Stress and Strain in Rocks Questions and Answers

Answer: B

The behaviour is expressed by Poisson's Ratio.

When a bar is stretched along its length it also contracts sideways. Poisson's ratio is the ratio of the lateral (transverse) strain to the longitudinal strain:

\[ \nu = -\frac{\Delta D / D}{\Delta L / L} \]

For most rocks it lies between about 0.2 and 0.35. The rigidity modulus relates shear stress to shear strain, tensile strength is the stress at which a material breaks in tension, and 'stretching constant' is not a standard elastic modulus, so none of them describes the paired lengthening and narrowing.

Answer: D

The correct match is a-4, b-3, c-2, d-1.

• The Schmidt hammer gives a rebound value from which an approximate uniaxial compressive strength is estimated (a-4).
• The point load test gives a strength index used to classify rock and to correlate with uniaxial compressive strength (b-3).
• The tilt test (printed 'Till test') tilts a block until its upper part slides, giving the friction angle of a discontinuity and the basic friction angle used in calculating the residual friction angle and joint roughness coefficient (c-2).
• The pressuremeter expands a probe against the walls of a borehole to measure the deformability of the rock (d-1).

Every other code mismatches at least two of these pairs.

Answer: B

The normal stresses are \(\sigma_{xx}\) and \(\sigma_{yy}\); the shear stresses are \(\sigma_{xy}\) and \(\sigma_{yx}\).

In index notation the first subscript names the plane (by its normal) and the second the direction of the stress component. When both indices are the same, the stress acts perpendicular to the plane, so it is a normal stress; when they differ, it acts along the plane, so it is a shear stress. In two dimensions (x, y) the stress tensor is therefore \(\sigma_{xx}, \sigma_{xy}, \sigma_{yx}, \sigma_{yy}\), with \(\sigma_{xy} = \sigma_{yx}\) for equilibrium.

The other options either swap normal and shear stresses or bring in a z index, which does not belong to a two-dimensional stress state.

Answer: D

The ultimate bearing capacity comes to 502.6 kPa.

A clay loaded quickly behaves undrained (\(\phi = 0\)), with cohesion half the unconfined compressive strength:

\[ c_{u} = \frac{q_{u}}{2} = \frac{150}{2} = 75\ \text{kPa} \]

For \(\phi = 0\) the bearing capacity factor is \(N_{c} = 5.14\) and the shape factor of a square footing is \(s_{c} = 1 + 0.2\,B/L = 1.2\). The overburden at the founding level is the total stress \(\gamma D_{f}\), because in an undrained analysis the pore pressure does not enter, even though the water table is at the surface:

\[ \begin{aligned} q_{ult} &= c_{u}N_{c}s_{c} + \gamma D_{f} \\ &= 75(5.14)(1.2) + 20(2) \\ &= 462.6 + 40 = 502.6\ \text{kPa} \end{aligned} \]

The other options (304.8, 403.9 and 700.9 kPa) cannot be reproduced from these data with a standard method.

This is the answer under the general bearing capacity equation without a depth factor. Skempton's chart value for a square footing at \(D_{f}/B = 1\) (\(N_{c} \approx 7.2\)) gives a larger figure of about 580 kPa, which is not among the options, so the question evidently intends the equation above.

Answer: C

This is a Mohr diagram.

In a Mohr diagram normal stress is plotted on the horizontal axis and shear stress on the vertical axis. The state of stress at a point is a circle whose centre is the mean stress and whose radius is the maximum shear stress; each point on the circle gives the normal and shear stress on one plane through the point. Adding a failure envelope shows whether the rock will fracture.

The Fry diagram is a method of measuring finite strain from the spacing of object centres. A phi-diagram plots angles in structural or textural analysis, and kinematic flow diagrams describe the flow of material, not the stresses on planes.

Answer: A

In progressive simple shear the fastest and slowest stretching directions are at 45° to the shear plane.

The instantaneous stretching axes (ISA) of a steady simple shear flow are fixed in orientation, at 45° and 135° to the shear plane, one being the direction of fastest stretching and the other of fastest shortening. The finite strain ellipse starts parallel to them and then rotates towards the shear plane as strain accumulates.

An angle of 0° would put them in the shear plane and 90° normal to it, which describes neither direction. A range of 0 to 30° describes the rotating finite strain axis at large strain, not the fixed instantaneous axes.

Answer: B

The second part is the deviatoric stress.

Any stress tensor can be written as the sum of an isotropic part, the mean stress (one third of the sum of the principal stresses) acting equally in all directions, and a deviatoric part, which is what remains after the mean is subtracted:

\[ \sigma_{ij} = \sigma_m \delta_{ij} + \sigma'_{ij} \]

The mean stress changes volume, while the deviatoric stress changes shape and drives distortion, folding and faulting.

Axial stress is stress along one axis, as in a uniaxial test. Non-axial stress and dialative stress are not standard names for this decomposition.

Answer: A

The volume change per unit pressure change is the compressibility.

Under a change in hydrostatic (all-round) stress a body changes volume but not shape. The compressibility is defined as

\[ \beta = -\frac{1}{V} \frac{dV}{dp} \]

and its reciprocal is the bulk modulus K.

Lithostatic pressure is the pressure exerted by the weight of overlying rock, a stress and not a response to it. Isostatic stress is not a standard term for this quantity, and hydrostatic deformation names the change itself, not the change per unit of pressure.

Answer: C

Statements 2 and 4 are true, so the answer is (2) and (4) are true; (1) and (3) are false.

In simple shear, material lines rotate faster than the axes of the strain ellipse, which rotate only gradually towards the shear plane (2). All lines rotate in the same sense, the sense of shear, whatever their orientation, unlike pure shear, where lines on either side of the principal axes rotate in opposite senses (4).

Statement 1 is false because lines lying in the shear plane neither rotate nor change length. Statement 3 is false because, on a Schmidt net, lines rotate along great circles towards the shear direction, not outward from it.

Answer: A

It is lithostatic stress.

Lithostatic stress, or lithostatic pressure, is the pressure at a depth due to the weight of the overlying rock, \(P = \rho g h\). Rocks at depth over long periods cannot sustain large differential stresses, so this pressure is taken to act equally in all directions, like hydrostatic pressure in a fluid; it rises by about 27 MPa per km in the crust.

Normal stress is the component of stress acting perpendicular to any plane, not specifically the weight of overlying rock. Homogeneous stress means stress that is the same throughout a body. 'Isotropic stress' describes a state equal in all directions but does not say what causes it; the specific term for the pressure from the overlying column is lithostatic.