SOIL MECHANICS

SECTION 4SOIL MECHANICSSOIL MECHANICS Composition of Soil Specific Weight of Soil Mass Analysis of Quicksand Conditions Measurement of Permeability by Falling-Head Permeameter Constructi

SECTION 4SOIL MECHANICS

SOIL MECHANICS

Composition of Soil

Specific Weight of Soil Mass

Analysis of Quicksand Conditions

Measurement of Permeability by Falling-Head Permeameter

Construction of Flow Net

Soil Pressure Caused by Point Load

Vertical Force on Rectangular Area Caused by Point Load

Vertical Pressure Caused by Rectangular Loading

Appraisal of Shearing Capacity of Soil by Unconfined Compression Test

Appraisal of Shearing Capacity of Soil by Triaxial Compression Test

Earth Thrust on Retaining Wall Calculated by Rankine's Theory

Earth Thrust on Retaining Wall Calculated by Coulomb's Theory

Earth Thrust on Timbered Trench Calculated by General Wedge TheoryThrust on a Bulkhead

Cantilever Bulkhead Analysis

Anchored Bulkhead Analysis

Stability of Slope by Method of Slices

Stability of Slope by ^-Circle Method

Analysis of Footing Stability by Terzaghi's Formula

Soil Consolidation and Change in Void Ratio

Compression Index and Void Ratio of a Soil

Settlement of Footing

Determination of Footing Size by Housel's Method

Application of Pile-Driving Formula

Capacity of a Group of Friction Piles

Load Distribution among Hinged Batter Piles

Load Distribution among Piles with Fixed Bases

Load Distribution among Piles Fixed at Top and Bottom

RECYCLE PROFIT POTENTIALS IN MUNICIPAL WASTES

Choice of Cleanup Technology for Contaminated Waste Sites

Cleaning up a Contaminated Waste Site via Bioremediation

Work Required to Clean Oil-Polluted Beaches

4.14.24.34.34.44.44.64.74.84.84.104.114.134.144.164.174.184.204.224.244.254.264.274.284.284.294.304.324.334.344.364.414.48

Soil Mechanics

The basic notational system used is c = unit cohesion; s = specific gravity; V = volume;

W = total weight; w = specific weight; 0 = angle of internal friction; T = shearing stress;

a = normal stress

In a three-phase soil mass, the voids, or pores, between the solid particles are occupied by

moisture and air A mass that contains moisture but not air is termed fully saturated; this constitutes a two-phase system The term apparent specific gravity (denotes the specific

gravity of the mass

Let the subscripts s, w, and a refer to the solids, moisture, and air, respectively Where

a subscript is omitted, the reference is to the

entire mass Also, let e = void ratio = (Vw + Va)IV 5 ; n = porosity = (V w + V0 )IV; MC = moisture content = WJW5 ; S = degree of sat- uration = VJ(VW + Va ).

Refer to Fig 1 A horizontal line sents volume, a vertical line represents specif-

ic gravity, and the area of a rectangle sents the weight of the respective ingredient

2 Compute the properties of the original mass

Thus, e = 100(18.0 + 7.9)741.1 = 63.0 percent; n = 100(18.0 + 7.9)767.0 = 38.7 percent;

MC = 100(18)7104 = 17.3 percent; 5= 100(18.0)1(18.0 + 7.9) + 69.5 percent The factor

of 100 is used to convert to percentage

Soil composition is important from an environmental standpoint Ever since the sage of the Environmental Protection Agency (EPA) Superfund Program by Congress,greater attention has been paid to soil composition by cities, states, and the federal gov-ernment

pas-The major concern of regulators is with soil contaminated by industrial waste andtrash Liquid wastes can pollute soil and streams Solid waste can produce noxious odors

in the atmosphere Some solid wastes are transported to "safe" sites for burning, wherethey may pollute the local atmosphere Superfund money pays for the removal and burn-ing of solid wastes

A tax on chemicals provides the money for Superfund operations Public and civic action to Superfund activities is most positive Thus, quick removal of leaking drums ofdangerous materials by federal agencies has done much to reduce soil contamination Fur-ther, the Superfund Program has alerted industry to the dangers and effects of carelessdisposal of undesirable materials

re-There are some 1200 dump sites on the Superfund Program agenda requiring

Solids

Moisture

Air

cleanup The work required at some sites ranges from excavation of buried waste to itseventual disposal by incineration Portable and mobile incinerators are being used forwastes that do not pollute the air Before any incineration can take place—either infixed or mobile incinerators—careful analysis of the effluent from the incinerator must

be made For all these reasons, soil composition is extremely important in engineeringstudies

SPECIFIC WEIGHT OF SOIL MASS

A specimen of sand has a porosity of 35 percent, and the specific gravity of the solids is2.70 Compute the specific weight of this soil in pounds per cubic foot (kilograms per cu-bic meter) in the saturated and in the submerged state

Calculation Procedure:

1 Compute the weight of the mass in each state

Set V= 1 cm3 The (apparent) weight of the mass when submerged equals the true weight

less the buoyant force of the water Thus, V w + V 0 = nV= 0.35 cm3 V 5 = 0.65 cm3 In the

saturated state, W= 2.70(0.65) + 0.35 = 2.105 g In the submerged state, W= 2.105 - 1 = 1.105 g; or W= (2.70 - 1)0.65 = 1.105 g.

2 Find the weight of the soil

Multiply the foregoing values by 62.4 to find the specific weight of the soil in pounds percubic foot Thus: saturated, w = 131.4 lb/ft3 (2104.82 kg/m3); submerged, w = 69.0 lb/ft3(1105.27 kg/m3)

ANALYSIS OF QUICKSAND CONDITIONS

Soil having a void ratio of 1.05 contains particles having a specific gravity of 2.72 pute the hydraulic gradient that will produce a quicksand condition

Com-Calculation Procedure:

1 Compute the minimum gradient causing quicksand

As water percolates through soil, the head that induces flow diminishes in the direction offlow as a result of friction and viscous drag The drop in head in a unit distance is termed

the hydraulic gradient A quicksand condition exists when water that is flowing upward

has a sufficient momentum to float the soil particles

Let / denote the hydraulic gradient in the vertical direction and i c the minimum ent that causes quicksand Equate the buoyant force on a soil mass to the submerged

gradi-weight of the mass to find i c Or

s s -l

For this situation, i = (2.72 - 1)7(1 + 1.05) = 0.84

MEASUREMENT OF PERMEABILITY

BY FALLING-HEAD PERMEAMETER

A specimen of soil is placed in a falling-head permeameter The specimen has a sectional area of 66 cm2 and a height of 8 cm; the standpipe has a cross-sectional area of0.48 cm2 The head on the specimen drops from 62 to 40 cm in 1 h 18 mm Determine thecoefficient of permeability of the soil, in centimeters per minute

cross-Calculation Procedure:

1 Using literal values, equate the instantaneous discharge

in the specimen to that in the standpipe

The velocity at which water flows through a soil is a function of the coefficient of ability, or hydraulic conductivity, of the soil By Darcy's law of laminar flow,

perme-v = ki (2) where / = hydraulic gradient, k = coefficient of permeability, v = velocity.

In a falling-head permeameter, water is allowed to flow vertically from a standpipethrough a soil specimen Since the water is not replenished, the water level in the stand-

pipe drops as flow continues, and the velocity is therefore variable Let A =

cross-section-al area of soil specimen; a = cross-sectioncross-section-al area of standpipe; h = head on specimen at given instant; H1 and H2 = head at beginning and end, respectively, of time interval T; L = height of soil specimen; Q = discharge at a given instant.

Using literal values, we have Q = Aki = -a dhfdt.

Substituting gives k = (0.48 x 8/66 x 78) In (62/40) = 0.000326 cm/in.

CONSTRUCTION OF FLOW NET

State the Laplace equation as applied to two-dimensional flow of moisture through a soilmass, and list three methods of constructing a flow net that are based on this equation

Calculation Procedure:

1 Plot flow lines and equipotential lines

The path traversed by a water particle flowing through a soil mass is termed a flow line, stream-line, or path of percolation A line that connects points in the soil mass at which the head on the water has some assigned value is termed an equipotential line A diagram consisting of flow lines and equipotential lines is called a flow net.

(a) Flow net

FIGURE 2

In Fig 2a, where water flows under a dam under a head H 9 lines AB and CD are flow lines and EF and GH are equipotential lines.

2 Discuss the relationship of flow and equipotential lines

Since a water particle flowing from one equipotential line to another of smaller head willtraverse the shortest path, it follows that flow lines and equipotential lines intersect atright angles, thus forming a system of orthogonal curves In a flow net, the equipotentiallines should be so spaced that the difference in head between successive lines is a con-stant, and the flow lines should be so spaced that the discharge through the space betweensuccessive lines is a constant A flow net constructed in compliance with these rules illus-trates the basic characteristics of the flow For example, a close spacing of equipotentiallines signifies a rapid loss of head in that region

3 Write the velocity equation

Let h denote the head on the water at a given point Equation 2 can be written as

dh

v = -k—- (2a) dL

where dL denotes an elemental distance along the flow line.

4 State the particular form of the general Laplace equation

Let jc and z denote a horizontal and vertical coordinate axis, respectively By investigatingthe two-dimensional flow through an elemental rectangular prism of homogeneous, isen-

tropic soil, and combining Eq Ia with the equation of continuity, the particular form of

the general Laplace equation

The seepage of moisture through soil may be investigated by analogy with either theflow of an electric current or the stresses in a body In the latter method, it is merely nec-

(b) Relaxation gridWater

Dam,

essary to load a body in a manner that produces identical boundary conditions and then toascertain the directions of the principal stresses.

5 Apply the principal-stress analogy

Refer to Fig 2a Consider the surface directly below the dam to be subjected to a uniform

pressure Principal-stress trajectories may be readily constructed by applying the ples of elasticity In the flow net, flow lines correspond to the minor-stress trajectoriesand equipotential lines correspond to the major-stress trajectories In this case, the flowlines are ellipses having their foci at the edges of the base of the dam, and the equipoten-tial lines are hyperbolas

princi-A flow net may also be constructed by an approximate, trial-and-error procedurebased on the method of relaxation Consider that the area through which discharge occurs

is covered with a grid of squares, a part of which is shown in Fig 2b If it is assumed that

the hydraulic gradient is constant within each square, Eq 5 leads to

h l +h 2 + h 3 + h 4 -4h Q = 0 (5)

Trial values are assigned to each node in the grid, and the values are adjusted until aconsistent set of values is obtained With the approximate head at each node thus estab-lished, it becomes a simple matter to draw equipotential lines The flow lines are thendrawn normal thereto

SOIL PRESSURE CAUSED BY POINT LOAD

A concentrated vertical load of 6 kips (26.7 IcN) is applied at the ground surface Computethe vertical pressure caused by this load at a point 3.5 ft (1.07 m) below the surface and 4

ft (1.2 m) from the action line of the force

Calculation Procedure:

1 Sketch the load conditions

Figure 3 shows the load conditions In Fig 3, O denotes the point

at which the load is applied, and A denotes the point under eration Let R denote the length of OA and r and z denote the length of OA as projected on a horizontal and vertical plane, re-

consid-spectively

2 Determine the vertical stress az at A

Apply the Boussinesq equation:

VERTICAL FORCE ON RECTANGULAR AREA

CAUSED BY POINT LOAD

A concentrated vertical load of 20 kips (89.0 kN) is applied at the ground surface mine the resultant vertical force caused by this load on a rectangular area 3 x 5 ft (91.4 x152.4 cm) that lies 2 ft (61.0 cm) below the surface and has one vertex on the action line

Deter-of the applied force

Calculation Procedure:

1 State the equation for the total force

Refer to Fig 4a, where A and B denote the dimensions of the rectangle, H its distance from the surface, and F is the resultant vertical force Establish rectangular coordinate axes along the sides of the rectangle, as shown Let C = A 2 + H 2 , D = B 2 + H 2 , E = A 2 + B 2

+ H2, S = sin-1 H(EfCD) 05 deg

The force dF on an elemental area dA is given by the Boussinesq equation as dF = [IPz 3 /(2TrR 5 )] dA, where z = H and R = (H 2 + x 2 + /)°5 Integrate this equation to obtain

an equation for the total force F Set dA = dx dy; then

charts have been devised to expedite the calculation of vertical soil pressure

FIGURE 4

Surfoce

VERTICAL PRESSURE CAUSED BY

RECTANGULAR LOADING

A rectangular concrete footing 6 x 8 ft (182.9 x 243.8 cm) carries a total load of 180 kips(800.6 kN), which may be considered to be uniformly distributed Determine the verticalpressure caused by this load at a point 7 ft (213.4 cm) below the center of the footing

Calculation Procedure:

1 State the equation for az

Referring to Fig 5, let/? denote the uniform pressure

on the rectangle abed and (T 2 the resulting verticalpressure at a point A directly below a vertex of therectangle Then

7 = 0 - 25 -W + ^(c + DJ (8)

2 Substitute given values and solve for a z

Resolve the given rectangle into four rectangles

hav-ing a vertex above the given point Then p =

180,000/[6(8)] = 3750 lb/ft2 (179.6 kPa) With A = 3

ft (91.4 cm); B = 4 ft (121.9 cm); H = 7 ft (213.4

FIGURE 5 cm); C = 58; D = 65; E = 74; O = snr1 0.9807 =

78.7°; (T 2 Jp = 4(0.25 - 0.218 + 0.051) = 0.332; (T 2 =3750(0.332) = 1245 lb/ft2 (59.6 kPa)

APPRAISAL OF SHEARING CAPACITY OF

SOIL BY UNCONFINED COMPRESSION TEST

In an unconfmed compression test on a soil sample, it was found that when the axialstress reached 2040 lb/ft2 (97.7 kPa), the soil ruptured along a plane making an angle of56° with the horizontal Find the cohesion and angle of internal friction of this soil byconstructing Mohr's circle

Calculation Procedure:

1 Construct Mohr's circle in Fig 6b

Failure of a soil mass is characterized by the sliding of one part past the other; the failure

is therefore one of shear Resistance to sliding occurs from two sources: cohesion of thesoil and friction

Consider that the shearing stress at a given point exceeds the cohesive strength It isusually assumed that the soil has mobilized its maximum potential cohesive resistanceplus whatever frictional resistance is needed to prevent failure The mass therefore re-mains in equilibrium if the ratio of the computed frictional stress to the normal stress isbelow the coefficient of internal friction of the soil

Loaded

area

(b) Mohr's diagram for unconfined compression lest

FIGURE 6

Consider a soil prism in a state of triaxial stress Let Q denote a point in this prism and

P a plane through Q Let c = unit cohesive strength of soil; a = normal stress at Q on

plane P; Cr1 and a 3 = maximum and minimum normal stress at Q 9 respectively; r =

shear-ing stress at Q, on plane P\ 6 = angle between P and the plane on which Cr1 occurs; <£ = gle of internal friction of the soil

an-For an explanation of Mohr's circle of stress, refer to an earlier calculation procedure;

then refer to Fig 6a The shearing stress ED on plane P may be resolved into the cohesive stress EG and the frictional stress GD Therefore, r = c + a tan a The maximum value of

a associated with point Q is found by drawing the tangent FH.

Assume that failure impends at Q Two conclusions may be drawn: The angle between

FH and the base line OAB equals ^, and the angle between the plane of impending

rup-ture and the plane on which Cr1 occurs equals one-half angle BCH (A soil mass that is on the verge of failure is said to be in limit equilibrium.)

In an unconfined compression test, the specimen is subjected to a vertical load without

being restrained horizontally Therefore, (T 1 occurs on a horizontal plane

Constructing Mohr's circle in Fig 66, apply these values: (T 1 = 2040 lb/ft2 (97 7

kPa)-(T = O; angle BCH= 2(56°) = 112°.

(a) Mohr's diagram for triaxial-stress condition

2 Construct a tangent to the circle

Draw a line through H tangent to the circle Let F denote the point of intersection of the tangent and the vertical line through O.

3 Measure OF and the angle of inclination of the tangent

The results are OF=C = 688 Ib/ft 2 (32.9 kPa); <£ = 22°

In general, in an unconfmed compression test,

c = Y 2 (T 1 = cot 0-' 0 = 20'-90° (9) where S' denotes the angle between the plane of failure and the plane on which Cr1 occurs

In the special case where frictional resistance is negligible, </> = O; c = VxT 1

APPRAISAL OF SHEARING CAPACITY OF

SOIL BY TRIAXIAL COMPRESSION TEST

Two samples of a soil were subjected to triaxial compression tests, and it was found that

failure occurred under the following principal stresses: sample 1, (T 1 = 6960 lb/ft2 (333.2kPa) and cr3 = 2000 lb/ft2 (95.7 kPa); sample 2, (T 1 = 9320 lb/ft2 (446.2 kPa) and V 3 =

3000 lb/ft2 (143.6 kPa) Find the cohesion and angle of internal friction of this soil, bothtrigonometrically and graphically

Calculation Procedure:

1 State the equation for the angle <f>

Trigonometric method: Let S and D denote the sum and difference, respectively, of the stresses (T 1 and (T 3 By referring to Fig 6<z, develop this equation:

D - S sin (f> = 2c cos (f> (10) Since the right-hand member represents a constant that is characteristic of the soil, D 1 - S1sin (/) = — S2 sin </>, or

sin<£=^-^ (11)

O2-O1where the subscripts correspond to the sample numbers

2 Evaluate <f> and c

By Eq 11, S1 = 8960 lb/ft2 (429.0 kPa); D 1 = 4960 lb/ft2 (237.5 kPa); S 2 = 12,320 lb/ft2

(589.9 kPa); D 2 = 6320 lb/ft2 (302.6 kPa); sin <£ = (6320 - 4960)/(12,320 - 8960); <£ =

23°53' Evaluating c, using Eq 10, gives c = Y 2 (D sec ^-Stan </>) = 729 lb/ft2 (34.9 kPa)

3 For the graphical solution, use the Mohr's circle

Draw the Mohr's circle associated with each set of principal stresses, as shown in Fig 7

4 Draw the envelope; measure its angle of inclination

Draw the envelope (common tangent) FHH', and measure OF and the angle of inclination

of the envelope In practice, three of four samples should be tested and the average value

of 0 and c determined.

EAATH THRUST ON RETAINING WALL

CALCULATED BY RANKINE'S THEORY

A retaining wall supports sand weighing 100 lb/ft3 (15.71 kN/m3) and having an angle ofinternal friction of 34° The back of the wall is vertical, and the surface of the backfill isinclined at an angle of 15° with the horizontal Applying Rankine's theory, calculate theactive earth pressure on the wall at a point 12 ft (3.7 m) below the top

Calculation Procedure:

1 Construct the Mohr's circle associated with the soil prism

Rankine's theory of earth pressure applies to a uniform mass of dry cohesionless soil.This theory considers the state of stress at the instant of impending failure caused by a

slight yielding of the wall Let h = vertical distance from soil surface to a given point, ft (m); p = resultant pressure on a vertical plane at the given point, lb/ft2 (kPa); <£ = ratio of

shearing stress to normal stress on given plane; 0 = angle of inclination of earth surface The quantity o may also be defined as the tangent of the angle between the resultant stress

on a plane and a line normal to this plane; it is accordingly termed the obliquity of the

In Fig 86, construct Mohr's circle associated with this soil prism Using a suitable

scale, draw line OD, making an angle Q with the base line, where OD represents p v Draw line OQ, making an angle <£ with the base line Draw a circle that has its center C on the base line, passes through D, and is tangent to OQ Line OD' represents/? Draw CM per- pendicular to OD.

(a) Resultant pressures (b)Mohr's circle

FIGURE 8

2 Using the Mohr's circle, state the equation for p

Thus,

_ [cos B - (cos2 00 - cos2 <fr)°-5]wh

P ~ cos 0+(cos2 6-cos2 <£)°-5 ^

By substituting, w = 100 lb/ft3 (15.71 kN/m3); A = 12 ft (3.7 m); O = 15°; <£ = 34°; p =

0.321(100)(12) = 385 lb/ft2 (18.4 kPa)

The lateral pressure that accompanies a slight displacement of the wall away from the retained soil is termed active pressure; that which accompanies a slight displacement of the wall toward the retained soil is termed passive pressure By an analogous procedure,

the passive pressure is

[cos B + (cos2 O - cos2 ft)0-5]wft

P~ c o s 0 - ( c o s 2 0 - c o s 2 < £ ) ° - 5 ( 1 3 )The equations of active and passive pressure are often written as

EARTH THRUST ON RETAINING WALL

CALCULATED BY COULOMB'S THEORY

A retaining wall 20 ft (6.1 m) high supports sand weighing 100 lb/ft3 (15.71 kN/m3) andhaving an angle of internal friction of 34° The back of the wall makes an angle of 8° withthe vertical; the surface of the backfill makes an angle of 9° with the horizontal The angle

of friction between the sand and wall is 20° Applying Coulomb's theory, calculate the tal thrust of the earth on a 1-ft (30.5-cm) length of the wall

to-Calculation Procedure:

1 Determine the resultant pressure P of the wall

Refer to Fig 9a Coulomb's theory postulates that as the wall yields slightly, the soil tends to rupture along some plane BC through the heel.

Let 8 denote the angle of friction between the soil and wall As shown in Fig 9b, the wedge ABC is held in equilibrium by three forces: the weight W of the wedge, the result- ant pressure R of the soil beyond the plane of failure, and the resultant pressure P of the

wall, which is equal and opposite to the thrust exerted by the each on the wall The forces

R and P have the directions indicated in Fig 9b By selecting a trial wedge and computing its weight, the value of P may be found by drawing the force polygon The problem is to

identify the wedge that yields the maximum value of P

In Fig 90, perform this construction: Draw a line through B at an angle <£ with the izontal, intersecting the surface at D Draw line AE, making an angle 8+ 4> with the back

hor-of the wall; this line makes an angle /3 - 8 with BD Through an arbitrary point C on the surface, draw CF parallel to AE Triangle BCF is similar to the triangle of forces in Fig 9b Then P = Wu/x, where W = w(area ABQ.

2 Set dP/dx = O and state Rebhann's theorem

This theorem states: The wedge that exerts the maximum thrust on the wall is that for

which triangle ABC and BCF have equal areas.

(a) Location of plane of failure (b) Free-body diagram

of sliding wedge

FIGURE 9

3 Considering BC as the true plane of failure, develop equations for x*, u, and P

Thus,

-^

4 Evaluate P, using the foregoing equations

Thus, <£ = 34°; 8 = 20°; 0 = 9°; p = 82°; LABD = 64°; LBAE = 54°; LAEB = 62°; LBAD

= 91°; ^4£>£ = 25°; AB = 20 esc 82° = 20.2 ft (6.16 m) In triangle ABD: BD = AB sin 91°/sin 25° = 47.8 ft (14.57 m) In triangle ABE: BE = AB sin 54°/sin 62° = 18.5 ft (5.64 m); AE = AB sin 64°/sin 62° = 20.6 ft (6.28 m); x 2 = 18.5(47.8); x = 29.7 ft (9.05 m); u = 20.6(47.8)7(29.7 + 47.8) = 12.7 ft (3.87 m); P = y2(100)(12.7)2 sin 62°; P = 7120 Ib/ft

(103,909 N/m) of wall

5 Alternatively, determine u graphically

Do this by drawing Fig 9a to a suitable scale.

Many situations do not lend themselves to analysis by Rebhann's theorem For stance, the backfill may be nonhomogeneous, the earth surface may not be a plane, a sur-charge may be applied over part of the surface, etc In these situations, graphical analysisgives the simplest solution Select a trial wedge, compute its weight and the surcharge it

in-carries, and find P by constructing the force polygon as shown in Fig 9b After several trial wedges have been investigated, the maximum value of P will become apparent.

If the backfill is cohesive, the active pressure on the retaining wall is reduced

Howev-er, in view of the difficulty of appraising the cohesive capacity of a disturbed soil, mostdesigners prefer to disregard cohesion

Calculation Procedure:

1 Start the graphical construction

Refer to Fig 10 The soil behind a timbered trench and that behind a cantilever retainingwall tend to fail by dissimilar modes, for in the former case the soil is restrained againsthorizontal movement at the surface by bracing across the trench Consequently, the soilbehind a trench tends to fail along a curved surface that passes through the base and isvertical at its intersection with the ground surface At impending failure, the resultant

FIGURE 10 General wedge theory applied to timbered trench.

force dR acting on any elemental area on the failure surface makes an angle </> with the

normal to this surface

The general wedge theory formulated by Terzaghi postulates that the arc of failure is a

logarithmic spiral Let V0 denote a reference radius vector and v denote the radius vector

to a given point on the spiral The equation of the curve is

Therefore, if the failure line is defined by Eq 20, the action line of the resultant force dR

at any point is a radius vector or, in other words, the action line passes through the center

of the spiral Consequently, the action line of the total resultant force R also passes

through the center

The pressure distribution on the wall departs radically from a hydrostatic one, and the

resultant thrust P is applied at a point considerably above the lower third point of the wall Terzaghi recommends setting the ratio BDIAB at between 0.5 and 0.6.

Perform the following construction: Using a suitable scale, draw line AB to represent

the side of the trench, and draw a line to represent the ground surface At middepth, draw

the action line of P at an angle of 12° with the horizontal.

On a sheet of transparent paper, draw the logarithmic spiral representing Eq 20,

set-ting cf> = 260SO' and assigning any convenient value to r0 Designate the center of the ral as O.

spi-Select a point C on the ground surface, and draw a line L through C at an angle $

Logarithmicspiral

with the horizontal Superimpose the drawing containing the spiral on the main drawing,

orienting it in such a manner that O lies on L and the spiral passes through B and C1 Onthe main drawing, indicate the position of the center of the spiral, and designate this point

as O1 Line ^C1 is normal to the spiral at C1 because it makes an angle </> with the radiusvector, and the spiral is therefore vertical at C1

2 Compute the total weight W of the soil above the failure line

Draw the action line of Wby applying these approximations:

Area of wedge = 2 A(AB)AC 1 c = 0.44C1 (21)

Scale the lever arms a and b.

3 Evaluate P by taking moments with respect to O 1

Since R passes through this point,

bW P=- (22)

4 Select a second point C 2 on the ground surface; repeat

the foregoing procedure

5 Continue this process until the maximum value of P is obtained

After investigating this problem intensively, Peckworth concluded that the distance AC to the true failure line varies between OAh and 0.5/j, where h is the depth of the trench It is

therefore advisable to select some point that lies within this range as the first trial positionofC

dry and submerged state The backfill carries a surclsarge of 320 lb/ft (15.3 kPa)

Apply-ing Rankine's theory, compute the total pressure on this structure between A and C.

Calculation Procedure:

1 Compute the specific weight of the soil in the submerged state

The lateral pressure of the soil below the water level consists of two elements: the sure exerted by the solids and that exerted by the water The first element is evaluated by applying the appropriate equation with w equal to the weight of the soil in the submerged state The second element is assumed to be the full hydrostatic pressure, as though the solids were not present Since there is water on both sides of the structure, the hydrostatic pressures balance one another and may therefore be disregarded.

pres-In calculating the forces on a bulkhead, it is assumed that the pressure distribution is hydrostatic (i.e., that the pressure varies linearly with the depth), although this is not strictly true with regard to a flexible wall.

Computing the specific weight of the soil in the submerged state gives w = 1 1 4 ( l 0.42)62.4 - 77.8 lb/ft 3 (12.22 kN/m 3 ).

-2 Compute the vertical pressure at A, B, and C caused by the surcharge and weight of solids

Thus, p A = 320 lb/ft2 (15.3 kPa); p B = 320 + 5(114) = 890 lb/ft2 (42.6 kPa); p c = 890 +

12(77.8) = 1824 lb/ft 2 (87.3 kPa).

3 Compute the Rankine coefficient of active earth pressure

Determine the lateral pressure at A 9 B and C Since the surface is horizontal, Eq 171

ap-plies, with </> = 34° Refer to Fig lib Then C 0 = tan 2 (450 - 170) = 0.283; p A =

0.283(320) = 91 lb/ft 2 (4.3 kPa);p 5 = 252 lb/ft 2 (12.1 kPa);/? c = 516 lb/ft 2 (24.7 kPa).

4 Compute the total thrust between A and C

Thus, P = 1 / 2(5)(91 + 252) + V 2 (12)(252 + 516) = 5466 Ib (24,312.7 N).

CANTILEVER BULKHEAD ANALYSIS

Sheet piling is to function as a cantilever retaining wall 5 ft (1.5 m) high The soil weighs

110 lb/ft 3 (17.28 kN/m 3 ) and its angle of internal friction is 32°; the backfill has a zontal surface Determine the required depth of penetration of the bulkhead.

hori-Calculation Procedure:

1 Take moments with respect

to C to obtain an equation for

the minimum value of d

Refer to Fig 12a, and consider a 1-ft

(30.5-cm) length of wall Assume that the

pres-sure distribution is hydrostatic, and apply

Rankine's theory.

The wall pivots about some point Z near

the bottom Consequently, passive earth (a > Cantilever (b) Assumed pressures and pressure is mobilized to the left of the wall bulkhead resultant forces

betwen B and Z and to the right of the wall FIGURE 12