Question 3 of 6: Shear Strength

Question 3: Shear Strength (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Test A — CD triaxial

Given. Cell (confining) pressure $\sigma_3=100\text{ kPa}$; no back pressure, so effective $=$ total; additional axial effective stress at failure (deviator stress) $\Delta\sigma_1=125\text{ kPa}$. The soil is a sand (given), so $c'=0$ is assumed.

Find. $c'$, $\phi'$; normal/shear stress and orientation of the actual failure plane; the plane of maximum shear stress and its available strength.

Approach. Compute $\sigma_1'$ at failure, then use the single-circle-through-the-origin construction (valid because the soil is cohesionless) to get $\phi'$, followed by the standard failure-plane stress and orientation formulas.

1. (a) Stresses at failure — the two Mohr circles. $\sigma_3'=100\text{ kPa}$ (initial, isotropic — the "initial" Mohr circle is a single point at $100$ kPa since cell pressure acts equally in all directions before shearing begins); $\sigma_1'=\sigma_3'+\Delta\sigma_1=100+125=225\text{ kPa}$. Circle centre $p=(\sigma_1'+\sigma_3')/2=162.5\text{ kPa}$, radius $R=(\sigma_1'-\sigma_3')/2=62.5\text{ kPa}$ (both circles plotted in the figure above).
2. (b)–(c) Failure envelope, $c'$ and $\phi'$. A cohesionless envelope through the origin tangent to the failure circle gives $$\sin\phi'=\frac{R}{p}=\frac{62.5}{162.5}=0.3846 \Rightarrow \boxed{\phi'=22.6^{\circ}, \quad c'=0}$$
3. (d) Normal and shear stress on the failure plane. Using $\sigma_{ff}=\dfrac{\sigma_1'+\sigma_3'}{2}-\dfrac{\sigma_1'-\sigma_3'}{2}\sin\phi'$ (matches the exam's own formula sheet) and $\tau_{ff}=\dfrac{\sigma_1'-\sigma_3'}{2}\cos\phi'$: $$\sigma_{ff}=162.5-62.5(0.3846)=\boxed{138.5\text{ kPa}} \qquad \tau_{ff}=62.5\cos(22.6^{\circ})=\boxed{57.7\text{ kPa}}$$ Check: $\sigma_{ff}\tan\phi'=138.5(0.4166)=57.7\text{ kPa}=\tau_{ff}$ ✓ (Mohr-Coulomb criterion satisfied exactly, as it must be at failure).
4. (e) Orientation of the failure plane. The failure plane always forms at $\theta_f=45^{\circ}+\phi'/2$ to the major principal plane (a general Mohr-circle theorem — the envelope is tangent to the circle at $2\theta_f=90^{\circ}+\phi'$ from the $\sigma_1'$ point): $$\boxed{\theta_f = 45^{\circ}+\tfrac{22.6^{\circ}}{2}=56.3^{\circ}\text{ from the horizontal (major principal) plane}}$$
5. (f) Plane of maximum shear stress. This is always the $45^{\circ}$ plane (from either principal plane), where $\sigma_n=p=162.5\text{ kPa}$ and $\tau=R=62.5\text{ kPa}$. Available strength there: $\tau_{\text{avail}}=\sigma_n\tan\phi'=162.5(0.4166)=67.7\text{ kPa}$, giving a local factor of safety $FoS=67.7/62.5=1.08$ — i.e. the $45^{\circ}$ plane carries slightly MORE available strength than demand, so the specimen actually fails on the weaker $56.3^{\circ}$ plane, not the $45^{\circ}$ plane of maximum shear. $$\boxed{\tau_{\max}=62.5\text{ kPa at }45^{\circ}, \quad \tau_{\text{avail,45}}=67.7\text{ kPa}}$$

Test B — direct shear

Given. Normal stress held at $\sigma_n=50\text{ kPa}$ throughout; at failure, $\sigma_n=50\text{ kPa}$, $\tau_f=45\text{ kPa}$ on the (horizontal) shear-box plane. (The stated "initial horizontal stress $=25$ kPa" is the pre-shear at-rest shear stress on the box before loading begins — it plays no role in the failure-state Mohr circle, which is fixed entirely by the failure-point $(\sigma_n,\tau_f)$.)

Find. Principal stresses at failure; orientation of the failure plane and of the major principal plane; the plane of maximum shear and its available strength; comparison with Test A part (f).

Approach. A direct shear test only ever measures one point on the failure envelope — the stress state on the (fixed, horizontal) shear-box plane at failure. Assuming $c'=0$ (consistent with Test A's cohesionless material and typical direct-shear interpretation), that point must lie exactly ON the envelope, so $\tan\phi'_B=\tau_f/\sigma_n$ directly; the full Mohr circle is then reconstructed from the tangency condition.

1. (a) Friction angle and principal stresses at failure. Since the failure point sits on the envelope $\tau=\sigma\tan\phi'$ through the origin: $$\boxed{\phi'_B=\arctan\!\left(\frac{45}{50}\right)=42.0^{\circ}}$$ For a circle tangent to $\tau=\sigma\tan\phi'_B$ with tangent point $(\sigma_n,\tau_f)$, the centre and radius satisfy $\sigma_n=p\cos^2\phi'_B$, giving $p=\sigma_n/\cos^2\phi'_B=50/\cos^2(42.0^{\circ})=90.5\text{ kPa}$ and $R=p\sin\phi'_B=90.5(0.6688)=60.5\text{ kPa}$: $$\boxed{\sigma_{1f}'=p+R=151.0\text{ kPa}, \qquad \sigma_{3f}'=p-R=30.0\text{ kPa}}$$ (check: $p\cos^2\phi'_B=50.0$ and $p\sin\phi'_B\cos\phi'_B=45.0$ — reproduces the given failure point exactly.)
2. (b) Orientation of the failure (shear-box) plane. By the same general theorem as Test A, the tangent/failure point sits at $\theta_f=45^{\circ}+\phi'_B/2$ from the major principal plane: $$\boxed{\theta_f = 45^{\circ}+21.0^{\circ}=66.0^{\circ}}$$ This is a genuinely useful result: since the box forces failure onto a horizontal plane whether or not that is the theoretically weakest orientation, the major principal plane in a direct shear test is NOT horizontal — it is tilted $66.0^{\circ}$ away from the shear plane, unlike the CD triaxial test where the principal planes are fixed by the apparatus (horizontal/vertical) throughout.
3. (c) Orientation of the major principal plane. Equivalently, the major principal plane is rotated $\theta_f=66.0^{\circ}$ from the horizontal shear-box plane (found via the pole method: the pole lies on the circle at the point diametrically reached by drawing a line through the failure point parallel to the horizontal plane it acts on; the chord from the pole to $\sigma_1'$ is then parallel to the major principal plane).
4. (d) Plane of maximum shear stress and its available strength. At the $45^{\circ}$ plane, $\sigma_n=p=90.5\text{ kPa}$, $\tau_{\max}=R=60.5\text{ kPa}$; available strength $=\sigma_n\tan\phi'_B=90.5(0.9)=81.5\text{ kPa}$, giving $FoS=81.5/60.5=1.35$ on that plane. $$\boxed{\tau_{\max}=60.5\text{ kPa at }45^{\circ}, \quad \tau_{\text{avail,45}}=81.5\text{ kPa}, \quad FoS=1.35}$$

The two tests give strikingly different $\phi'$ (22.6° vs 42.0°) on what the question implies is comparable material, and Test B's $45^{\circ}$-plane factor of safety ($1.35$) is well above Test A's ($1.08$) — the direct-shear result reads as a materially STRONGER soil, for reasons developed in the discussion below.

Which test do you trust more?

Trust the CD triaxial result (Test A) more. The direct shear test forces failure onto a pre-determined horizontal plane defined by the shear box halves, regardless of whether that is actually the plane of least resistance in the specimen; the CD triaxial specimen is free to fail on whichever plane the Mohr-Coulomb criterion first reaches, so its $\phi'$ reflects the true weakest orientation rather than an apparatus-imposed one. Direct shear also has a non-uniform stress and strain distribution across the specimen (stress concentrations at the box edges, progressive rather than simultaneous rupture across the shear plane), no direct measurement or control of pore pressure — so a "drained" assumption cannot be verified as it can in a triaxial cell with volume-change/pore-pressure instrumentation — and the principal stress directions rotate continuously during shearing in ways the simple Mohr-circle back-calculation above does not capture exactly. These effects typically bias direct-shear friction angles upward (as seen here), which is consistent with Test B reporting the higher, less conservative $\phi'$.

CD triaxial test steps. (1) Trim and mount a saturated cylindrical specimen inside a rubber membrane in the triaxial cell; apply cell pressure $\sigma_3$ all around via the confining fluid. (2) Saturation: apply back pressure in increments (raising both cell and pore pressure together) until the pore-pressure parameter $B=\Delta u/\Delta\sigma_3\approx 1$, confirming full saturation (this exam's Test A explicitly skips back pressure, so saturation is instead assumed from sample preparation). (3) Consolidation: with drainage valves open, hold $\sigma_3$ constant and allow the specimen to consolidate fully — pore pressure dissipates to zero and volume change stabilizes, fixing the effective confining stress $\sigma_3'$. (4) Shearing: increase the axial load slowly, with drainage valves kept OPEN throughout and the loading rate slow enough that excess pore pressure stays at (or very near) zero at all times — this is what makes the test "drained." (5) Record axial load, axial strain, and volume change continuously; identify failure at the peak deviator stress (or at critical state for a soil with no clear peak). (6) Compute $\sigma_1'=\sigma_3'+\Delta\sigma_{1,f}$ at failure and repeat at 2–3 different $\sigma_3'$ values to draw a full Mohr-Coulomb envelope (a single test, as given here, only fixes $\phi'$ by assuming $c'=0$).