Numerical Methods in Soil Mechanics 04.PDF Numerical Methods in Geotechnical Engineering contains the proceedings of the 8th European Conference on Numerical Methods in Geotechnical Engineering (NUMGE 2014, Delft, The Netherlands, 18-20 June 2014). It is the eighth in a series of conferences organised by the European Regional Technical Committee ERTC7 under the auspices of the International Society for Soil Mechanics and Geotechnical Engineering (ISSMGE). The first conference was held in 1986 in Stuttgart, Germany and the series has continued every four years (Santander, Spain 1990; Manchester, United Kingdom 1994; Udine, Italy 1998; Paris, France 2002; Graz, Austria 2006; Trondheim, Norway 2010). Numerical Methods in Geotechnical Engineering presents the latest developments relating to the use of numerical methods in geotechnical engineering, including scientific achievements, innovations and engineering applications related to, or employing, numerical methods. Topics include: constitutive modelling, parameter determination in field and laboratory tests, finite element related numerical methods, other numerical methods, probabilistic methods and neural networks, ground improvement and reinforcement, dams, embankments and slopes, shallow and deep foundations, excavations and retaining walls, tunnels, infrastructure, groundwater flow, thermal and coupled analysis, dynamic applications, offshore applications and cyclic loading models. The book is aimed at academics, researchers and practitioners in geotechnical engineering and geomechanics.
Anderson, Loren Runar et al "SOIL MECHANICS" Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 Figure 4-1 Vertical soil pressure under one pair of dual wheels of a single axle HS-20 truck load, acting on a pipe buried at depth of soil cover, H, in soil of 100 pcf unit weight Pressure is minimum at or ft of cover ©2000 CRC Press LLC CHAPTER SOIL MECHANICS An elementary knowledge of basic principles of soil stresses is essential to understanding the structural performance of buried pipes These principles are explained in standard texts on soil mechanics A few are reviewed in the following paragraphs because of their special application to buried pipes VERTICAL SOIL PRESSURE P For the analysis and design of buried pipes, external soil pressures on the pipes must be known Vertical soil pressure at the top of the pipe is caused by: dead load, Pd , the weight of soil at the top of the pipe; and live load, Pl , the effect of surface live loads at the the top of the pipe Figure 4-1 shows these vertical soil pressures at the top of the pipe as functions of height of soil cover, H, for an HS=20 truck axle load of 32 kips, and soil unit weight of 100 pcf Similar graphs are found in pipe handbooks such as the A ISI Handbook of Steel Drainage and Highway Construction Products Soil unit weight can be modified as necessary Also other factors must be considered What if a water table rises above the top of the pipe, or the pipe deflects, or the soil is not compacted, or is overcompacted? For these and other special cases, the following fundamentals of soil mechanics may be useful If the embedment about a buried pipe is densely compacted, vertical soil pressure at the top of the pipe is reduced by arching action of the soil over the pipe, like a masonry arch, that helps to support the load To be conservative, arching action is usually ignored However, soil arching provides an added margin of safety If the soil embedment is loose, vertical soil pressure at the top of the pipe may be increased by pressure concentrations due to the relatively noncompressible area within the ring in loose, compressible soil Pressure concentrations due to loose embedment cannot be ignored For design, either a pressure ©2000 CRC Press LLC concentration factor is needed, or minimum soil density should be specified Over a long period of time, pressure concentrations on the pipe may be reduced by creep in the pipe wall (plastic pipes), earth vibrations, freeze-thaw cycles, wet-dry cycles, etc The most rational soil load for design is vertical soil pressure at the top of the pipe due to dead weight of soil plus the effect of live load with a specification that the soil embedment be denser than critical void ratio Critical void ratio, roughly 85% soil density (AASHTO T-99), is the void ratio at such density that the volume of the soil skeleton does not decrease due to disturbance of soil particles For design, the total vertic al soil pressure at the top of the pipe is: P = Pd + Pl (4.1) where (see Figure 4-2) P = total vertical soil pressure at the level of the top of the pipe Pd = dead load pressure due to weight of the soil (and water content) Pl = vertical live load pressure at the level of the top of the pipe due to surface loads This is a useful concept in the analysis of buried pipes Even rigid pipes are designed on this basis if a load factor is included See Chapter 12 In fact, P is only one of the soil stresses At a given point in a soil mass, a precise stress analysis would consider three (triaxial) direct stresses, three shearing stresses, direct and shearing moduli in three directions and three Poisson's ratios — with the additional condition that soil may not be elastic The imprecisions of soil placement and soil compaction obfuscate the arguments for such rigor Elastic analysis may be adequate under some few circumstances Superposition is usually adequate without concern for a combined stress analysis involving triaxial stresses and Poisson ratio Basic soil mechanics serves best Figure 4-2 Vertical soil pressure P at the level of the top of a buried pipe where P = Pl + Pd , showing live load pressure Pl and dead load pressure Pd superimposed Figure 4-3 A single stratum of saturated soil with water table at the top (buoyant case) showing vertical stress at depth H ©2000 CRC Press LLC S e Dead Load Vertical Soil Pressure Pd Dead load is vertical pressure due to the weight of soil at a given depth H In the design of buried pipes, H is the height of soil cover over a pipe Total pressure Pd is the weight of soil, including its water content, per unit area See Figure 4-3 Intergranular (or effective) pressure Pd is the pressure felt by the soil skeleton when immersed in water The total and intergranular vertical stresses at the bottom of a submerged stratum can be related by the following stress equation: _ s=s-u where _ s = s u = = gt gw = (4.2) intergranular vertical soil stress (felt by the soil when buoyed up by water), total vertical soil stress = g tH, pore water pressure = g wH, = total unit weight of soil and water, unit weight of water = 62.4 pcf Pd = Sgt H H = (4.3) total unit weight (wet weight) of soil in a given stratum, and height of the same stratum Values of H for each soil stratum are provided by soil borings Values of g t are simply the unit weights of representative soil samples including the water content If the soil samples are not available, from soil mechanics, g t = (G+Se) g w /(1+e) where G = degree of saturation = when saturated, void ratio, from laboratory analysis, unit weight of water Table 4-1 is a summary of dead load soil stresses from which dead load pressure Pd can be found and combined with live load pressure Pl Live load pressure is found from techniques described in the paragraphs to follow Intergranular vertical soil pressure P, at the bottom of multiple soil strata, is: _ P = P - u = P - g wh (4.5) _where Now consider more than one stratum of soil as shown in Figure 4-4 The total vertical dead load soil pressure Pd at the bottom of the strata is the sum of the loads imposed by all of the strata; i.e., where gt = gw = = = (4.4) specific gravity of soil grains, about 2.65, ©2000 CRC Press LLC P P h = = = vertical intergranular soil pressure, total dead plus live load pressures, height of water table above the pipe Total pressure is used to calculate ring compression stress Intergranular soil pressure is used to calculate ring deflection which is a function of soil compression As the soil is compressed, so is the pipe compressed — and in direct ratio But soil compression depends only on intergranular stresses See Chapter Live Load Vertical Soil Pressure Pl Live load soil pressure Pl is the vertical soil pressure at the top of the buried pipe due to surface loads See Figure 4-5 For a single concentrated load W on the surface, vertical soil pres sure at point A at the top of the pipe is: s = NW/H2 where W = H = R = N = N = (4.6) concentrated surface load (dual-wheel) height of soil cover over the top of the pipe horizontal radius to stress, s , Boussinesq coefficient from the line of action of load W, Boussinesq coefficient = 3(H/R) 5/2p Figure 4-4 Multiple strata (three strata with the clay stratum divided into two at the water table) showing the total vertical dead load soil pressure Pd at the bottom (level of the top of the pipe) Figure 4-5 Vertical soil pressure at depth H (at the level of the top of a pipe) and at radius R from the line of action of a concentrated surface load W (After Boussinesq) ©2000 CRC Press LLC ©2000 CRC Press LLC Figure 4-6 Chart for evaluating the vertical stress s at a depth H below the corner A of a rectangular surface area loaded with a uniformly distributed pressure q (After Newmark) ©2000 CRC Press LLC For a single wheel (or dual wheel) load, the maximum stress smax at A occurs when the wheel is directly over the pipe; i.e R = 0, for which smax = 0.477 W/H2 (4.7) Load W can be assumed to be concentrated if depth H is greater than the maximum diameter or length of the surface loaded area For multiple wheel (or dual) loads, the maximum stress at point A, due to effects of all loads must be ascertained The trick is to position the wheel loads so that the combined stress at A is maximum This can be done by trial The effect of a uniformly distributed surface load can be found by dividing the loaded surface area into infinitesimal areas and integrating to find the sum of their effects at some point at depth H See Figure 46 Newmark performed such an integration and found the vertical stress s at a depth H below corner A of a rectangular area of greater length L and lesser breadth B, loaded with uniform pressure q His neat solution is: s = Mq (4.8) where M is a coefficient which can be read on the chart of Figure 4-6 by entering with arguments L/B and B/H If the stress due to pressure on an area is desired below some point other than a corner, the rectangular area can be expanded or subdivided such that point A is the common corner of a number of areas The maximum stress under a rectangular area occurs below the center See Figure 4-7 The rectangle is subdivided into four identical rectangles of length L and breadth B as shown The stress at point A is 4Mq, where M is found from the Newmark chart, Figure 4-6 Alternatively, the Boussinesq equation can be used with less than five percent error if the concentrated load, Q = qBL, for each of the quadrants is assumed to act at the center of each quadrant For this case, R = (L2+B2)/2 The resulting stress at A is 4NQ/H2 from Equation 4.6 ©2000 CRC Press LLC Example What is the stress at point A below A' of Figure 48? The vertical stress s at depth H below surf a c e point A is, by superposition: s = s ' - S s'" + s " where s' = S s'" = s" = stress at corner A due to loaded area L'xB' sum of stresses at corner A due to loaded areas L'B" and L"B' stress at corner A due to loaded area L"B' Clearly, s " due to area L"B" was subtracted twice, so must be added back once An occlusion in the soil mass, such as a pipe, violates Boussinesq's assumptions of elasticity, continuity, compatibililty, and homogeneity The pipe is a hard spot, a discontinuity Soil is not elastic, nor homogeneous, nor compatible when shearing planes form Nevertheless, the Boussinesq assumptions are adequate for most present-day installation techniques For most buried pipe design, it is sufficient, and conservative, to solve for Pl at the top of the pipe due to a single wheel load W at the surface by using the Boussinesq equation with the radius R = For additional wheel loads, simply add by superposition the influence of other wheel loads at their radii R Example What is the maximum vertical soil stress at a depth of 30 inches due to the live load of a single axle HS20 truck? Neglect surface paving See Figures 4-9 and 4-15 By trial, it can be shown that the point of maximum stress is point A under the center of one tire print The rectangular tire prints are subdivided as shown for establishing a common corner A' The effects of the left tire print and the right tire print are analyzed separately, then combined The length L and breadth B of each tire print are based on 104 psi tire pressure Use Newmark because H is less than 3L Figure 4-7 Procedure for subdividing a rectangular surface area such that the stress below the center at depth H is the sum of the stresses below the common corners A' of the four quadrants Figure 4-8 Subdivision of the loaded surface area, LxB, for evaluation of vertical stress, P, at depth, H, under point A ©2000 CRC Press LLC Figure 4-9 Single axle HS-20 truck load showing typical tire prints for tire pressure of 104 psi, and showing the Newmark subdivision for evaluating vertical soil stress under the center of one tire print ©2000 CRC Press LLC Figure 4-10 Infinitesimal soil cube B and the corresponding Mohr circle which provides stresses on any plane through B Note the stresses sq and t q shown on the q -plane At soil slip, the circle is tangent to the strength envelopes described below Figure 4-11 Shearing stress t as a function of normal stress s , showing a series of Mohr circles at soil slip, and the strength envelopes tangent to the Mohr circles ©2000 CRC Press LLC Given: W = q = B = L = H = SOIL STRENGTH 32 k for HS-20 truck load (single axle), 104 psi, inches, 22 inches, 30 inches For the left tire print: sL = 4Mq, L/B = 11/3.5 = 3.14, B/H = 3.5/30 = 0.12, From Figure 4.6, M = 0.018 and sL = 1.078 ksf Failure of a buried pipe is generally associated with failure of the soil in which the pipe is buried The classical, two-dimensional, shear-strength soil model is useful for analysis Analysis starts with an infinitesimal soil cube on which stresses are known and the orientation is given The model comprises three elements, the Mohr stress circle, orientation diagram, and strength envelopes Mohr Stress Circle For the right tire print: sR = 2(M'-M")q, where, L'/B' = 83/3.5 = 23.7, B'/H' = 3.5/30 = 0.12, From Figure 4.6, M' = 0.02 (extrapolated) L"/B" = 61/3.5 = 17.4, B"/H" = 3.5/30 = 0.12, From Figure 4.6, M" = 0.02 (extrapolated) sR = r(M'- M")q = 0: sR = ksf At point A, s = s L + s R; s = 1.08 ksf A rough check by Boussinesq is of interest because the results are conservatively higher and are more easily solved Given: W = 16 kips at the center of each tire print, H = 2.5 ft, RL = 0, RR = ft, R/H = 2.4 From Figure 4.5, N = 0.004 sL = 0.477 W/H2 = 1.22 ksf sR = NW/H2 = 0.01 ksf At point A, s = sL + s R ; s = 1.23 ksf The Boussinesq solution is in error by 13.9%, but on the high (conservative) side Of interest is the small (negligible) effect of the right wheel load ©2000 CRC Press LLC The Mohr stress circle is a plot of shearing stress, t , as a function of normal stress, s , on all planes at angle q through an infinitesimal soil cube B See Figures 4-10 and 4-11 The sign convention is compressive normal stress positive ( + ) and counterclockwise shearing stress positive ( + ) The center of the circle is always on the s -axis Two additional points are needed to determine the circle They are (sx, t xy ) on a y-plane and (s y , ty x) on an x-plane These are known stresses on cube B An origin of planes always falls on the circle t xy = -ty x from standard texts on solid mechanics Any plane from the origin intersects the Mohr circle at the stress coordinates acting on that plane— which is correctly oriented if the following procedure is followed Orientation Diagram Figure 4-10 shows infinitesimal cube B with the xplane and y-plane identified and with the soil stresses acting on each plane Cube B and its axes of orientation can be superimposed on the Mohr circle such that stress coordinates (where each plane intersects the Mohr circle) are the stresses on that plane With cube B located on the Mohr circle as the origin of axes, and with the axes correctly oriented, any plane through B will intersect the Mohr circle at the point whose stress coordinates are the stresses acting on that plane, and all planes are correctly oriented with respect to the original soil cube B Figure 4-12 Trigonometry for analysis of stresses at soil slip on shear planes, q f in cohesionless soil which has a soil friction angle of j ©2000 CRC Press LLC Another cube B' is shown with shearing stresses acting on it For most pipe-soil interaction, only principal stresses are required as on cube B Strength Envelopes Tangents to a series of Mohr circles plotted from shear strength data are called strength envelopes Shear strength circles are plotted from laboratory tests to failure (soil slip) See Figure 4-11 With the strength envelopes known, the stresses at soil slip can be evaluated Suppose normal and shearing stresses are known on a specific plane at a specific point in a soil mass These stresses, s and t, are the coordinates of a point on the stress diagram If the point falls between the strength envelopes, soil does not slip on that plane But if the point falls outside of the strength envelopes, the soil slips on that plane For any cube B, If Mohr circle intersects the strength envelopes, soil slips on planes through the origin and points of intersection Planes are correctly oriented The soil represented by Figure 4-11 has both cohesion (glued soil grains) and frictional resistance to shearing stress Cohesion is the y-intercept, c Even at zero normal stress the glue offers resistance to shearing stress But to this cohesion must be added the frictional resistance which is normal stress s times the coefficient of friction The shearing strength of the soil is the sum: Strength = c + s (tanj) where j is the soil friction angle and tan j is the internal coefficient of friction of the soil Cohesionless Soil Failure For most buried pipes, the embedment around the pipe is specified to be cohesionless soil such as sand or gravel For cohesionless soil, c = and the strength envelopes are idealized by straight lines as shown in Figure 4-12 At soil friction angle j, as indicated, if any Mohr circle is tangent to the strength envelopes, the soil slips at stress coordinates, t f and s f , at the point of tangency Assume ©2000 CRC Press LLC that an infinitesimal cube of soil is located at some point O in a pipe embedment with y-axis vertical and x-axis horizontal The axes, shown dotted on Figure 4-12, are planes through the cube The shearing stresses on the vertical y-plane and the horizontal xplane are zero Therefore the normal stresses on these planes are principal stresses, the minimum, sx, acting on the y-plane and the maximum, sy, on the x-plane Any q -plane through O (at angle q ) intersects Mohr circle at a point whose coordinates are the normal stress sq and shearing stress tq on that q -plane The planes on which failure (soil slip) occurs are those planes from O that intersect the Mohr circle at the points of soil slip; i.e., where the Mohr circle intersects the strength envelopes Angle q f on the circumference intercepts the same arc as central angle 2qf But 2q f = 90o + j Therefore, q f = 45o + j/2 q f is the angle of the failure plane, i.e., the plane of soil shear The minimum principal stress is s3 the furthest point to the left on the Mohr circle It acts on a vertical y-plane shown dotted The maximum principal stress is s1 the furthest point to the right on the Mohr circle and acts on a horizontal x-plane As expected, the shearing stresses are zero on the planes of principal stresses s2 is an intermediate principal stress at right angles to s1 and s Criticality is related to s1 and s3, not s2 and s1 or s2 and s3 because the s 1-s Mohr circle is the largest and approaches closer to the strength envelopes than either of the other Mohr circles Because the planes through O are oriented, the failure planes in a triaxial test specimen of cohesionless soil loaded as shown, would develop shearing planes called Lueders' lines at angles q f Note that the x and y axes are correctly oriented and O is correctly located for the triaxial test specimen For most soil analyses, the principal stresses are horizontal or vertical For analysis of failure of the embedment around a buried pipe, the relationship of the principal stresses, s1 and s at soil slip becomes pertinent See Figure 4-12 from which, s1 = X + Figure 4-13 Analysis of stresses from the Mohr circle for cohesive soil (clay) with no frictional resistance (worst case) for which j = Figure 4-14 Strength envelopes for cohesionless soil with the maximum principal stress acting horizontally on a y-plane, etc., and for which the infinitesimal soil cube is located at the right side of the Mohr circle ©2000 CRC Press LLC Xsin j, and s = X - Xsin j Eliminating X between the two equations, pear s1 = s 3(1+sin j)/(1-sin j) For convenience, let K = (1+sin j)/(1-sin j), then Soil Stress and Strength Analysis K = s1 /s (4.9) which is the ratio of maximum to minimum principal stresses at soil slip in cohesionless soil The soil slips if s1 > Ks s is called passive resistance — the soil retreats from high pressure The soil slips if s3 < s /K s is called active resistance — soil advances against low pressure Suppose that in cohesionless soil, at point O, the maximum principal stress s1 is horizontal and the minimum principal stress s3 is vertical See Figure 4-14 The vertical and horizontal planes on which these stresses act require that the infinitesimal cube be superimposed on the Mohr circle with origin O at the right side of the circle At soil slip, s1 /s = (1+sin j)/(1-sin j) and the shear planes are oriented as shown, i.e., q f = 45o - j/2 Cohesive Soil Failure Under some circumstancess, pipes are buried in cohesive soil such as clay A saturated fat clay has a negligibly small friction angle j, but does have significant cohesion c The strength envelopes are idealized by two horizontal straight lines spaced at 2c as shown in Figure 4-13 When the Mohr stress circle is tangent to the strength envelopes, its diameter is 2c The relationship between maximum and minimum principal stresses at soil slip, called deviator stress, is, s1 - s = 2c (4.10) where s1 = maximum principal stress s2 = intermediate principal stress, not critical s3 = minimum principal stress c = cohesion in shearing force per unit area Because principal stresses are horizontal and vertical, cube O is oriented and superimposed on the left side of the Mohr circle The orientation is such that vertical stress s1 acts on a horizontal x-plane and the horizontal stress s3 acts on a vertical y-plane The shear planes (slip), at points of tangency of the Mohr circle to the strength envelopes, are at q f = 45 o However, failure in clay is general shear; i.e., viscous or plastic flow Lueders' lines not ap- ©2000 CRC Press LLC In fact, the origin O may be located anywhere on the Mohr circle depending on the orientation of the infinitesimal soil cube on which the stresses act For the purposes of this text, the principal stresses act on horizontal and vertical planes It must be remembered that soil strength is based on effective (intergranular) soil stresses If the water table is above the point at which soil strength is to be calculated, the water pressure u must be subtracted from total vertical stresses Example Consider a point in cohesionless soil at the spring line of a flexible pipe g = 110 lb/ft3 = soil unit weight, j = 30o = soil friction angle, c = = cohesion, z = 10 ft = depth of soil at pipe spring lines a) What is the minimum stress sx of the pipe wall against the sidefill soil that would be required to prevent inward collapse of the pipe? Because sy is the maximum principal stress and s x is the minimum principal stress, at soil shear, s1 /s = K = (1+sin j)/(1-sin j) = But s1 = 10 ft(110 lb/ft3) = 1100 lb/ft2, so s = s x = 1100/3 = 36.7 lb/ft2 = horizontal stress against the soil needed to prevent soil slip It is called the active resistance of soil as the soil advances into the pipe If the pipe wall were to advance against the soil, maximum sx at soil slip would be s x = 1100(3) = 3300 lb/ft2 Because the soil retreats before the advancing pipe wall, its resistance is passive resistance of soil b) Assume the soil backfill is saturated clay g = 139 lb/ft3, j = 0, c = 420 lb/ft2, z = 10 ft = depth of soil at spring line of the pipe What is the minimum stress, sx of the pipe against the soil in order to prevent inward collapse of the pipe wall? Because the soil is clay, s1 - s = 2c = 840 lbft2 But s1 = 130(10) = 1300 lb/ft2 At active soil resistance, s3 = s x = 1300 - 840 = 460 lb/ft2 4-4 What is the total pressure at the top of a buried pipe if height of soil cover is ft, unit weight of soil is 140 pcf, and live load is an HS-20 truck (W = 16 kips)? See Figure 4-15 (P = 1.3 ksf) 4-5 In Problem 4-4, an external water table rises to the ground surface What is the effective (intergranular) vertical soil pressure at the top of the buried pipe? (P = 1.1 ksf) 4-6 Due to internal vacuum, the spring line of a buried flexible pipe bears against the sidefill sand with stress of 3200 lb/ft2 What is minimum height of soil above the spring line to prevent vertical pipe collapse? Soil friction angle is 35o and unit weight is 115 lb/ft3 (H = 7.5 ft) 4-7 Suppose that a water table rises to the ground surface in Problem 4-6 What is minimum soil cover to prevent vertic al pipe collapse? Assume that the specific gravity of the soil grains is 2.65 Soil friction angle is 35o (H = 16.5 ft) At passive soil resistance, sx = 1300 + 840 = 2140 lb/ft2 4-8 If horizontal soil pressure is 3200 psf at passive soil resistance, c = 0, and j = 35o, what is the angle q f of the slip planes? (q f = 27.5o) PROBLEMS 4-9 Complete Table 4-2 for maximum vertical effective stresses at depths of 5, 10, and 20 meters for dead loads and for each of two surface live loads, concentrated Q-load and uniformly distributed q-load, both of which are 100 kN loads The surface area is a rectangle x 10 meters 4-1 What is the vertical effective (intergranular) soil pressure at the level of the top of a pipe if the water table is 17 ft above the top of the pipe and the soil surface is 47 ft above the top of the pipe? The soil is dune sand for which: G = 2.7 = specific gravity of grains, e = 0.7 = void ratio, S = 0.1 = degree of saturation of soil above the water table (4.11 ksf) 4-2 What is the maximum principal stress s1 acting vertically at soil slip in granular soil if horizontal stress is 600 gr/cm2 and if j = 30o ? (1800 gr/cm2) 4-3 What is the maximum principal stress acting vertically at failure in clay if j = 0, s = 4000 Pa acting horizontally, and cohesion is c = 2800 Pa? (s1 = 9.6 kPa) ©2000 CRC Press LLC 4-10 In Problem 4-9, what are the vertical live load stresses at 5, 10 and 20 m below the surface, at a horizontal radius of meters from the vertical line of action of the concentrated Q-load? 4-11 In Problem 4-9, what are the vertical live load stresses at 5, 10, and 20 m below the q-load, but at points laterally meters outboard from mid-length of the long side? 4-12 What is the error in Figure 4-1 for an H-20 live load below the center of a 36x40-inch surface area at a height of cover of ft? (< 10%) ©2000 CRC Press LLC ... point in cohesionless soil at the spring line of a flexible pipe g = 110 lb/ft3 = soil unit weight, j = 30o = soil friction angle, c = = cohesion, z = 10 ft = depth of soil at pipe spring lines... ©2000 CRC Press LLC Xsin j, and s = X - Xsin j Eliminating X between the two equations, pear s1 = s 3(1+sin j)/(1-sin j) For convenience, let K = (1+sin j)/(1-sin j), then Soil Stress and Strength... found by dividing the loaded surface area into infinitesimal areas and integrating to find the sum of their effects at some point at depth H See Figure 46 Newmark performed such an integration