Question 6 of 6: Angle of shearing resistance from a drained triaxial series
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Engineers Canada / PEO National Examinations, May 2016 —
07-Str-A3 Geotechnical Materials and Analysis. Closed book, three hours, 100 marks,
one approved Casio or Sharp calculator, drawing instruments required. All six questions are
compulsory and are weighted 20 / 10 / 10 / 20 / 20 / 20. The paper carries its own appendix
(pages 8–11): a formula sheet, the rectangular-loading m–n influence chart
and a Newmark influence chart whose influence value is printed on the formula sheet as
$\sigma_z = 0.005\,N q$, i.e. $I_N = 0.005$ over 200 elements. Values quoted below are taken
from those sheets.
Reference texts.
B. M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. —
Ch. 2–3 (grain size, phase relations), Ch. 6 (compaction), Ch. 7–8 (permeability
and seepage), Ch. 9 (stresses in a soil mass), Ch. 11 (consolidation), Ch. 12 (shear strength),
Ch. 13 (lateral earth pressure).
R. F. Craig / J. A. Knappett, Craig’s Soil Mechanics, 8th ed. — Ch. 2
(seepage and flow nets), Ch. 3 (effective stress), Ch. 4 (consolidation, $C_\alpha$), Ch. 5
(shear strength and the choice of test), Ch. 6 (stress distribution), Ch. 7 (lateral earth
pressure).
M. E. Harr, Groundwater and Seepage — Ch. 4 (the conformal solution for a
single sheet pile in a stratum of finite depth, used here to audit the drawn flow net).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM),
4th ed. — Ch. 4 (site investigation), Ch. 10 (settlement of organic soils and peat),
Ch. 27–28 (lateral earth pressure and support systems). The Canadian reference for
practice, sampling classes and terminology.
K. Terzaghi, R. B. Peck & G. Mesri, Soil Mechanics in Engineering Practice,
3rd ed. — Art. 16–17 (seepage, piping and the exit prism), Art. 19–20 (shear
strength), Art. 25 (secondary compression and $C_\alpha/C_c$).
Check — three readings of the printed paper, carried as stated.
(1) Question 1 is headed “(4 x 5 = 20 marks)” but prints five
lettered parts (i)–(v). All five are answered and the header total of 20 marks is kept
as printed; the mismatch is a printing slip in the source, not a missing part.
(2) Figure 1 (part (ii)) prints no dimension for the depth of the permeable stratum.
The dam base is dimensioned $L = 15$ m and the section is drawn to a single horizontal scale,
so the stratum has been scaled off the drawing at $T \approx 5.6$ m ($T/L = 0.37$). The
answer to (ii) does not depend on that number — it is read from the printed flow net
— but the independent numerical check below does, and the sensitivity is stated where
it is used.
(3) In Figure 4 the stem of Question 5(b) says point $A$ is at the “back of
the piling”, while the dot is drawn on the upstream (headwater) face of the
sheet pile, 1 m below ground. The drawn face is answered, and the value on the downstream face
at the same elevation is reported alongside it.
Question 6: Angle of shearing resistance from a drained triaxial series (Value: 20 marks)
Given. Four consolidated-drained tests on the same sand at the same
porosity. The all-round (cell) pressure is the minor effective principal stress at failure,
$\sigma_3'$, because the test is drained and the pore pressure is zero; the principal stress
difference is the deviator stress at failure, $(\sigma_1'-\sigma_3')_f$.
Find. $\phi'$ analytically, and a Mohr diagram that confirms it.
Approach. The soil is a sand, so the effective-stress envelope passes through
the origin and $c' = 0$; each test then gives $\phi'$ on its own from the geometry of its Mohr
circle, and the four values are compared to test whether a single straight envelope is
justified.
Form the major principal stress for each test. Adding the deviator stress
to the cell pressure,
$$\sigma_1' = \sigma_3' + (\sigma_1'-\sigma_3')_f,$$
which gives 552, 1108, 2210 and 4424 kN/m2 for the four cell pressures in turn.
Write the failure condition for a soil with no cohesion. For a Mohr circle
of centre $\tfrac12(\sigma_1'+\sigma_3')$ and radius $\tfrac12(\sigma_1'-\sigma_3')$ touching a
straight envelope through the origin, the perpendicular from the origin to the tangent point
gives directly
$$\sin\phi' = \frac{\sigma_1'-\sigma_3'}{\sigma_1'+\sigma_3'}.$$
This is the analytical route the question asks for; it needs no chart and no drawing.
Evaluate it test by test. Substituting each pair,
$$\sin\phi' = \frac{452}{652} = 0.69325,\quad \frac{908}{1308} = 0.69419,\quad
\frac{1810}{2610} = 0.69349,\quad \frac{3624}{5224} = 0.69372,$$
so that $\phi' = 43.89^\circ,\ 43.96^\circ,\ 43.91^\circ$ and $43.93^\circ$.
Assess the scatter before averaging. The four values span
$0.07^\circ$ — far tighter than the repeatability of a triaxial test — so a single
linear envelope through the origin is fully justified and the mean is meaningful:
$$\phi' = \tfrac14(43.89 + 43.96 + 43.91 + 43.93),$$
$$\boxed{\phi' = 43.9^\circ \quad\text{with}\quad c' = 0}$$
Confirm through the stress-ratio form. The formula sheet gives the
alternative statement $\sigma_1' = \sigma_3'\tan^2(45^\circ + \phi'/2)$. Rearranged,
$$\phi' = 2\left[\arctan\sqrt{\frac{\sigma_1'}{\sigma_3'}} - 45^\circ\right]
= 2\left[\arctan\sqrt{5.52} - 45^\circ\right] = 43.89^\circ$$
for the first test, reproducing the same value; the ratio $\sigma_1'/\sigma_3'$ lies between 5.52 and 5.54 for
all four tests, which is another way of saying the envelope is straight through the origin.
Confirm through the stress-point plot. The formula sheet also offers the
$t$–$s'$ construction, plotting $t = \tfrac12(\sigma_1'-\sigma_3')$ against
$s' = \tfrac12(\sigma_1'+\sigma_3')$ and recovering $\phi' = \arcsin(\tan\alpha)$. A
least-squares line through the origin fitted to the four stress points has
$\tan\alpha = 0.6937$, so $\alpha = 34.75^\circ$ and
$$\phi' = \arcsin(\tan 34.75^\circ) = 43.9^\circ,$$
identical to the direct result. The stress-point plot is the more robust construction when the
data scatter, because it fits a line rather than relying on a hand-drawn tangent.
Draw the Mohr circles. Each circle is drawn on the $\sigma'$ axis between
$\sigma_3'$ and $\sigma_1'$, and a single straight line from the origin at $43.9^\circ$ is found
to touch all four. The tangent points move up and to the right as the confining stress rises,
and the failure plane in each specimen is inclined at
$$\theta = 45^\circ + \frac{\phi'}{2} = 45^\circ + 21.96^\circ = 66.96^\circ$$
to the horizontal — the plane on which the tangent point sits.
The four Mohr circles at failure, with a single straight envelope from the origin at 43.9°. The horizontal and vertical scales are equal, so the tangency can be read directly off the drawing.
Test
$\sigma_3'$ (kN/m2)
$(\sigma_1'-\sigma_3')_f$
$\sigma_1'$
$\sigma_1'/\sigma_3'$
$\sin\phi'$
$\phi'$
1
100
452
552
5.52
0.69325
43.89°
2
200
908
1108
5.54
0.69419
43.96°
3
400
1810
2210
5.53
0.69349
43.91°
4
800
3624
4424
5.53
0.69372
43.93°
Angle of shearing resistance (mean of four tests)
43.9°
Effective cohesion intercept
$c' = 0$
Stress-point regression, $\alpha = 34.75^\circ$
43.9°
Inclination of the failure plane, $45^\circ + \phi'/2$
66.96°
Two remarks close the answer. A $\phi'$ of nearly $44^\circ$ is high, and it is consistent
with the stem: the specimens were all prepared at the same porosity, and a value in the low
forties implies that porosity was a dense one — a loose sand of the same mineralogy would
give something nearer $30$–$33^\circ$. And the perfect constancy of $\sigma_1'/\sigma_3'$
across a range of confining pressure from 100 to 800 kN/m2 is an idealisation: a real
dense sand shows a slightly curved envelope, because at high confining stress the
dilatancy that supplies much of the peak strength is suppressed and the grains begin to crush,
so $\phi'$ falls by a few degrees over that range. Reporting a single $\phi'$ is correct for
these data, but the secant value should always be quoted at the stress level of the works.
