ₁¹ – σ₃¹))/ c¹ cot ϕ¹ + 0.5 (σ₁¹ + σ₃¹) (but c = 0)

sin ϕ¹ = (0.5 (σ₁¹ – σ₃¹))/ 0.5 (σ₁¹ + σ₃¹) = AB/OA

sin ϕ¹ = (σ₁¹ – σ₃¹))/ (σ₁¹ + σ₃¹) = (552 – 276)/ (552 – 276) = 0.333

ϕ¹ = sin^-1 0.333 = 19.45^o

(b) θ = 45 + ϕ¹/2 = 45 + 19.45/2 = 54.725^o

(c) σ₁ (on the failure plane) = ((σ₁¹ + σ₃¹)/2) + ((σ₁¹ – σ₃¹)/2) cos2θ

τ_f = ((σ₁¹ – σ₃¹)/2) sin2θ

σ₁ = ((552 + 276)/2) + ((552 – 276)/2) cos (2 x 54.725^o) = 368.048 kN/m²

τ_f = ((552 – 276)/2) sin (2 x 54.725^o) = 130.125 kN/m²

 

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

σ¹ = ((σ₁¹ + σ₃¹)/2) + ((σ₁¹ – σ₃¹)/2) cos2θ = ((552 + 276)/2) + ((552 – 276)/2) cos (2 x 45^o) = 414 kN/m²

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

τ_f = σ¹ tan ϕ¹ = 414 tan 19.45 = 146.2 kN/m²

However, the shear stress induced on that plane is

τ = ((σ₁¹ – σ₃¹)/2) sin2θ = ((552 – 276)/2) sin (2 x 45) = 138 kN/m²

Because τ = 138 kN/m² ˂ 146.2 kN/m² = 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 σ₃ is 196.1 kN/m². At failure, (σ₁ – σ₃) = 274.6 kN/m² and u = 176.5 kN/m². 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

σ₃ = 196.1 kN/m²; σ₁ – σ₃ = 274.6 kN/m²; u = 176.5 kN/m²; θ = 57^o

σ₁ = σ₃ + (σ₁ – σ₃) = 196.1 + 274.6 = 470.7 kN/m²

» σ₁¹ = σ₁ – u = 470.7 – 176.5 = 294.2 kN/m²

σ₃¹ = 196.1 – 176.5 = 19.6 kN/m²

σ₁¹ – σ₃¹ = σ₁ – σ₃ = 274.6 kN/m²

σ₁¹ + σ₃¹ = 294.2 + 19.6 = 313.8 kN/m²

σ¹ = ((σ₁¹ + σ₃¹)/2) + ((σ₁¹ + σ₃¹)/2) cos 2θ = (313.8/2) + (274.6/2) cos (2 x 57) = 101.055 kN/m²

τ = ((σ₁¹ – σ₃¹)/2) sin 2θ = (274.6/2) sin (2 x 57) = 125.43 kN/m²

Maximum shear stress (this would occur at θ = 45^o) = τ = ((σ₁¹ – σ₃¹)/2) sin 2θ = 137.3 sin (2 x 45) = 137.3 kN/m²

 

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/m²

Shear stress, τ = 40 kN/m²

(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, ϕ¹, tan ϕ¹ = 40/100, ϕ¹ = 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 + ϕ¹/2 = 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/m² = σ₃

Line BC ≈ 43 (from 116.5 – 73.5)

» 116.5 + 43 = 159.5 kN/m²

(c) Analytical approach

Length OA = √ 100² + 40² = 107.7

AC = OA tan ϕ¹ = 107.7 x tan 21.80^o = 43.08 kN/m²

OC = OA sec ϕ¹ = 107.7 sec 21.8 = 115.993 » σ¹ ≈ 116 kN/m²

OD = OC + AC = 116 + 43.08 = 159.08 kN/m²

OB = OC – AC = 116 – 43.08 = 72.92 » σ₃ ≈ 73 kN/m²

(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/m²) 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/m².

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 (πd²/4)-surface area x (2/3 x d/2)-moment arm] x 2 -(for the two ends)

T = ((πdh) C_u) x (d/2) + [(C_u x (πd²/4) x (2/3 x d/2)] x 2

T = (C_u x (πd²h)/2 + (C_u x (πd³ x 4))/ 24

T = πC_u [((πd²h)/2) + ((πd³ x 4)/ 24)]

C_u = T/ [π [((d²h)/2) + (d³ / 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 [((0.08² x 0.16)/2) + (0.08³ / 24)]] = 35 x 10^-3/ 0.00187657 = 18.65107397 ≈ 18.65 kN/m²

(b) ϕ = 15^o; Total axial stress, σ₁ = 207 kN/m²

σ₁ = σ₃tan² (45 + ϕ/2) + 2 C_u tan (45 + ϕ/2)

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

σ₃ = (σ₁ – 2 C_u tan (45 + ϕ/2))/ tan² (45 + ϕ/2) = (207 – 2 x 18.651 tan 52.5)/ tan² 52.5 = 93.256845 ≈ 93.3 kN/m².

 

10. An embankment is to be constructed on reclaimed area with a soil whose properties are c¹ = 51 kN/m²; ϕ¹ = 25^o (in effective stress terms) and γ = 17.0 kN/m². 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: c¹ = 51 kN/m², ϕ¹ = 25^o; γ = 17.0 kN/m²

Pore pressure parameters, A = 0.5; B = 0.9

Assumed: Δσ₃ = (3/4) Δσ₁

σ₁ = 0; Δσ₁ = 6 x 17 = 102 kN/m²

Δσ₃ = (3/4) Δσ₁ = ¾ x 102 = 76.5 kN/m²

Δu = B [Δσ₃ + A (Δσ₁ – Δσ₃)] = 0.9 [76.5 + 0.5 (102 – 76.5)] = 80.325 kN/m²

Total stress, σ = σ₁ + Δσ₁ = 0 + (6 x 17) = 102 kN/m²

Effective stress, σ¹ = σ – u = 102 – 80.325 = 21.675 kN/m²

Shear strength, τ = c¹ + σ¹ tan ϕ¹ = 51 + 21.675 tan 25 = 61.107 kN/m²

 

Thank you for reading! You can share to students of geotechnical engineering