Prova d'esame svolta Progettazione geotecnica e strutturale

GEOTECHNICAL DESIGN MODULE – PART 1

05/09/2025

Name, Surname and ID student number: ______________________________

The soil conditions adjacent to a sheet pile wall are given in the following figure, a surcharge pressure of 50kPa being carried on the surface behind the wall. The soil properties for the Coarse Sand are: c'=0, φ'=36°, γ_dn=16 kN/m³, γ_sat=18 kN/m³ and k_sat=10^-5 m/s. The soil properties for the O.C. Clay are: c'=8 kPa, φ'=26°, OCR=4, c_u=40 kPa, γ_dn=17 kN/m³, γ_sat=19 kN/m³ and k_sat=10^-9 m/s. A smooth contact can be assumed between the soil and the wall.

A water unit weight γ_w=10 kN/m³ should be considered for simplicity, the water tables being at different elevations on the two sides of the wall. Given the significant difference in permeability between the two materials, it can be assumed that all the hydraulic head difference is dissipated within the O.C. Clay layer (therefore, hydrostatic conditions can be assumed within the Coarse Sand layer).

Assuming that the wall tends to move to the left until failure, you are asked to calculate active and passive (total) pressures at the points shown in the figure (with reference to both short and long term conditions, hence evaluating appropriate undrained or drained responses for the materials in short and long term conditions).

GEOTECHNICAL DESIGN MODULE – PART 1

Name, Surname and ID student number: ___________________________________________05/09/2025

The soil conditions adjacent to a sheet pile wall are given in the following figure, a surcharge pressure of 50kPa being carried on the surface behind the wall. The soil properties for the Coarse Sand are: c’=0, φ’=36°, γ_dn=16 kN/m³, γ_sat=18 kN/m³ and k_sat=10^-5 m/s. The soil properties for the O.C. Clay are: c’=8 kPa, φ’=26°, OCR=4, c_u=40 kPa, γ_dn=17 kN/m³, γ_sat=19 kN/m³ and k_sat=10^-9 m/s. A smooth contact can be assumed between the soil and the wall.

A water unit weight γ_w=10 kN/m³ should be considered for simplicity, the water tables being at different elevations on the two sides of the wall. Given the significant difference in permeability between the two materials, it can be assumed that all the hydraulic head difference is dissipated within the O.C. Clay layer (therefore, hydrostatic conditions can be assumed within the Coarse Sand layer).

Assuming that the wall tends to move to the left until failure, you are asked to calculate active and passive (total) pressures at the points shown in the figure (with reference to both short and long term conditions, hence evaluating appropriate undrained or drained responses for the materials in short and long term conditions).

DRAFT

GEOTECHNICAL DESIGN EXAM - PART 1

5-9-25

SAND: c' = 0, φ' = 36°

γ_dry = 16 kN/m³; γ_sat = 18 kN/m³; k_sat = 10^-5 m/s

CLAY: c' = 8 kPa, φ' = 26°, OCR = 4

s_u = 40 kPa

γ_dry = 17 kN/m³

γ_tot = 19 kN/m³

k_tot = 10^-9 m/s

MURO INCLINATO F

γ_w = 10 kN/m³

H_max = 30 m

ΔH = 4 m

N_L = 10

ACTIVE AND PASSIVE PRESSURE COEFFICIENTS

• k₀^s = 1 - sin φ'
• k₀^s = 1 - sin 36° = 0.26
• k_a^s = 1 - sin φ'
• k_a^s = 1 - sin 36° = 0.26
• k_p^s = 1 - sin φ'
• k_p^s = 1 + sin 36° = 3.85
• k₀^c = 1 - sin φ'
• k₀^c = 1 - sin 26° = 0.33
• k_p^c = 1
• k_p^c = 2.56

DRAFT

ACTIVE - SHORT TERM

POINT A: n = kqv' + mu = 0.26 * 50 = 13 kPa

POINT B: n = kqv' + mu = 0.26 * (50 + 16 * ) = 21,32 kPa

POINT C': qv' = qv - m = (50 + 16 * + 12 * ) - (12 * 10) = 17 kPa

POINT C: n = kqv' + mu = 0.26 * 17 + 120 = 166,2 kPa

qv = (2 * 16) + (12 * ) + 50 = 238 kPaqv = qv - 2 cm = 238 - 2 = 236 kPa

POINT D: qv' = 50 + (2 * 16) + (12 * ) + (5 * ) = 393 kPaqv = qv - 2 cm = 393 - 2 = 40 = 313 kPa

POINT E: n = (2 * 16) + (12 * ) + (10 * ) + 50 = 40 kPaqv = qv - 2 cm = 40 - 2 * 40 = 40 kPa

PASSIVE - SHORT TERM

POINT F: qv = 0

POINT G: qv' = (4 * ) - (4 * 19) = 32 kPan = kqp' - m = (3.85 * 32) + (4 * ) = 163.2 kPa

POINT G': qv = 4.18 = 72 kPa; qv = qv + 2 cm = 72 + (2 * 40) = 152 kPa

POINT H: qv = (4 * ) + (5 * 19) = 167 kPaqv = qv + 2 cm = 167 + (2 * 40) = 247 kPa

POINT I: qv = (4 * ) + (10 * ) = 262 kPaqv = qv + 2 cm = 262 + 80 = 342 kPa

ACTIVE - LONG TERM

POINTS A, B, C, A AS ABOVE

POINT C: qv' = qv - m = (50 + 2 * 16 + 4 * ) - (12 * 10) = 174 kPan = kq' - 2cf = = (0.30 * 174) - (2 * ) + (12 * 10) = 173,43 kPa

POINT D: nb = 130 - 15 - (d / 10) = 28 / mnb = , nbnb = (nb ) = (d * ) - 10 = 150 fnqv = 50 + (2 * 16) + (12 * ) + (5 * ) = 393 kPan = kq' - 2cf = = (2)= 0.34 * (393 - 150) - (2 * ) + 150 = 239,6 kPa

DRAFT

Point E: h_e = 30 - 4 = 26 m

u_e = (h_e - z_g) × ȣ_w = (26 - _) × 10 = 168 kPa

🟢_v = 50 + (2 ∙ 6) + (12 ∙ 10) + (10 ∙ 3) = 408 kPa

σ_v = K_c ∙ 🟢_v - 2c | K_F + μ

∙ = 0.39 ∙ (🟢_AF - 🟢_AF) - (2 ∙ √0.39) + 10 = 295 kPa

Passive - Long Term Points F and G1 as above

Point G1: 🟢_v = "v – μ = (4 + _) – (4 + 10) = 32 kPa

σ_R = K_P ∙ 🟢_v + 2c | K_F + μ

= (2.56 ∙ 32) + (2 ∙ √2.56) + 40 = 147.52 kPa

Point H: h_H = 30 - 7 = 23.2 m

u_H = (h_H - z_H) ∙ ȣ_w = (23.2 - 13) ∙ 10 = 102 kPa

🟢_v = (4 + _) + (5 + 13) = 167 kPa

🟢_v’ = 🟢_v - "v = 167 - 102 = 65 kPa

σ_R = K_P ∙ 🟢_v’ + 2c | K_F + μ

= (2.56 ∙ 65) + (2 ∙ √2.56) + 10 = 254 kPa #

Point I: h_I = 30 - 6 = 25.2 m

μ = (h_I - z_g) ∙ ȣ_w = (25.2 - _) ∙ 10 = 172 kPa

🟢_v = (4 + _) + (10 + 13) = 262 kPa

σ_v = 🟢_v – μ = 262 - 172 = 90 kPa

σ_R = K_P ∙ 🟢_v’ + 2c | K_F + μ

= (2.56 ∙ 90) + (2 ∙ √2.56) + 172 = 427 kPa

GEOTECHNICAL DESIGN MODULE – PART 2

05/09/2025

Name, Surname and ID student number:

You are asked to verify the design of a gravity wall formed of bricks (see figure) according to EC7 with reference to the following 3 Ultimate Limit States (ULS): 1) toppling; 2) sliding; 3) bearing capacity (where appropriate, adopting the Design Approach 1b with the combination A2-M2-R1).

The soil adjacent to the wall is a Loose Coarse Sand having the following properties: c'_k=0, φ'_k=32°, γ_d=20 kN/m³, γ_sat=22 kN/m³ and k_sat=10⁻⁴ m/s. The unit weight of the wall can be considered equal to: γ_w,k=25 kN/m³. For earth pressure calculations: a) the friction angle at the wall-soil interface, δ'_k, can be considered equal to 0.5 φ'_k; b) the coefficient of earth pressure in active conditions, K_a, can be assumed equal 0.34. As shown in the figure, the water table is located several metres below the ground levels (if relevant, hydrostatic conditions can be assumed).

DRAFT

GEOTECHNICAL DESIGN EXAM

PART 2 - GRAVITY WALL

IT IS REQUESTED TO CHECK THE DESIGN OF THE GRAVITY WALL FORMED OF BRICKS (SEE FIGURE) ACCORDING TO ECT WITH REFERENCE TO THE FOLLOWING 3 ULTIMATE LIMIT CASES: (1) TOPPLING, (2) SLIDING, (3) BEARING CAPACITY (WHERE APPROPRIATE ADOPTING THE DESIGN APPROACH & COMBINATION A2-R1-M2).

SOIL PROPERTIES:

• φ’_k = 32°
• c’_k = 0
• γ_k = 20 KN/m³

NO WATER

WALL: γ_WK = 25 KN/m³

ASSUMING: δ₁ = 0.5 ; K_a = 0.34 FOR EACH PRESSURE CALCULATIONS

(1) TOPPLING: THE RESULTANT T SHOULD HAVE A LINE OF ACTION FALLING IN THE MIDDLE THIRD OF THE BASE

DESIGN MATERIAL PROPERTIES:

• γ_d = 1 → γ_dk = 20 KN/m³ → γ_wd = 25 KN/m³
• φ_d = tan^-1 (tan φ_k / γ₁) = tan^-1 (tan 32° / 1.25) = 26.6°
• γ_d = 1.25 BUT c' = 0
• δ_d = tan^-1(tan φ_d/2) = tan^-1(tan 16 / 1.25) = 12.9°

DRAFT

WALL SELF WEIGHT:

V_wk = (q_s * 2,1) * 25 = 236,25 kN/m

l_w (0) = 0,25m

HORIZONTAL VERTICAL FORCES DUE TO SOIL WEIGHT:

P_a (z) = K_a * n’(z) = K_a * v’(z) = K_a * γ₁ * z

H_a,wk = 1/2 (K_a * γ_d * h) * h = 1/2 * 0,34 * 20 * 4,5 * 4,5 = 6,05 kN/m

V_a,wk = H_a,wk * tan δ’_d = 6,05 * tan (12,3) = 15,77 kN/m

l_a (0) = 1/3 4,5 = 1,5 m , l_a (0) = 2/3 2,1 = 1,4 m

ACCORDING TO ANNEX "A" FOR "EQU":

PERMANENT UNFAVOURABLE -> γ_{g, dst} = 1,1

          "         FAVOURABLE -> γ_{g, stb} = 0,9

BENDING MOMENT AROUND "0":

H_a,wk * l_ha * γ_{g, dst} ≤ V_a,wk * l_va * γ_{g, stb} + V_wk * l_w * γ_{g, stb} →

→ 60,25 * 1,5 * 1,1 ≤ 15,77 * 1,4 * 0,9 + 236,25 * 0,35 * 0,9 →

→ 113,6 ≤ 94,3 NOT OK

(2) SLIDING: γ_h SAME AS BEFORE

  -γ_h = 1 (ALSO FOR BEARING)

  -PERMANENT UNFAVOURABLE: γ_g,u = 1,1

    "   FAVOURABLE: γ_g,f 1,0