Geotechnical Engineering

n = e / (1 + e)   S = w G_s / e

γ_d = G_sγ_w / (1 + e)   γ_sat = (G_s + e)γ_w / (1 + e)   γ = γ_d(1 + w)

Sanity checks that catch arithmetic slips: S can never exceed 100%, e and n move together, and γ_d < γ < γ_sat for a partially saturated soil. S = 1 marks full saturation, S = 0 marks perfectly dry soil.

PI = LL − PL   LI = (w − PL) / PI

C_u = D₆₀/D₁₀   C_c = D₃₀² / (D₁₀ D₆₀)

USCS in one pass: more than 50% retained on the No. 200 sieve → coarse-grained (G gravel / S sand); second letter W (well-graded), P (poorly), M (silty), C (clayey). Otherwise fine-grained (M silt / C clay / O organic / Pt peat), placed by LL and PI against the A-line, PI = 0.73(LL − 20). Well-graded sand needs C_u ≥ 6 and 1 ≤ C_c ≤ 3 (gravel: C_u ≥ 4). AASHTO (highway soils) runs A-1 (best) to A-8 (organic, worst) with a group index that grows as fines, LL, and PI grow.

RC = γ_d,field / γ_{d,max (lab)}

γ_zav = G_sγ_w / (1 + w G_s)

The compaction curve peaks at the optimum moisture content: dry of optimum the soil is stiff and hard to compact; wet of optimum the water starts carrying the compactive effort. A measured field density that plots above the zero-air-voids curve is impossible — recheck the numbers.

σ′ = σ − u

i_c = γ_sub / γ_w = (G_s − 1) / (1 + e)

Seepage shifts effective stress: downward flow adds seepage force and increases σ′; upward flow subtracts it. At the quick condition the sand boils and bearing capacity vanishes — check FS = i_c/i on any upward-seepage question.

q = k i A   v = k i   v_s = v / n

q = k h (N_f / N_d)   per unit length of structure

Flow-net rules worth memorising: flow lines and equipotentials cross at right angles, each “square” is a curvilinear square, and head drops by h/N_d across each equipotential band. Uplift under a dam comes from the pore pressure at the base of the structure, read off the net.

s_c = H₀1 + e₀  C_c log₁₀σ′_v0 + Δσσ′_v0

T_v = c_v t / H_dr²

The log is base 10 — the classic trap is reaching for ln. Overconsolidated clays (OCR = σ′_p/σ′_v0 > 1) settle far less while stresses stay below σ′_p, because C_r is a small fraction of C_c.

τ = c + σ′ tan φ

s_u = q_u / 2

Test types: UU (unconsolidated-undrained) → total-stress, φ = 0 for saturated clay; CU (consolidated-undrained) → gives both total and effective parameters; CD (consolidated-drained) → slow enough that u = 0, so c′, φ′ directly. Match the parameters to the drainage conditions of the problem.

K₀ = 1 − sin φ   K_a = 1 − sin φ1 + sin φ = tan²(45° − φ/2)   K_p = 1/K_a = tan²(45° + φ/2)

σ′_h = K σ′_v

φ goes into the tan² formula in degrees — a calculator in radian mode silently gives a wrong K. Compute the water pressure separately and add it to the effective lateral stress; water has no K.

q_ult = s_c c N_c + γD_fN_q + s_γ ½ γB N_γ   q_all = q_ult / FS

Use the submerged unit weight γ′ for soil below the water table in the third term. Net ultimate capacity subtracts the overburden: q_net = q_ult − γD_f. Try the bearing capacity calculator to check your hand work.

FS = c + γz cos²β tan φγz sin β cos β

Seepage parallel to the slope is the usual exam twist: it adds a seepage force down the slope, so replace γ with the submerged weight and add the flow force — FS drops. For general slip surfaces the exam expects the concept of slices (method of slices: moment equilibrium about the circle centre), not a full slice table by hand.