Prova d'esame svolta Progettazione geotecnica e strutturale

σ'a = Ka · σ'n - 2c'√(Ka) + u

σ'p = Kp · σ'n + 2c'√(Kp) + u

LONG TERM

σ'a = σ'n - 2cm

σ'p = σ'n + 2cm

SHORT TERM

GEOTECHNICAL DESIGN EXAM - PART 1

03-02-25

SAND: c'=0 ; φ' = 36° ; γdry = 16 kN/m³ ; γsat = 16 kN/m³ ; Ksat = 10⁵ μm/s

CLAY: c' = 4 kPa ; φ' = 26° ; OCR = 4

50 kPa

Eu = 40 kPa

γdry = 17 kN/m³

γsat = 19 kN/m³

Ksat = 10⁻⁹ m/s

MURO INCLINED γm = 10 kN/m³

ACTIVE AND PASSIVE PRESSURE COEFFICIENTS:

x Ka⁰ = (1 - sin φ') / (1 + sin φ') = (1 - sin 36°) / (1 + sin 36°) = 0.26

x Kp = 1/Ka⁰ = 3.85

x Kac = (1 - sin φ') / (1 + sin φ') = (1 - sin 26°) / (1 + sin 26°) = 0.33

x Kp = 1/Kac = 2.56

σʹ₀ = K₀ * γʹ * u

σʹ₀ = σʹv - 2c√K₀ + u

σʹp = σʹv + 2c

LONG TERM

SHORT TERM

GEOTECHNICAL DESIGN EXAM - PART 1 03-02-23

SAND: cʹ=0; φʹ=36°; γdry=16 KN/m³; γsat=16 KN/m³; Ksat=10⁻⁵ m/s

CLAY: cʹ=8 kPa; φʹ=26°; OCR=4; 50 kPa

σv=40 kPa

γdry=17 KN/m³

γtot=19 KN/m³

Ku=10⁻⁵ m/s

MURD LINGO γ=10 KN/m³

ACTIVE AND PASSIVE PRESSURE COEFFICIENTS:

ACTIVE - SHORT TERM

POINT A:  _h = K_a·_v·_w + _u = 0.26·50 = 13 KPa

POINT B:  _h = K_a·_v·_w + _u = 0.26·(50 + 16 - 2) = 21,32 KPa

POINT C¹:  _v = _v - _w - _u = (50 + 16 - 2 + 12·10) - (12·10) = 17 _v

POINT C²:  _h = K_a·_v + _u = 0.26·17 + 120 = 166.2d KPa

 _v = (2·16) + (12·1d) + 50 = 238 KPa

POINT D:  _v = 50 + (2·16) + (12·1d) + (5·1g) = 393 KPa

 _v = _v - 2_m = 393 - 2_m = 3/3 KPa

POINT E:  _v = (2·16) + (12·1d) + (10·1g) + 50 = 466 KPa

 _v = _v - 2_m = 46g - 2·40 = 40d KPa

PASSIVE - SHORT TERM

POINT F:  _v = 0

POINT G¹:  _v = (4.18) - (4 - 1g) = 32 KPa

 _h = K_p·_v·_w + _u = (3.85·32) + (4·1g) = 163.2 KPa

POINT G²:  _v = 4·18 = 72 KPa  _v = _v + 2_m = 72 + (2·40) = 152 KPa

POINT H:  _v = (4·1d) + (5·1g) = 167 KPa

 _v = _v + 2_cm = 167 + (2·40) = 2.47 KPa

POINT I:  _v = (4·1d) + (10·1g) = 262 KPa

 _v = _v + 2_cm = 262 + d0 = 342 KPa

ACTIVE - LONG TERM

POINTS A, B, C¹ AS ABOVE

POINT C¹:  _v = _v - _w = (50 + 2·16 + 1d·12) - (12·10) = 17d KPa

 _h = K_a·_v + _u =

 (0,30 x1 7d) - (2 ·¹ 0,39) + (12·10) = 13,43 KPa

POINT D:  _b = ^{h  U} _m  ^b ₀ = 130 - 15... (d,/10) = 2d/m

 = z_μB = ^h _m  ^h _m = ₀^d0 = 150kPa

POINT E:  _v = 50 + (2·16) + (12·1d) + (5·1g) = 333 KPa

 _h = K_a·_v - 2_cm ⁰  _v + _u =

 =0,33 d(333 - 15d) - (2·⁰ 023《

POINT E: h_E = 30 - 4 - ^φ/₁₀ = 26,9 m

u_E = (h_E - z_E)⋅_w = (26,9 - 8)⋅10 = 180 kPa

_v = 50 + (2⋅16) + (12⋅18) + (10⋅13) = 400 kPa

σ_v = K_c⋅σ'_v - 2c '[ K_c + ]

= 0,39 ⋅ (48P - AP) - (2⋅d⋅0,939) + 10ff = 295 kPa

PASSIVE - LONG TERM POINTS F AND G AS ABOVE

POINT G: σ'_v - = (4 + 16) - (4 + 10) = 32 kPa

_R = K_p⋅σ'_v + 2c [ K_p + =

= (2,56⋅32) + (2⋅d⋅√2,56) + 40 = 147,52 kPa

POINT H: h_H =30-^φ/₁₀=23,2m

u_H = (h_H - z_H)⋅_w = (23,2 - 13)⋅10 = 102 kPa

_v = (4 + d) + (5 - 19) = 167 kp

σ' = σ'_v - _H = 167 - 102 = 65 kPa

σ_R = K_c⋅σ'_v + 2c [ K_p + =

= (2,56⋅65) + (2⋅d⋅√2,56) + 10,2 = 254 kPa #

POINT I: h_I = 30 - 6 - ^φ/₁₀ = 25,2 m

u_I = (h_I - z_I)⋅_w = (25,2 - 8)⋅10 = 172 kPa

σ_v = (4 + d) + (10 - 13) = 262 kPa

σ'_v - = 262 - 172 = 90 kPa

σ_R = K_p⋅σ'_v + 2c [ K_p + =

= (2,56⋅90) + (2⋅d⋅√2,56) + 17,2 = 424 kPa

σ'_R IN POINTS "D" AND "H" IN LONG TERMIN IF WALL DOESN'T MOVE !

K_σ = K_c ocrⁿ = (1 - rim 26) . 14 = 1,12

σ'_R (D) = σ'_v (D) . 1,12 = 263,2 kPa

σ'_R (H) = σ'_v (H) . K_c = 72 / 72 kPa

σ_R (D) = σ'. 263,2 + 158 = 421,2 kPa *

σ_R (H) = σ'_R (H) + 62 = 92 + 10,2 = 102 || 102 || 102 || 72 kPa #

EC7 -> DA/1 -> A2 - R1 - M2

BEARING CAPACITY AND SLIDING

GEOTECHNICAL DESIGN EXAM - PART 2

B=6m

(L=7m)

B=6m

_0.5m

_0.5m

γ_kn = 19 kN/m³

o.c. CLAY

b' = 2d'; c'_f = 4R_p; φ'_c = 26°; c_c=0

C_μ = 45 kPa; OCR = 6

LOOSE SAND

γ_cd = 35 ; c_c = 0 ; γ_df = 20 kN/m³

WATER TABLE NOT VARYING WITH TIME (γ_w = 10 kN/

PERCENT LOADS: F_k = 1000 kN ; H_k = 400 kN

VARIABLE LOADS: M_k1 = 250 kNm ; M_k2 = 800 kNm (U.T. ONLY)

× A2 ->

1. γ_Q^f = 1 ; γ_E = 1 ; γ_E^f = 1 ; γ_G^f = 1.3 ; γ_E = 1 + δ^G = 0 x R
2. a R_qk = γ_Emf

× M2 ->

1. + δ_E = 1.25 ; δ_c = 1.25 ; γ_c = 1.4 ; δ_E = 1
2. γ_cd = 19 kN/m³ ; φ_f = tan^-1 (tan φ') = tan ∞ (tan 26°)
3. = 23° ; c_c = c^p ; _cp / _cc ; 3R_c.
4. γ_cd = 180/
5. f (tan 26°) - 2.4° ;
6. c_μ = 45 R
7. γ_sf = 45 ; a

γ_sub=2.4 γ_

LONG TERM - BEARING

× V_k = (6.7 + 1) - 24 = 1.00 kN ; W_L = W_L ; φ_P

× Vk = 1000 kN ; j K_ad = V_e ; φ_c

× V_k = (6×7)² = 2.10 kN

× U_Lk = F