Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 — 04-Geol-A6 Soil Mechanics. Three-hour, closed-book exam; a Casio or Sharp approved calculator, compass and ruler are permitted. Six questions constitute the complete 100-mark paper: Questions 1–5 are compulsory; Question 6 offers five 5-mark optional items of which the source asks for three.
Approach. Compute $\sigma_1$ at failure for each test, fit the common tangent (Mohr–Coulomb envelope) to both failure circles, then use $\theta_f=45^{\circ}+\phi'/2$ and the tangent-point stresses for the remaining parts.
(a) Failure stresses. Test A: $\sigma_{1f}=\sigma_3+\Delta\sigma_f=150+600=750$ kPa. Test B: $\sigma_{1f}=600+2550=3150$ kPa. Initial (isotropic, pre-shear) states plot as points (zero-radius circles) at $\sigma=150$ kPa and $\sigma=600$ kPa respectively.
(a) Circle geometry. Test A: centre $=\tfrac{750+150}{2}=450$ kPa, $R_A=\tfrac{750-150}{2}=300$ kPa. Test B: centre $=\tfrac{3150+600}{2}=1875$ kPa, $R_B=\tfrac{3150-600}{2}=1275$ kPa.
(b) Common-tangent envelope (shear strength). Solving $R=c'\cos\phi'+(\text{centre})\sin\phi'$ simultaneously for both circles gives $\sin\phi'=\dfrac{R_B-R_A}{\text{centre}_B-\text{centre}_A}=\dfrac{1275-300}{1875-450}=0.684\Rightarrow\phi'=\boxed{43.2^{\circ}}$, and $c'=-10.8$ kPa. A small negative intercept from a two-point fit is expected data scatter, not a real cohesion — dense, dry, cohesionless sand has $c'\equiv0$ physically, so the reported strength is $\boxed{c'\approx0,\ \phi'=43.2^{\circ}}$ (individual per-circle friction angles, forcing $c'=0$, bracket it: $\phi'_A=41.8^{\circ}$, $\phi'_B=42.8^{\circ}$).
(c) Shear stress on the failure plane. At the tangent point, $\tau_{ff}=R\cos\phi'$: Test A $\tau_{ff}=300\cos(43.2^{\circ})=\boxed{218.8\text{ kPa}}$; Test B $\tau_{ff}=1275\cos(43.2^{\circ})=\boxed{929.8\text{ kPa}}$ (normal stress on that plane, $\sigma_{ff}=\text{centre}-R\sin\phi'$: 244.7 kPa and 1002.6 kPa).
(d) Failure-plane orientation. $\theta_f=45^{\circ}+\phi'/2=45+21.6=\boxed{66.6^{\circ}}$ from the plane on which $\sigma_3$ acts (i.e. from horizontal), identical for both tests since both share the same envelope.
(e) Major principal plane. In a standard triaxial compression test $\sigma_1$ is the axial (vertical) stress, so the plane it acts on is normal to the vertical axis, i.e. $\boxed{\text{horizontal }(0^{\circ})}$, for both tests.
(f) Plane of maximum shear. Always $45^{\circ}$ from both principal planes → $\boxed{45^{\circ}\text{ from horizontal}}$, for both tests (a purely geometric result, independent of $\phi'$).
(g) Direct simple shear response. Both specimens are dense sand ($\phi'\approx42$–$43^{\circ}$, well above a typical critical-state friction angle of $\sim32$–$34^{\circ}$ for quartz sand); to shear, dense particle packings must ride up and over neighbouring grains, so the soil would $\boxed{\text{dilate}}$ (volume increase) in a DSS test.
Fig. Q3 — initial (points) and failure (semicircles) Mohr circles, Tests A and B, with the common-tangent Mohr–Coulomb envelope ($c'\approx0$, $\phi'=43.2^{\circ}$).
