Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2015 — 04-Geol-06, Soil Mechanics. Three-hour, closed-book exam; one of two approved calculators permitted, plus a compass and ruler. Candidates must answer Questions 1 to 5 and choose three of the five sub-questions in Question 6 (5 marks each); all five Question-6 parts are answered here as a complete study resource.
Reference texts: Das, Principles of Geotechnical Engineering — USCS classification (Q1), phase relations (Q2), consolidation theory (Q4), Boussinesq/Newmark stress-influence factors (Q4), flow nets and seepage (Q5), sand-cone compaction (Q6d); Craig, Craig's Soil Mechanics — Mohr-circle shear strength and limit-equilibrium slope stability (Q3), effective stress and capillarity (Q6b, Q6c); EGBC Geoscience Professional Practice Guidelines for assumption-disclosure conventions.
Check — reading notes. (1) Table Q1's percent-finer values are read as 99% and 98%; the sieve-size row labels "420 μm", "150 μm", "75 μm" are the No. 40, No. 100 and No. 200 sieves. (2) Figure Q5's flow net (equipotential/flow-line count, and the relative positions of points A–D on specific equipotentials) was read from the exam's printed figure — Nf=3 flow channels and Nd=8 equipotential drops, with A on drop 1, point 1 on drop 2, point 2/B on drop 3, point 3 on drop 4, point 4/C on drop 5, point 5 on drop 6, D on drop 7, and the downstream boundary at drop 8; this is a reading of the figure rather than printed exam text. (3) The elevation-head datum for Q5.2 is taken at the impervious layer (z=0), so the ground surface/dam base sits at z=H=10 m and the headwater surface at z=H+ht=20 m — this datum choice is the one that makes the given zA,zB,zC,zD values produce physically valid (positive) pore pressures at all four points, and is disclosed since the exam figure is schematic, not to scale. (4) Q4's Boussinesq corner-influence factors were computed directly from Newmark's closed-form integral rather than read from the exam's Table 8B.
Given. Sieve analysis (percent finer) for Soil A and Soil B (Table Q1); Soil A: $w_L=32\%$, $w_P=25\%$; Soil B: $w_L=52\%$, $w_P=32\%$.
Table Q1 — Percent Finer
Metric Sieve
US Sieve
Soil A (%)
Soil B (%)
75 mm
3 in
100
100
50 mm
2 in
99
100
25 mm
1 in
98
100
19 mm
0.75 in
96
100
9.5 mm
0.375 in
—
100
4.76 mm
No. 4
77
100
2.38 mm
No. 8
—
96
0.84 mm
No. 20
55
94
0.42 mm
No. 40
—
73
0.15 mm
No. 100
30
—
0.075 mm
No. 200
18
55
Find. The USCS group symbol for Soil A and for Soil B, with the gradation and plasticity data supporting each.
Approach. Split each soil into gravel/sand/fines using the No. 4 and No. 200 sieves; a soil with ≤50% passing No. 200 is coarse-grained (classify by gradation, or by the plasticity chart if fines >12%), while a soil with >50% passing No. 200 is fine-grained (classify directly on the plasticity chart against the A-line).
Figure Q1 — grain-size distribution curves for Soil A and Soil B (semi-log). $D_{30,A}=0.15$ mm lands exactly on the No. 100 sieve point.
Soil A — split into gravel, sand and fines. Percent finer than the No. 4 sieve (4.76 mm) is 77%, so gravel (retained on No. 4) is $100-77=23\%$. Percent finer than the No. 200 sieve is 18% (fines). Sand is the remainder.
$$\text{Gravel}=23\%,\qquad \text{Sand}=77-18=\boxed{59\%},\qquad \text{Fines}=\boxed{18\%}$$
Sand (59%) exceeds gravel (23%), so Soil A is a sand (S). Fines = 18% exceeds the 12% threshold, so the plasticity chart — not the gradation coefficients — governs the qualifier.
Soil A — plasticity chart. $I_P=w_L-w_P=32-25=\boxed{7\%}$. The A-line at $w_L=32$ is
$$I_{P,A\text{-line}}=0.73(w_L-20)=0.73(12)=\boxed{8.76\%}$$
Since $I_P=7\%<8.76\%$, the fines plot below the A-line (silty, non-plastic-leaning fines) → qualifier M.
$$\boxed{\text{Soil A} = SM \text{ (silty sand)}}$$
Soil B — grain-size split. Percent finer than the No. 200 sieve is 55%, which exceeds 50% → Soil B is fine-grained; classification is read directly off the plasticity chart (gravel/sand split is not needed).
Soil B — plasticity chart. $I_P=w_L-w_P=52-32=\boxed{20\%}$. The A-line at $w_L=52$ is
$$I_{P,A\text{-line}}=0.73(52-20)=0.73(32)=\boxed{23.36\%}$$
$I_P=20\%<23.36\%$ plots below the A-line, and $w_L=52\ge50$ places it in the high-plasticity band.
$$\boxed{\text{Soil B} = MH \text{ (elastic silt of high plasticity)}}$$
$D_{60}$/$D_{30}$ for Soil A (log-linear interpolation, illustrative — not needed for the final symbol since fines control). $D_{30}=0.15$ mm lands exactly on the No. 100 point (30% finer). $D_{60}$ interpolates between the No. 4 point (4.76 mm, 77%) and the No. 20 point (0.84 mm, 55%):
$$D_{60}=10^{\left[\log(4.76)+\frac{60-77}{55-77}\big(\log(0.84)-\log(4.76)\big)\right]}\approx\boxed{1.25\text{ mm}}$$
$D_{10}$ would need hydrometer data below the No. 200 sieve (18% finer is already below the 10% mark) and is not determinable from the sieve data alone — consistent with fines >12% making $C_u$/$C_c$ inapplicable to the final Soil-A symbol anyway.
