The figure below picturizes a rock slope containing a planar sliding surface and avertical tension crack. The tension crack is partly-filled with water due to the watertable intersecting the slope face where the sliding surface daylights on the face. Thewater pressures in the tension crack and on the sliding surface are approximated bythe triangular force diagrams illustrated in the figure. The slope has been reinforcedby a tensioned rock anchor. Assume that;A: the area of the slidingsurface.TROTc: the cohesion of the slidingsurfacep: the internal friction angle ofthe sliding surfaceIyp: the dip of the sliding surface.YT: anchor installation anglefrom the horizon.hw: the height of water column inthe tension crack.Yw: the unit weight of waterW: the weight of the blockV and U: the water forces actingin the tension crack,and on the slidingplane, respectively.So, by taking this geometrical, physical, andmechanical circumstances into the account,derive the governing equation step by step,which defines the factor of safety (Fs) of thisanchored, partially saturated slope.T: the tension in the anchor.L: Length of the sliding surface1: the width of the block517

Answered: Image /qna-images/answer/18ba4f24-2204-4f9e-b6f1-cb1a08d36f98.jpg

The figure below picturizes a rock slope containing a planar sliding surface and a vertical tension crack. The tension crack is partly-filled with water due to the water table intersecting the slope face where the sliding surface daylights on the face. The water pressures in the tension crack and on the sliding surface are approximated by the triangular force diagrams illustrated in the figure. The slope has been reinforced by a tensioned rock anchor. Assume that; TIRTRIT TR A: the area of the sliding surface. c: the cohesion of the sliding surface : the internal friction angle of the sliding surface yp: the dip of the sliding surface. YT: anchor installation angle from the horizon. hw: the height of water column in the tension crack. Yw: the unit weight of water W: the weight of the block V and U: the water forces acting in the tension crack, and on the sliding plane, respectively. So, by taking this geometrical, physical, and mechanical circumstances into the account, derive the governing equation step by step, which defines the factor of safety (Fs) of this anchored, partially saturated slope. T: the tension in the anchor. L: Length of the sliding surface 1: the width of the block

The figure below picturizes a rock slope containing a planar sliding surface and a vertical tension crack. The tension crack is partly-filled with water due to the water table intersecting the slope face where the sliding surface daylights on the face. The water pressures in the tension crack and on the sliding surface are approximated by the triangular force diagrams illustrated in the figure. The slope has been reinforced by a tensioned rock anchor. Assume that; TIRTRIT TR A: the area of the sliding surface. c: the cohesion of the sliding surface : the internal friction angle of the sliding surface yp: the dip of the sliding surface. YT: anchor installation angle from the horizon. hw: the height of water column in the tension crack. Yw: the unit weight of water W: the weight of the block V and U: the water forces acting in the tension crack, and on the sliding plane, respectively. So, by taking this geometrical, physical, and mechanical circumstances into the account, derive the governing equation step by step, which defines the factor of safety (Fs) of this anchored, partially saturated slope. T: the tension in the anchor. L: Length of the sliding surface 1: the width of the block

Principles of Foundation Engineering (MindTap Course List)

9th Edition

ISBN:9781337705028

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter16: Lateral Earth Pressure

Section: Chapter Questions

Problem 16.3P: The soil profile at a site is shown Figure P16.3. Find the total horizontal normal stresses at A and...

See similar textbooks

Related questions

Q: The vertical stress at point P due to the point load Q on the ground surface as shown in figure is…

Q: please solve the problem in the picture, final answer is 233KN/m

A: Step 1: Determine maximum moment, M ( For simply supported Maximum moment is located at midspan so…

Q: Situation 5 - A bar is subjected to a uniaxial stress. The resulting stress is plotted as Ishown. In…

Q: A wooden block of length 400 mm, width 50 mm and depth 100 mm is subjected to uniaxial load as shown…

A: Given data, For wooden block length = 400 mm width =50 mm Depth =100 mm To find the ratio of normal…

Q: 90 MPa For the shown 2D stress element, sketch Mohr circle then, use Mohr circle to find the normal…

A: As per the given data we have to determine the value of normal stresses and shear stresses at a…

Q: 2. A thin-walled steel tube has an outer diameter of 100mm and a material thickness of Smm. The…

A: Given Data Do=100mm, tm=5mm, Fs=2, δu=250Mpa, δu'=520 Now inter-Diameter Di=100-5=95mm Now, by…

Q: Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses…

Q: The plastic hexagonal tube shown in Figure (3), is subjected to a torque of 150 N. m. Determine the…

Q: 1. A tubular beam with circular cross section has an outer diameter do 300 mm and wall thickness t =…

Q: For the stress block shown in the figure below. Determine the following by drawing Mohr’s circle.…

A: Radius of the circleR=(σx-σy2)2+τxy2R=(-160+1252)2+202R=34.73 MPAσAVG=σx+σy2=-160-1252σAVG=-142.5mpa

Q: 1. (Problem 7.3 from the text book) For the given state of stress, determine the normal and shearing…

A: step 1-

Q: A homogeneous bar with a cross-sectional area of 400 mm2 is attached to fixed supports as shown in…

A: Given: The cross-sectional area is 400 mm2. The lateral force P1 is 20 kN. The lateral force P2 is…

Q: EXAM QUESTION 2 The state of plane stress at a point is represented by the element shown in the…

Q: The solid circular section EF rod in the figure is fixed to the wall at point F. Forces P1 = 1 kN,…

A: The solid circular cross-section is fixed at FFind:-the stress states occurring at points A and B…

Q: For the bar subjected to the uniform line load in the axial direction shown in Figure P3-55,…

Q: 1. The MC describes the state of stress of our point in the field. That is, the MC shows us the…

A: "Since you have posted a question with multiples parts, I will solve first 3 for you as per…

Concept explainers

Principles of Seepage

In hydrology, when water flows down in a pore or crack of unsaturated soil due to gravity, the deep layer of soil is called seepage. It is the slow movement of water in the soil layer. Seepage is mainly found near hydraulic structures and is responsible for the stability of earthen-type hydraulic structures.

Question

The figure below picturizes a rock slope containing a planar sliding surface and a
vertical tension crack. The tension crack is partly-filled with water due to the water
table intersecting the slope face where the sliding surface daylights on the face. The
water pressures in the tension crack and on the sliding surface are approximated by
the triangular force diagrams illustrated in the figure. The slope has been reinforced
by a tensioned rock anchor. Assume that;
A: the area of the sliding
surface.
TROT
c: the cohesion of the sliding
surface
p: the internal friction angle of
the sliding surface
I
yp: the dip of the sliding surface.
YT: anchor installation angle
from the horizon.
hw: the height of water column in
the tension crack.
Yw: the unit weight of water
W: the weight of the block
V and U: the water forces acting
in the tension crack,
and on the sliding
plane, respectively.
So, by taking this geometrical, physical, and
mechanical circumstances into the account,
derive the governing equation step by step,
which defines the factor of safety (Fs) of this
anchored, partially saturated slope.
T: the tension in the anchor.
L: Length of the sliding surface
1: the width of the block
517

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

The soil profile at a site is shown Figure P16.3. Find the total horizontal normal stresses at A and B, assuming at-rest conditions.
Refer to Figure P6.4. A strip load of q = 900 lb/ft2 is applied over a width B = 36 ft. Determine the increase in vertical stress at point A located z = 15 ft below the surface. Given: x = 27 ft. Figure P6.4
EB and FG are two planes inside a soil element ABCD as shown in Figure 10.50. Stress conditions on the two planes are Plane EB: EB = 25 kN/m2; EB = +10 kN/m2 Plane FG: FG = 10 kN/m2; FG = 5 kN/m2 (Note: Mohrs circle sign conventions for stresses are used above) Given ; = 25, determine: a. The maximum and minimum principal stresses b. The angle between the planes EB and FG c. The external stresses on planes AB and BC that would cause the above internal stresses on planes EB and FG
Refer to Figure 10.43. A strip load of q = 1450 lb/ft2 is applied over a width with B = 48 ft. Determine the increase in vertical stress at point A located z = 21 ft below the surface. Given x = 28.8 ft. Figure 10.43
Two line loads q1 and q2 of infinite lengths are acting on top of an elastic medium, as shown in Figure P8.6. Find the vertical stress increase at A.
A 9 ft wide and infinitely long flexible strip load of 800 lb/ft2 is placed on an elastic medium as shown in Figure P8.7. Find the vertical stress increase at points A, B, and C located 3 ft below the surface.
Refer to Figure 10.46. A flexible circular area of radius 6 m is uniformly loaded. Given: q = 565 kN/m2. Using Newmarks chart, determine the increase in vertical stress, z, at point A. Figure 10.46
Repeat Problem 10.12 for q = 700 kN/m2, B = 8 m, and z = 4 m. In this case, point A is located below the centerline under the strip load. 10.12 Refer to Figure 10.43. A strip load of q = 1450 lb/ft2 is applied over a width with B = 48 ft. Determine the increase in vertical stress at point A located z = 21 ft below the surface. Given x = 28.8 ft. Figure 10.43
Refer to Figure 8.24. Determine the vertical stress increase, , at point A with the following values: q1 = 100 kN/m x1 = 3 m z = 2 m q2 = 200 kN/m x2 = 2 m FIG. 8.24 Stress at a point due to two line loads
Use Eq. (6.14) to determine the stress increase () at z = 10 ft below the center of the area described in Problem 6.5. 6.5 Refer to Figure 6.6, which shows a flexible rectangular area. Given: B1 = 4 ft, B2 = 6 ft, L1, = 8 ft, and L2 = 10 ft. If the area is subjected to a uniform load of 3000 lb/ft2, determine the stress increase at a depth of 10 ft located immediately below point O. Figure 6.6 Stress below any point of a loaded flexible rectangular area
For the same line loads given in Problem 10.8, determine the vertical stress increase, z, at a point located 4 m below the line load, q2. Refer to Figure 10.41. Determine the vertical stress increase, z, at point A with the following values: q1 = 110 kN/m, q2 = 440 kN/m, x1 = 6 m, x2 = 3 m, and z = 4 m. Figure 10.41
A 10 ft diameter flexible loaded area is subjected to a uniform pressure of 1200 lb/ft2. Plot the variation of the vertical stress increase beneath the center with depth z = 0 to 20 ft. In the same plot, show the variation beneath the edge of the loaded area.