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 a0 of the acceleration of gravity g (as = a0g, typically a0 = 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 Vo to V1 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 W1 at the toe of the slide, where cohesion c1 and friction angle φ1 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/m3), 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) φr = 38◦
5 cohesion (rock) cr = 1000 psi (6.90MPa);
6 friction angle (joint) φj = 27◦;
7 cohesion (joint) cj = 10 psi (0.069MPa);
8 specific weight γ = 158pcf (25.0kN/m3);
9 tension crack depth hc = 0.0 ft (m);
10 water table depth hw = 0.0 ft (m);
11 seismic coefficient ao = 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(107) 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/m3);
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/m3).
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) φr = 33◦;
5 cohesion (rock) cr = 2580 psi (17.79 MPa);
6 friction angle (joint) φj = 33◦;
7 cohesion (joint) cj = 0.0 psi (0.0 MPa);
8 specific weight γ = 162pcf (25.63kN/m3);
9 tension crack depth hc = 0.0 ft (m);
10 water table depth hw= 0.0 ft (m);
11 seismic coefficient ao = 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? 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/m3), 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. 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. 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 (WT1) is 30 ft (9.14 m) below the ground surface.
Water table 2 (WT2) is 100 ft (30.5 m) below the ground surface.
Areas ft2 (m2):
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/m3). 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
Co = 22,000 psi To = 1,200 psi
γ = 144pcf E = 5 × 106 psi
G = 2.0 × 106 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 Wo/Ho 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 σxx = 1,141, σyy = 2,059, σzz = 1,600, τxy = 221, τyz = 0, τzx = 0, with compression positive, x=east, y=north, and z =up, in psi. Rock properties are: E = 4.5 × 106 psi, v = 0.20, Co = 15,000psi, To = 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 σxx = 7.9, σyy = 14.2, σzz = 11.0, τxy = 1.5, τyz = 0, τzx = 0, with compression positive, x=east, y=north, and z =up, in MPa. Rock properties are: E = 31.0GPa, v = 0.20, Co = 103 MPa, To = 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 Sv = 1.2h, Sh = 120 + 0.5h, SH = 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?
(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)?
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.