1 Million+ Step-by-step solutions

Given a cylindrical sandstone laboratory test specimen of diameter D and height L subjected to a vertical load F (force), find

(a) Te forces N and T acting normal and tangential to a surface inclined at an angle θ to the vertical and

(b) Stresses, that is, the forces per unit area, σ and τ on the inclined surface in terms of the applied stress σo (F/A), where A is the area acted on by F. Illustrate with sketches. Note: The inclined surface is elliptical and has an area πab where a and b are the semi-axes of the ellipse.

(a) Te forces N and T acting normal and tangential to a surface inclined at an angle θ to the vertical and

(b) Stresses, that is, the forces per unit area, σ and τ on the inclined surface in terms of the applied stress σo (F/A), where A is the area acted on by F. Illustrate with sketches. Note: The inclined surface is elliptical and has an area πab where a and b are the semi-axes of the ellipse.

Given a rectangular prism of edge length D and height L subjected to a vertical load F (force) applied to the prism ends and a horizontal pressure p applied to the sides of the prism, find

(a) the forces N and T acting normal and tangential to a surface inclined at an angle θ to the vertical, measured in a plane parallel to a side of the prism and

(b) stresses, that is, the forces per unit area, σ and τ on the inclined surface in terms of the applied stress σo (F/A), and pressure p where A is the area acted on by F. Illustrate with sketches.

(a) the forces N and T acting normal and tangential to a surface inclined at an angle θ to the vertical, measured in a plane parallel to a side of the prism and

(b) stresses, that is, the forces per unit area, σ and τ on the inclined surface in terms of the applied stress σo (F/A), and pressure p where A is the area acted on by F. Illustrate with sketches.

Is the following equation a physical law, kinematic relationship, or material law?

Ms = ∫v rdm

Here M =mass in volume V, dm = a mass element, r = position vector of dm, and s is the position of the mass center.

Ms = ∫v rdm

Here M =mass in volume V, dm = a mass element, r = position vector of dm, and s is the position of the mass center.

Is the following equation a physical law, kinematic relationship, or material law?

˙σ = ∂σ / ∂t

˙σ = ∂σ / ∂t

Darcy’s law for fluid flow in porous media relates the seepage velocity v to the hydraulic gradient h through a constant k (hydraulic conductivity). Thus,

v = kh

Is this relationship a physical law, kinematic relation, or material law?

v = kh

Is this relationship a physical law, kinematic relation, or material law?

Consider a “two-dimensional” state of stress in the x–y plane characterized by:

σxx = 2,500 σyy = 5,200 τxy = 3,700

σxx = 2,500 σyy = 5,200 τxy = 3,700

Given the stress state in Problem 1.11, find: the magnitude and direction of the maximum shear stress and illustrate with a sketch.

Consider a “two-dimensional” state of stress in the x–y plane characterized by:

σxx = 17.24 σyy = 35.86 τxy = 25.52

Where units are MPa and tension is positive. Find: the magnitude and direction of the principal stresses and illustrate with a sketch

σxx = 17.24 σyy = 35.86 τxy = 25.52

Where units are MPa and tension is positive. Find: the magnitude and direction of the principal stresses and illustrate with a sketch

Given the stress state in Problem 1.13, find: the magnitude and direction of the maximum shear stress and illustrate with a sketch

Suppose that

σxx = 2,500

σyy = 5,200

τxy = 3,700

and the z-direction shear stresses (τzx, τzy) are zero, while the z-direction normal stress (σzz) is 4,000 psi. Find: the major, intermediate, and minor principal stresses.

σxx = 2,500

σyy = 5,200

τxy = 3,700

and the z-direction shear stresses (τzx, τzy) are zero, while the z-direction normal stress (σzz) is 4,000 psi. Find: the major, intermediate, and minor principal stresses.

With reference to Problem 1.15, find the magnitude of the maximum shear stress and show by sketch the orientation of the plane it acts on.

Suppose that

σxx = 17.24 σyy = 35.86 τxy = 25.52 in MPa and the z-direction shear stresses (τzx, τzy) are zero, while the z-direction normal stress (σzz) is 27.59 MPa. Find: the major, intermediate, and minor principal stresses.

σxx = 17.24 σyy = 35.86 τxy = 25.52 in MPa and the z-direction shear stresses (τzx, τzy) are zero, while the z-direction normal stress (σzz) is 27.59 MPa. Find: the major, intermediate, and minor principal stresses.

With reference to Problem 1.17, find the magnitude of the maximum shear stress and show by sketch the orientation of the plane it acts on.

Consider a “two-dimensional” state of stress in the x–y plane characterized by: σxx = 5,200 σyy = 2,500 τxy = −3,700 where units are psi and tension is positive. Find: the magnitude and direction of the principal stresses and illustrate with a sketch.

Consider a “two-dimensional” state of stress in the x–y plane characterized by:

σxx = 35.86 σyy = 17.24 τxy = −25.52

Where units are MPa and tension is positive. Find: the magnitude and direction of the principal stresses and illustrate with a sketch.

σxx = 35.86 σyy = 17.24 τxy = −25.52

Where units are MPa and tension is positive. Find: the magnitude and direction of the principal stresses and illustrate with a sketch.

Consider a three-dimensional state of stress characterized by:

σxx = 3,000 σyy = 2,000 τxy = 0.

σzz = 4,000 τzx = 0. τyz = 2,500

where units are psi and compression is positive. Find: the magnitude and direction of the principal stresses and illustrate with a sketch

σxx = 3,000 σyy = 2,000 τxy = 0.

σzz = 4,000 τzx = 0. τyz = 2,500

where units are psi and compression is positive. Find: the magnitude and direction of the principal stresses and illustrate with a sketch

Given the stress state in Problem 1.21, find the state of stress relative to axes
abc(σaa, . . . , τca) rotated 30◦ counterclockwise about the z-axis

Consider a three-dimensional state of stress characterized by:

σxx = 20.69 σyy = 13.79 τxy = 0.0

σzz = 27.59 τzx = 0.0 τyz = 17.24

Where units are MPa and compression is positive. Find: the magnitude and direction of the principal stresses and illustrate with a sketch.

σxx = 20.69 σyy = 13.79 τxy = 0.0

σzz = 27.59 τzx = 0.0 τyz = 17.24

Where units are MPa and compression is positive. Find: the magnitude and direction of the principal stresses and illustrate with a sketch.

Given the stress state in Problem 1.23, find the state of stress relative to axes abc(σaa, . . . , τca) rotated 30◦ counterclockwise about the z-axis.

Show in two dimensions that the mean normal stress σm = (σxx +σyy)/2 is invariant under a rotation of the reference axes and is equal to one-half the sum of the major and minor principal stresses in the considered plane. In the context of this problem, invariant means independent of the magnitude of the rotation angle.

Show in two dimensions that the maximum shear stress τm = {[(σxx − σyy)/2]2 +
(τxy)2}(1/2) is invariant under a rotation of the reference axes and is equal to one-half
the difference of the major and minor principal stresses in the considered plane.

Consider an NX-size drill core (2-1/8 in., 5.40 cm diameter) with an L/D ratio (length to diameter) of two under an axial load of 35,466 pounds (158.89 kN). Find:

1 The state of stress within the core sample relative to a cylindrical system of coordinates (rθz)

2 The state of stress relative to an abc set of axes associated with a plane that bears due north and dips 60◦ to the east. The c-axis is normal to this plane, the b-axis is parallel to strike and the a-axis is down dip. Use the equations of transformation.

3 Show by a direct resolution of force components and calculation of area that the normal stress (σcc) on the plane in part (2) is the same as that given by the equations of transformation.

4. Is the shear stress (τac) acting on the plane in part (2) the same using forces and areas as that given by the equations of transformation? Show why or why not.

1 The state of stress within the core sample relative to a cylindrical system of coordinates (rθz)

2 The state of stress relative to an abc set of axes associated with a plane that bears due north and dips 60◦ to the east. The c-axis is normal to this plane, the b-axis is parallel to strike and the a-axis is down dip. Use the equations of transformation.

3 Show by a direct resolution of force components and calculation of area that the normal stress (σcc) on the plane in part (2) is the same as that given by the equations of transformation.

4. Is the shear stress (τac) acting on the plane in part (2) the same using forces and areas as that given by the equations of transformation? Show why or why not.

Strain measurements are made on a wide, flat bench in a dimension stone quarry using a 0-45-90 rosette. The 0-gauge is oriented N60E. Specific weight of the stone (a granite) is 162 pcf (25.6 kN/m3); Young’s modulus E = 12.7(10)6 psi (87.59 GPa) and Poisson’s ratio ν = 0.27. Tension is considered positive. Measured strains in micro units per unit are:

(0) = −1480, (45) = −300, (90) = −2760 Find (where x = east, y = north and z = up):

1 the strains xx, yy, xy (tonsorial shear strain);

2 the stresses σzz, τyz, τzx (in psi or MPa);

3 the stresses σxx, σyy, and τyx (in psi or MPa);

4 the strain zz (micro inches per inch or mm/m);

5 the strains yz and zx (tonsorial shear strains);

6 the direction of the true principal stresses σ1, σ2, σ3, Where σ1 ≥ σ2 ≥ σ3 (tension is positive) and sketch;

7 the magnitudes of the principal stresses σ1, σ2, σ3; Where σ1 ≥ σ2 ≥ σ3 (tension is positive);

8 the magnitudes of the principal strains 1, 2, 3;

9 the directions of the principal strains; and sketch;

10 the change v in volume per unit volume that would occur if the stresses were Entirely relieved, that is reduced to zero, for example, by over coring the rosette.

(0) = −1480, (45) = −300, (90) = −2760 Find (where x = east, y = north and z = up):

1 the strains xx, yy, xy (tonsorial shear strain);

2 the stresses σzz, τyz, τzx (in psi or MPa);

3 the stresses σxx, σyy, and τyx (in psi or MPa);

4 the strain zz (micro inches per inch or mm/m);

5 the strains yz and zx (tonsorial shear strains);

6 the direction of the true principal stresses σ1, σ2, σ3, Where σ1 ≥ σ2 ≥ σ3 (tension is positive) and sketch;

7 the magnitudes of the principal stresses σ1, σ2, σ3; Where σ1 ≥ σ2 ≥ σ3 (tension is positive);

8 the magnitudes of the principal strains 1, 2, 3;

9 the directions of the principal strains; and sketch;

10 the change v in volume per unit volume that would occur if the stresses were Entirely relieved, that is reduced to zero, for example, by over coring the rosette.

Given the stresses in psi:
σxx = 1,500 σyy = −2,000 σzz = 3,500
τxy = 600 τyz = −300 τzx = −500
Where x = east, y = north, z = up, compression is positive, find: the secondary principal stresses in the zx-plane and sketch

Given the stresses in MPa:
σxx = 10.35 σyy = −13.79 σzz = 24.14
τxy = 4.14 τyz = −2.07 τzx = −3.45
where x = east, y = north, z = up, compression is positive, find: the secondary
principal stresses in the zx-plane and sketch.

Given the strains:
εxx = 2,000 εyy = 3,000 εzz = 4,500
γxy = −200 γyz = 300 γzx = 225
Where x = east, y = north, z = up, compression is positive, the units are micro inches per inch, Young’s modulus E = 5.0(10)6 psi, and shear modulus
G = 2.0(10)6 psi, find the corresponding stresses for a linear, homogeneous, isotropic elastic response.

Consider a cylindrical test specimen under a confining pressure of 3,000 psi and an axial stress of 3,000 psi with compression positive, so that in cylindrical coordinates
σzz = 3,000 σrr = 3,000 σθθ = 3,000
τrz = 0.0 τzθ = 0.0 τθr = 0.0
Find:
1 εrr , εzz , εθθ , γrz , γzθ , γθr ;
2 The axial stress σzz required to maintain a zero axial strain;
3 The strain energy and strain energy density, if the test specimen is an NX core with a height to diameter ratio of two.
Note: E = 2.4 × 106 psi, ν = 0.20, and the sample is isotropic.

Given the strains:
εxx = 2,000 εyy = 3,000 εzz = 4,500
γxy = −200 τyz = 300 γzx = 225
Where x = east, y = north, z = up, compression is positive, the units are micro inches per inch, Young’s modulus E = 34.48 GPa, and shear modulus G = 13.79 GPa psi, find the corresponding stresses for a linear, homogeneous, isotropic elastic response.

Consider a cylindrical test specimen under a confining pressure of 20.69 MPa and an axial stress of 20.69 MPa with compression positive, so that in cylindrical coordinates

σzz = 20.69 σrr = 20.69 σθθ = 20.69

τrz = 0.0 τzθ = 0.0 τθr = 0.0

Find:

1 εrr , εzz , εθθ , γrz , γzθ , γθr ;

2 the axial stress σzz required to maintain a zero axial strain;

3 the strain energy and strain energy density, if the test specimen is an NX core with a height to diameter ratio of two. Note: E = 16.55 GPa, ν = 0.20, and the sample is isotropic.

σzz = 20.69 σrr = 20.69 σθθ = 20.69

τrz = 0.0 τzθ = 0.0 τθr = 0.0

Find:

1 εrr , εzz , εθθ , γrz , γzθ , γθr ;

2 the axial stress σzz required to maintain a zero axial strain;

3 the strain energy and strain energy density, if the test specimen is an NX core with a height to diameter ratio of two. Note: E = 16.55 GPa, ν = 0.20, and the sample is isotropic.

Show that under complete lateral restraint and gravity loading only, that the vertical normal stress in a homogeneous, isotropic linearly elastic earth at a depth z measured from the surface with compression positive is γ z and that the horizontal normal stresses are equal to ν/(1 − ν) times the vertical stress where ν is Poisson’s ratio.

Suppose an NX-size test cylinder with an L/D ratio of two has a Young’s modulus of 10 million psi, a Poisson’s ratio of 0.35 and fails in uniaxial compressive at 0.1% strain.

Find:

1 The axial load (kN) and stress (unconfined compressive strength, psi) at failure;

2 The relative displacement of the cylinder ends at failure (in inches).

Find:

1 The axial load (kN) and stress (unconfined compressive strength, psi) at failure;

2 The relative displacement of the cylinder ends at failure (in inches).

Suppose an NX-size test cylinder with an L/D ratio of two has a Young’s modulus of 68.97 GPa, a Poisson’s ratio of 0.35, and fails in uniaxial compressive at 0.1% strain.

Find:Br> 1 The axial load (kN) and stress (unconfined compressive strength, MPa) at failure;

2 The relative displacement of the cylinder ends at failure (in mm).

Find:Br> 1 The axial load (kN) and stress (unconfined compressive strength, MPa) at failure;

2 The relative displacement of the cylinder ends at failure (in mm).

Consider gravity loading under complete lateral restraint in flat, stratified ground where each stratum is homogeneous, isotropic, and linearly elastic. Assume compression is positive, ν is Poisson’s ratio, z is depth, and γ is average specific weight to any particular depth. A 250 ft thick, water-bearing sandstone is encountered at a depth of 1,300 ft. Water pressure at the top of the sandstone is 80 psi. Estimate the total and effective stresses at the center of the sandstone.

Consider gravity loading under complete lateral restraint in flat, stratified ground where each stratum is homogeneous, isotropic, and linearly elastic. Assume compression is positive, ν is Poisson’s ratio, z is depth, and γ is average specific weight to any particular depth. A 76.2 m thick, water-bearing sandstone is encountered at a depth of 396 m. Water pressure at the top of the sandstone is 552 kPa. Estimate the total and effective stresses at the center of the sandstone.

Consider gravity loading only under complete lateral restraint in flat strata with properties given in Table 1.2. Vertical stress at the top of the geologic column given in Table 1.2 is 1,000 psi. Compression is positive, γ is specific weight, h is thickness, E is Young’s modulus, and G is shear modulus. Find the horizontal stress at the bottom of the sandstone and the top of the limestone

Consider gravity loading only under complete lateral restraint in flat strata with properties given in Table 1.3. Vertical stress at the top of the geologic column given in Table 1.3 is 6.9 MPa. Compression is positive, SG is specific gravity, h is thickness,

Tables included in the Solution

Tables included in the Solution

A stress measurement made in a deep borehole at a depth of 500 m (1,640 ft) indicates a horizontal stress that is three times the vertical stress and that the vertical stress is consistent with an estimate based on overburden weight alone, that is, 25 kPa per meter of depth (1.1 psi per ft of depth). Estimate the portions of horizontal stress associated with weight alone and with other forces.

Identify the three major categories of equations that form the “recipe” for calculating rock mass motion. Give an example of each in equation form.

If ρ is mass density, explain why or why not the conservation of mass may be expressed as
M = ∫v ρdV
For a body of mass M occupying volume V

Name the two types of external forces recognized in mechanics and give an example of each.

A huge boulder is inadvertently cast high into the air during an open pit mine blast. Using the definition of mass center and Newton’s second law of motion, show that the center of mass of the boulder travels the same path even if it disintegrates into 1,001 pieces of various sizes and shapes.

Suppose a static factor of safety is defined as the ratio of resisting to driving forces, that is, FS = R/D. Show that a factor of safety less than one implies acceleration is impending

Derive an expression for the factor of safety FS for a planar block slide with a tension crack behind the crest. The slope height is H and has a face angle β measured from the horizontal to the slope face; the failure surface is inclined at α degrees to the horizontal. Tension crack depth is h. Only the force of gravity acts on the slide mass; specific weight is γ. A Mohr–Coulomb failure criterion applies to the failure surface (cohesion is c, angle of internal friction is φ). Note: The slide mass extends b units into the plane of the page. Show schematically a plot of FS (y-axis, range from 0.0 to 4.0) as a function of the reciprocal of the slope height H−1 (x-axis, range from 0.0 to 0.001) and discuss briefly the form and significance of the plot.

Modify the expression for the safety factor from Problem 2.1 to include the effect of a water force P acting on the inclined failure surface. Note that the water table is below the bottom of the tension crack.

With reference to Problem 2.2, suppose the water pressure p increases linearly with depth (according to p = γwz where z = depth below water table and γw = specific weight of water) from the water table elevation down to a point one-half the vertical distance to the slide toe and then decreases linearly from the halfway point to the toe.

Consider a seismic load effect S that acts horizontally through the slide mass center with an acceleration as given as a decimal fraction ao of the acceleration of gravity g (as = aog, typically ao = 0.05 to 0.15, depends on seismic zone). Modify the safety factor expression from Problem 1 to include seismic load effect S.

Given the planar block slide data shown in the sketch where a uniformly distributed surcharge σ is applied to the slope crest over an area (bl), first find the slope safety factor without a surcharge and then find the magnitude of the surcharge possible for a slope safety factor of 1.1 against translational sliding. Note that b = 25 ft (7.62 m).

Consider the planar block slide in the sketch. If no surcharge is present, find the maximum depth H of excavation possible before failure impends.

Suppose the slope in the sketch is cable bolted. No surcharge is present. Bolt spacing is 50 ft (15.2 m) in the vertical direction and 25 ft (7.6 m) in the horizontal direction. Bolts assemblages are composed of 12 strands of Type 270 cable (495,600 lbf, ultimate strength or 2.22 MN) and are installed in down holes (5◦). Design tension is 60% of the ultimate strength. Find the safety factor obtained by bolting and therefore the improvement in the safety factor obtained (difference between the bolted and unbolted slope safety factors).

Consider the planar block slide in sketch without surcharge, seismic load, and bolt reinforcement and suppose the water table rises to the top of the slide. Find the safety factor of the “flooded” slide mass.

Suppose the cohesion of the slide mass shown in the sketch decreases to zero and no surcharge, seismic load, or water is present. Determine whether the slide mass will accelerate, and if so, the magnitude and direction of the acceleration of the slide mass center.

Some data for a possible planar block slide are given in the sketch. If α = 35◦, β = 45◦, and H = 475 ft (145 m), what cohesion c (psf, kPa) is needed to give a safety factor of at least 1.5?

Consider the planar block slide in the sketch.

(a) Show algebraically that reduction of the slide mass volume from V_{o} to V_{1} by excavating a relieving bench near the crest necessarily increases the safety factor, provided the water table is lowered below the toe.

(b) Show that placing a berm of weight W_{1} at the toe of the slide, where cohesion c_{1} and friction angle φ_{1} are mobilized at the berm bottom, necessarily increases the safety factor, other factors remaining the same.

With reference to the sketch, if no tension crack is present, α = 29^{◦}, γ = 156pcf (24.7kN/m^{3}), joint persistence = 0.87, no crest relieving bench and no toe berm are being considered, determine

(a) The maximum slope height H when β = 50◦ and the water table is at the crest, and

(b) When the slope is completely depressurized. Rock and joint properties are given in Table 2.6.

A generic diagram of a slope in a jointed rock mass that is threatened by a planar block slide is shown, in the sketch. Although not shown, bench height is 55 ft (16.76 m).

Given that Mohr–Coulomb failure criteria apply, the clay-filled joints constitute 93%

of the potential shear failure surface, no tension cracks have yet appeared and:

1 slope height H =? ft (m);

2 failure surface angle α = 32◦;

3 slope angle β = 49◦;

4 friction angle (rock) φ_{r} = 38◦

5 cohesion (rock) c_{r }= 1000 psi (6.90MPa);

6 friction angle (joint) φ_{j} = 27◦;

7 cohesion (joint) c_{j} = 10 psi (0.069MPa);

8 specific weight γ = 158pcf (25.0kN/m^{3});

9 tension crack depth h_{c} = 0.0 ft (m);

10 water table depth h_{w} = 0.0 ft (m);

11 seismic coefficient a_{o} = 0.00;

12 surcharge s = 0.0 psf (kPa)

Find the maximum pit depth possible without drainage.

A generic diagram of a slope in a jointed rock mass that is threatened by a planar block slide is shown in the sketch. A minimum safety factor of 1.05 is required. Although not shown bench height is 60 ft (18.3 m). Determine if FS =1.05 is possible, given that Mohr–Coulomb failure criteria apply, the joints constitute 79% of the potential shear failure surface, slide block weight is 1.351(10^{7}) lbf/ft (198.6 MN per meter) of thickness and:

1 slope height H = 540 ft (165 m);

2 failure surface angle α = 32◦;

3 slope angle β = 45◦;

4 friction angle (rock) φr = 33◦;

5 cohesion (rock) cr = 2870 psi (19.8 MPa);

6 friction angle (joint) φj = 28◦;

7 cohesion (joint) cj = 10.0 psi (0.069 MPa);

8 specific weight γ = 158pcf (25.0kN/m^{3});

9 tension crack depth hc = 50.0 ft (15.2 m);

10 water table depth hw = 60.0 ft (18.3 m);

11 seismic coefficient ao = 0.15;

12 surcharge σ = 0.0 psf (0.0 kPa).

If FS < 1.05 will drainage allow the safety factor objective to be achieved?

Given the following planar block slide data: Mohr–Coulomb failure criterion applies

1 slope height *H *= 613 ft (187.8 m);

2 failure surface angle *α *= 34^{◦};

3 slope angle *β *=?^{◦};

4 friction angle *φ *= 30^{◦};

5 cohesion *c *= 1440 psf (0.069 MPa);

6 specific weight *γ *= 162pcf (26.63kN/m^{3}*)*.

Find the maximum slope angle *β *possible, when the water table is drawn down below the toe of the slope. Note: No tension crack forms.

A generic diagram of a slope in a jointed rock mass that is threatened by a planar block slide is shown in the sketch. With neglect of any seismic load, find the slope height (pit depth) possible with a safety factor of 1.15, given that Mohr–Coulomb failure criteria apply, the joints constitute 86% of the potential shear failure surface and:

1 slope height H =? ft (m);

2 failure surface angle α = 37◦;

3 slope angle β = 48◦;

4 friction angle (rock) φ_{r} = 33◦;

5 cohesion (rock) c_{r} = 2580 psi (17.79 MPa);

6 friction angle (joint) φ_{j} = 33◦;

7 cohesion (joint) c_{j} = 0.0 psi (0.0 MPa);

8 specific weight γ = 162pcf (25.63kN/m^{3});

9 tension crack depth h_{c} = 0.0 ft (m);

10 water table depth h_{w}= 0.0 ft (m);

11 seismic coefficient a_{o} = 0.15;

12 surcharge σ = 0.0 psf (kPa).

With reference to the sketch of the potential slope failure shown in the sketch, find:

(a) The factor of safety of a cable bolted slope when the water table is drawn down 100 ft (30.5 m), bench height is 40 ft (12.2 m) (vertical bolt spacing), horizontal bolt spacing is 20 ft (6.1 m), the bolting angle is 5◦ down, and bolt loading is 700 kips (3.14 MN) per hole;

(b) The factor of safety of the same slope but without bolts when the water table is drawn down below the toe. The design trade-off here is between drainage and bolting, any reasons for preferring one or the other? Note: A tension crack 37 ft (11.3 m) deep is observed at the crest behind the face. The slope is 320 ft (97.5 m) high, failure surface angle is 32◦, slope angle is 40◦, specific weight is 158pcf (25.0kN/m^{3}*)*, cohesion is 1440 psf (0.069 MPa), and the slide surface friction angle is 28◦.

Given the data in Table 2.7, determine the direction angles of the normal vectors to joint planes A andB.

Calculate the dip and dip direction of the line of intersection of joint planes A and B, then sketch the result in compass coordinates.

Given that the vertical distance between points a and d is 120 ft (36.6 m), determine the areas of joint planes A and B, assuming no tension crack is present.

Use computer programs to verify the area data in Table 2.7 with a tension crack present and a rock specific weight of 158pcf (25.0kN/m3). What is the wedge safety factor when the water table is below the toe of the slope? Note: Tension crack offset (ae) is 90 ft (27.4m).

With reference to the generic wedge in the sketch and the specific data in Table 2.8 if the safety factor with respect to sliding down the line of intersection is greater than 1.10, then safety is assured. The water table is at the crest of the slope (but may be lowered to meet the safety criterion), vertical distance between a and g is 85 ft (25.9 m), point e is 48 ft (14.6 m) from point a, rock specific weight is 158pcf (25.0kN/m3) and active wedge volume is 3,735 yd3 (2,856 m3). What range of face orientations (dip directions) is of concern in an open pit mine where benches 95 ft (28.96 m) high areplanned?

Two joints K1 and K2 are mapped in the vicinity of a proposed surface mine. Joints in set K1 have dip directions of 110◦ and dips of 38◦; joints in set K2 have dip directions of 147◦ and dips of 42◦. Determine if potential wedge failures may form, and if so, what the dip direction and dip of the lines of intersection would be, and the range of bench face azimuths that should be examined more closely for slope stability. Sketch the data and results. Note: Formula for a joint plane normal (nx ny nz): (sin (δ) sin (α), sin (δ) cos (α), cos (δ)). Also, the components of a unit normal vector to a joint in set K1 are (0.5785, −0.2106, 0.7880).

With reference to the generic wedge shown in the sketch and the data in Table 2.9, the water table is below the toe, vertical distance between A and O is 85 ft (25.9 m), point T is 40 ft from point A, rock specific weight is 158pcf (25.0kN/m3), and active wedge volume is 3,850 yd3 (2,944 m3).

(a) Find the dip direction (azimuth) and dip of the line of intersection of joint planes A and B.

(b) Find the range of intersection line azimuths that preclude sliding down the line of intersection.

(c) Find the length of the line of intersection between the face and plane A. Note: The foreland isflat.

With reference to the wedge shown in the sketch and the Table 2.10 data, if the safety factor with respect to sliding down the line of intersection is greater than 1.10, then safety is assured. Can this objective be met? Yes, no, maybe? Explain. Note: The water table is below the toe of the slope, vertical distance between A and O is 100 ft, point T is 150 ft from point A and wedge volume is 308,0003.

Given the wedge data in Table 2.11, if the vertical distance between crest and the point where the line of intersection between the joint planes intersects the face is 120 ft (36.6 m), find the dip direction, dip, and length of the line formed by the intersection between the joint plane A and the face F.

With reference to the generic wedge shown in the sketch and the data in Table 2.12, the water table is below the toe, vertical distance between A and O is 25.9 m (85 ft),point T is 12.2 m (40 ft) from point A, rock specific gravity is 2.46, and active wedge volume is 2,944 m3 (3,850 yd3).

(a) Find the dip direction (azimuth) and dip of the line of intersection of joint planes A and B.

(b) Find the range of intersection line azimuths that preclude sliding down the line of intersection.

(c) Find the length of the line of intersection between the face and plane A. Note: The foreland isflat.

Consider a bank 40 ft (12.2 m) high with a slope of 30◦ that may fail by rotation on a slip circle of radius 60 ft (18.3 m) with center at (20, 48 ft) or (6.10, 14.63 m) relative to coordinates in ft with origin at the bank toe. In this case, the clip circle dips below the toe. Cohesion and friction angle are 720 psf (34.5 kPa) and 18◦. Table 2.13 gives widths of nine slices to be used for analysis. Numbering proceeds from the left near the toe to the right at the slope crest.

(a) Use a spreadsheet to analyze the slope stability and to show that the factor of safety is 2.164. Verify the reliability of this result using the computer program Slide or similar method of slices program. Attach a print of your spreadsheet and also attach prints of the slope and slip circle from the Interpreter and the Info viewer of Slide. The water table is below the bottom of the slip circle, that is, dry case.

(b) Repeat the analysis using your spreadsheet and then verify with Slide or equivalent using a water table that coincides with the toe-slope face-crest profile. What are the safety factors from your spreadsheet and from Slide? Again, attach prints of your spreadsheet and two prints from Slide.

With reference to the sketch and tables, find the safety factors for the four cases posed using the method of slices. Include all important calculations. If you use a spreadsheet, include suitable data prints. Complete a table for each of the conditions following:

1 With water table (WT) 1 and no relieving benches

2 With WT 2 and no relieving benches

3 With WT 2 and a 140×60 ft (42.7×18.3 m) relieving bench at the crest

4 With WT 1, a 140×60 ft relieving bench at the crest, and a relieving Bench 2 at the toe required to give a factor of safety of 1.05. Compute the length L of Bench 2 in this case.

Factor of safety = ________________________

Notes: Units are feet-ft (m).

Water table 1 (WT_{1}) is 30 ft (9.14 m) below the ground surface.

Water table 2 (WT_{2}) is 100 ft (30.5 m) below the ground surface.

Areas ft^{2} (m^{2}):

Parts of relief bench 1: slices:

A = 118 (10.96) 1 = 470 (43.7)

B = 1,682 (156.3) 2 = 1,937 (180)

C = 4,800 (445.9) 3 = 5,451 (506.4)

D = 1,800 (167.2) 4 = 12,246 (1,137.7)

Notes

a. A blank means zero (0). Drift is cohesion less, clay and interfaces are frictionless. The difference between dry and wet specific weight is neglected in this problem.

With reference to the sketch illustrating a rotational slide being considered for a method of slices stability analysis, show that the inclusion of seismic forces as quasistatic horizontal forces results in the expression and present a formula for the effective normal force in this expression. Illustrate with a slice force diagram and be sure to state importantassumptions.

Given the potential one-foot thick slope failure along the circular arc (R = 300 ft, 91.44 m) shown in the sketch, the data in Table 2.14, a slope height = 120 ft (36.6 m); slope angle = 29◦, specific weight = 95pcf (15.0kN/m3), cohesion = 367 psf (17.6 kPa), and angle of internal friction= 16◦, find:

1. Seismic force on slice 7, if the seismic coefficient is 0.15;

2. Water force at the bottom of slice 7;

3. Force safety factor of slice 7 alone taking into account the seismic and water forces;

4. Safety factor for the considered slip surface with no seismic or water forces acting.

Given the potential one-foot thick slope failure along the circular arc (R = 300 ft, 91.44 m) shown in the sketch and the data in Table 2.14, find the safety factor possible with drawdown of the water table below the failure surface. Slope height is 120 ft (36.6 m); slope angle is 29◦ and specific weight is about 95pcf (15kN/m3). Cohesion is 367 psf (17.6 kPa) and the angle of internal friction of the material is 16◦.

Given the potential one-foot thick slope failure along the circular arc shown in the sketch and the data in Table 2.14, find the safety factor for slice 2 and for slice 7, then show algebraically the safety factor for the slope. Slope height is 120 ft (36.6 m), slope angle is 29◦, and specific weight is about 95pcf (15kN/m^{3}). Cohesion is 367 psf (17.6 kPa) and the angle of internal friction of the material is 16◦.

Given the circular failure illustrated in the sketch, find algebraically the factor of safety. Assume that the material properties and geometry of slope and slices areknown.

Consider the possibility of a rock slide starting at point A and ending at B, as shown in the sketch, and suppose the slide is driven by gravity and resisted by friction. Derive expressions for acceleration, velocity, and distance moved down slope by the mass center of the slide over any segment of constant inclination from the horizontal. Be sure to identify all terms andassumptions.

Given the slope profile in vertical section shown in the sketch and an angle of friction φ = 15◦, if the mass center of the slide is initially at 1, determine the height of the mass center after sliding. This is the distance of 5 above the valley floor.

Toppling of tall blocks is possible when one corner tends to lift off from the base supporting the block, as shown in the sketch. Show that the block is stable provided tan(α)

A rock block with a square base rests on a 28◦ slope. The block base plane has a friction angle of 32◦. There is no adhesion between block and base plane. Find the base dimensions necessary to just prevent toppling.

A 20 foot diameter circular shaft is planned in a massive rock. Laboratory tests on core from exploration drilling show that

C_{o} = 22,000 psi T_{o} = 1,200 psi

γ = 144pcf E = 5 × 10^{6} psi

G = 2.0 × 10^{6} psi

Depth is 3,000 ft. Site measurements show that no tectonic stresses are present. Determine the factors of safety with respect to failure in compression and tension.

If the opening in Problem 1 is in an in situ stress field such that σh = σv, what are the safety factors?

If the opening in Problem 1 is in an in situ stress field such that σh = 0.01σv, is there a possibility of failure? Justify your answer.

A 6 m diameter circular shaft is planned in massive rock. Laboratory tests on core from exploration drilling show that

Co = 152 MPa To = 8.3MPa

γ = 23kN/m3 E = 34.5GPa

G = 13.8GPa

Depth is 915 m. Site measurements show that no tectonic stresses are present. Determine the factors of safety with respect to failure in compression and tension.

If the opening in Problem 4 is in an in situ stress field such that σh = σv, what are the safety factors?

If the opening in Problem 4 is in an in situ stress field such that σh = 0.01σv, is there a possibility of failure? Justify your answer.

Which is preferable from the rock mechanics viewpoint: an elliptical, ovaloidal, or rectangular opening of W_{o}/H_{o} of 2 for a stress field M = 1/3? M = 0? Justify your choices.

A rectangular shaft 10 ft wide by 20 ft long is sunk vertically in ground where the state of premining stress is σ_{xx} = 1,141, σ_{yy} = 2,059, σ_{zz} = 1,600, τ_{xy} = 221, τ_{yz} = 0, τ_{zx} = 0, with compression positive, x=east, y=north, and z =up, in psi. Rock properties are: E = 4.5 × 10^{6} psi, v = 0.20, C_{o} = 15,000psi, T_{o} = 900 psi. Find

(a) The most favorable orientation of the shaft and

(b) The safety factors in this orientation. Indicate by sketch where the peak stresses occur in cross section.

A rectangular shaft 3 m wide by 6 m long is sunk vertically in ground where the state of premining stress is σ_{xx} = 7.9, σ_{yy} = 14.2, σ_{zz} = 11.0, τ_{xy} = 1.5, τ_{yz} = 0, τ_{zx} = 0, with compression positive, x=east, y=north, and z =up, in MPa. Rock properties are: E = 31.0GPa, v = 0.20, C_{o} = 10^{3} MPa, T_{o} = 6.2 MPa. Find

(a) The most favorable orientation of the shaft and

(b) The safety factors in this orientation. Indicate by sketch where the peak stresses occur in cross section.

A 12 ft by 24 ft rectangular shaft is sunk to a depth of 3,000 ft in ground where the premining stress field is given by formulas S_{v} = 1.2h, S_{h} = 120 + 0.5h, S_{H} = 3, 240 + 0.3h where h is depth in feet and the stresses are in psi. Show by sketch, the best orientation of the shaft relative to the directions of h and H. Is this the best orientation at all depths to 3,000 ft?

A 3.66 m by 7.32 m rectangular shaft is sunk to a depth of 914 m in ground where the premining stress field is given by formulas Sv = 27.2h, Sh = 826 + 11.3h, SH = 22, 345 + 6.8h where h is depth in m and the stresses are in kPa. Show by sketch, the best orientation of the shaft relative to the directions of h and H. Is this the best orientation at all depths?

Thin overburden is scrapped to expose fresh bedrock at the site of a planned shaft. A 0-45-90 strain gauge rosette is bonded to the bedrock after a five inch square portion of the surface is ground smooth. The 0-gauge is oriented N30E. A 6-inch diameter coring bit is then used to relieve the in situ strains by over coring the gauge. Final readings after over coring are: ε(0) = 1,000 μ in./in., ε(45) = 500 μ in./in., ε(90) = 3,000μin./in. where the strain meter reads tension as positive.

(a) If Young’s modulus is 2.4 million psi and Poisson’s ratio is 0.20, what is the state of strain and the state of stress at the gauge site relative to compass coordinates? Note: The in situ strains are opposite in sign to the relieved strains.

(b) What is the best orientation of a 10 × 20 ft rectangular shaft at this site?

(c) What unconfined compressive and tensile strengths (Co, To) are needed to ensure safety factors of 2 and 4 in compression and tension, respectively?

With reference to Problem 12, if the state of stress at the surface continues to depth in addition to gravity load, and the specific weight of rock is 28kN/m3, what strengths are needed at a depth of 1,234 m for the same safety factors?

Thin overburden is scrapped to expose fresh bedrock at the site of a planned shaft. A 0-45-90 strain gauge rosette is bonded to the bedrock after a 12.7 cm square portion of the surface is ground smooth. The 0-gauge is oriented N30E. A 15.2 cm diameter coring bit is then used to relieve the in situ strains by over coring the gauge. Final readings after over coring are: ε(0) = 1,000 μ in./in., ε(45) = 500 μ in./in., ε(90) = 3,000μ in./in. where the strain meter reads tension as positive.

(a) If Young’s modulus is 16.55GPa and Poisson’s ratio is 0.20, what is the state of strain and the state of stress at the gauge site relative to compass coordinates? Note: The in situ strains are opposite in sign to the relieved strains.

(b) What is the best orientation of a 3 × 6 m rectangular shaft at this site?

(c) What unconfined compressive and tensile strengths (Co, To) are needed to ensure safety factors of 2 and 4 in compression and tension, respectively?

With reference to Problem 14, if the state of stress at the surface continues to depth in addition to gravity load, and the specific weight of rock is 28kN/m3, what strengths are needed at a depth of 1,234 m for the same safety factors?

With reference to Problem 14, if the state of stress at the surface continues to depth in addition to gravity load, and the specific weight of rock is 28kN/m3, what strengths are needed at a depth of 1,234 m for the same safety factors? Discuss.

If the premining stress state in Problem 16 where hydrostatic (3D), what shape would be more favorable, an ellipse, rectangle, or ovaloid (with semi-axes ratio of 2)? Discuss.

If the premining stress state in Problem 16 where hydrostatic (3D), what shape would be more favorable, an ellipse, rectangle, or ovaloid (with semi-axes ratio of 2)?

If the premining stress state in Problem 18 where hydrostatic (3D), what shape would be more favorable, an ellipse, rectangle, or ovaloid (with semi-axes ratio of 2)?

A rectangular shaft 18 ft by 24 ft is planned for a depth of 4,800 ft where the premining stresses relative to compass coordinates (x = east, y = north, z = up) are given by: σxx = 250 + 0.5h, σyy = 800 + 0.2h, σzz = 1.1h, where h is depth and compression is positive. Find the best shaft orientation (explain), then determine if shaft wall support is needed at some depth. Wall rock properties are: E = 6.2 million psi, ν = 0.33, c (cohesion) = 5,600 psi, φ = 52◦, γ = 158pcf.

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