Numerical Methods in Soil Mechanics 30.PDF

Numerical Methods in Soil Mechanics 30.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 "ECONOMICS OF BURIED PIPES AND TANKS " Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 CHAPTER 30 ECONOMICS OF BURIED PIPES AND TANKS The engineer is responsible for cost effectiveness Cost effectiveness of any project is the extent to which the returns on investment outweigh the costs Costs are money expenditures; but costs also include time, effort, overhead, insurance, warranty, public relations, etc Returns include money, but also reputation, etc All costs and returns must be reduced to monetary equivalents And for analysis, all monetary equivalents must be reduced to the same basis — a present worth, P; or a periodic payment (annually), R; or a future lump sum, S For planning and design, it is customary to compare the monetary equivalents of costs for all possible alternatives after reducing them to the same basis of monetary equivalence — usually equal periodic payment, R, (or sometimes present worth, P) For sale of a project, or settlement of claims, present worth, P, is usually the best monetary equivalent The present worth of a project that is to continue forever by repeated replacements is called capitalized cost, P The interest on capitalized cost provides the periodic payments Monetary equivalence depends upon time and interest rates because the value of money is not constant VALUE OF MONEY Nomenclature P S R P i n = present worth, an amount of money now, = sum, the value after n periods of time at interest rate i, = equal periodic payments over n periods at interest rate i, that accumulates sum S or repays present worth P, = capitalized cost (interest = R), = interest rate, compounded at the end of each pay period, = number of pay periods of time (usually in years) ©2000 CRC Press LLC Derivation of Equations The interrelationships of money values, P, S, and R, for n periods at interest rate i compounded at the end of each period, are as follows S = P + Pi + i(P+Pi) + i[P+Pi+i(P+Pi)] + S = P(1+i) + Pi(1+i) + Pi(1+i) + Pi(1+i) + + Pi(1+i) n-1 (30.1) Rewriting, S = P[(1+i) + i(1+i) + i(1+i) + i(1+i) + + i(1+i) n-1] (30.2) Multiplying both sides of Equation 30.1 by (1+i) S(1+i) = P[(1+i) + i(1+i) + i(1+i) + i(1+i) + + i(1+i) n] (30.3) Subtracting Equation 30.2 from Equation 30.3, S = P(1+i) n (30.4) Consider each R of a series of periodic payments to be a separate P For this analysis, let R be due at the beginning of each pay period Then, evaluating and adding up the sums of all of the R's in the series, while working backward from the last R-payment to the first from one year past the last R, S = R(1+i) + R(1+i) + R(1+i) + + R(1+i) n (30.5) Multiplying both sides by (1+i), S(1+i) = R(1+i) + R(1+i) + + R(1+i) n+1 (30.6) Subtracting Equation 30.5 from Equation 30.6 yields Equation 30.7 Eliminating S between Equations 30.4 and 30.7 yields Equation 30.8 If R is at the beginning of each pay period, Si = R[(1+i) n+1 - (1+i)] (30.7) Pi(1+i) n = R[(1+i) n+1 - (1+i)] (30.8) S = P(1+i) n (30.9) P = R(1+1/i) (30.10) year repayment of purchase cost, P = $1,000,000 from Equation 30.12 in n = years at i = 7% interest R = $100,000 + $1,000,000(0.07)(1.07)4/[(1.07)4 - 1] R = $100,000 + $295,228 = $395,228 The three variables P, R, and S, can be interrelated by the three Equations, 30.7, 30.8, and 30.9, for any given interest rate, i, and any number of pay periods, n Equation 30.10 is capitalized cost, P , which is the present worth of all series payments, R, if the project is to continue on forever by replacements; i.e., for n = oo If R is at the end of each pay period, Si = R[(1+i) n - 1] (30.11) Pi(1+i) n = R[(1+i) n - 1] (30.12) Find the present worth, P, if R = $395,228 From Equation 30.12, P = R[(1+i) n - 1]/i(1+i) n = $395,228[(1.07)4 - 1]/(0.07)(1.07)4 P = $1,338,721 If the life of the tanks is 50 years, and payments are made over the 50-year life rather than years, what is the annual end-of-year payment? From Equation 30.12, if P = $1,338,721, i = 7%, and n = 50 pay periods, R = $1,338,721(0.07)(1.07)50/[(1.07)50 - 1] S = P(1+i) n (30.13) R = $97,003.50 P = R/i (30.14) The three variables P, R, and S, can be interrelated by the three Equations, 30.11, 30.12, and 30.13 for any given interest rate, i, and any number of pay periods, n Equation 30.14 is capitalized cost, P , which is the present worth of all periodic payments, R, if the projec t is to continue on forever; i.e for infinite time, n = oo In general, R is used as a basis for comparing alternatives for least cost P is used as a basis for sale or for settlement of a claim Example The purchase cost of tanks in a buried tank farm is $1,000,000 Installation costs are $100,000 per year at year end for years What is the annual end-ofperiod payment R to pay off the project in four years? R is annual installation cost plus the end-of- ©2000 CRC Press LLC If the tank farm must be replaced every 50 years, find the capitalized cost From Equation 30.14, P = R/i, where R = $97,003.50 and i = % P = $97,003.50/0.07 = $1,385,764.32 P = $1,385,764 MONETARY EQUIVALENCE Monetary equivalence includes overhead and all direct costs such as purchases and installation But monetary equivalence also includes maintenance and the cost of risk Risk includes public relations, insurance, and legal counsel, in addition to the cost of replacement Example Purchase price of a buried tank is $10,000 Installation cost is $40,000 Maintenance is $500 per year The cost of a leak (replacement, damage, and liability) is estimated to be $100,000 For this analysis, assume that insurance covering leaks can be obtained for $2000 per year The tank is designed for 50 years of service If the interest rate is 8%, what is the equivalent annual R at beginningof-pay periods for the project? This R can be compared with the R's for alternatives such as more expensive (dual-containment) tanks with less probability of a leak For beginning-of-pay periods, utilizing Equation 30.8, R = $500 + $2,000 + $50,000(0.08)(1.08)50/[(1.08)51 - 1.08] R = $500 + $4,000 + $3784 = $6284 R = $6284 What is the present worth, P, of the project in the event of sale or legal action? Already known from Example is R = $6284 From Equation 30.8, P(0.08)(1.08)50 = $6284[(1.08)51 - 1.08] P(3.7521) = $6284[49.5737] = $311,521.39 P = $83,025 COST OF FAILURE The cost of failure could be staggering, depending on damage and liabilities, but the probability of failure may be remote The monetary equivalence of failure in any pay period is the approximate cost of failure times the probability of failure Two techniques for finding the monetary equivalence of failure follow Probability of a 100-Year Event: A 100-year event (or any time-indexed event) is defined as the average period of time between occurrences of an event of magnitude greater than some given magnitude If there is no periodicity, an event can occur at any time For example, the periodicity of storms, floods, and earthquakes has not yet been predicted with confidence, despite promising new techniques Available only are ©2000 CRC Press LLC records of previous events with magnitudes and times noted The best monetary equivalent of the time-indexed magnitude of an event is an insurance premium — a series payment, Rc r, which is equal to the cost of the event times the probability that the event will occur in any one pay period A reasonable profit for the insurance carrier is added For an insurance carrier, with many portfolios, the probability that a 100-year event will occur in any one year is 1/100 The unpredictable timing of each is balanced out in the totality of portfolios The monetary equivalent of the magnitude of an event is Pc r The probability is 1/nc r The periodic payment to cover the event i s Rcr = Pc r/nc r , where nc r is the time index (100 years, etc.) Example The tank of Example will float out of its embedment if it is empty when a 50 year flood occurs The cost of flotation is tank replacement at $50,000 plus public liability estimated to be $100,000 What insurance premium would be justified if reasonable profit for the carrier is 30%? Assuming flotation of an empty tank, from Example 2, the cost of the disaster is the purchase price plus installation, plus liability; i.e., Pcr = $10,000 + $40,000 + $100,000 = $150,000 But the probability of a flood that could cause flotation in any one time period is only one in 50 years; i.e., 1/ncr = 0.02 R = ($150,000)(0.02) = $3000 Including a 30% profit for the insurance carrier, the insurance premium is $3900 This is based on the assumption that the tank is always empty But what is the probability that the tank is empty if it is used for gasoline storage at a service station? Example Find the probability, Pe, that a tank will float if it is not always empty The subscript, e, distinguishes probability from present worth If the tank is used to store gasoline at a service station, it will be refilled as soon as it is empty Assume that gasoline is pumped out at a constant rate between refillings The first problem is that the tank may float with some gasoline in it What is the least amount of gasoline in the tank that acts as ballast and Figure 30-1 Buried tank (top) showing the level of the contents (cross-hatched) lower than which the tank will float; and (bottom) the volume of contents as a function of time between refills, showing (cross-hatched) the time, between refills, during which flatation can occur Figure 30-2 Systems diagram of a buried tank under worst-case conditions of external loads ©2000 CRC Press LLC keeps the tank from floating? See the cross-hatched areas of Figure 30-1 Assume saturated soil with flood level at or above ground surface The tank volume is 12,000 gallons D = ft, L = 42 ft, WT = kips = weight of the tank, w T = 190 lb/ft = tank wt/unit length, H = ft = height of soil cover, g s = 57.6 pcf = unit weight of soil, w s = 1980 lb/ft = buoyant soil wt/unit length, w = 2170 lb/ft = total wt/ft acting down, w = 2400 lb/ft = buoyant force acting up, Dw = 230 lb/ft = wt/ft of gasoline at a level lower than which the tank floats, w = 1616 lb/ft = wt/ft of gasoline in the tank when the tank is full (unit weight of gasoline = 42 pcf), T = time between refillings, DT = time during which the tank can float (with less than 230 lb/ft of gasoline ballast) What is the probability, Pe, that the tank will float considering gasoline is in the tank between refillings? P e = DT/T , where DT = (230/1616)T = 0.1423T Pe = 0.1423 System Diagram The first step is a diagram of the system See Figure 30-2 The soil cover, H, is minimum for an HS-20 dual-wheel load It is understood that the tank meets specifications for quality Fault Tree Construction A fault tree is a trunk and root system The trunk is the event (failure), and the roots are the faults (causes of failure) Figure 30-3 is a table of symbols In this example, the failure is a fuel LEAK It is shown at the top of the fault tree of Figure 30-4 in a rectangle indicating that it is an event The two causes (faults) considered in this example, are shown in a horizontal row below the event They are connected to LEAK through an and gate (bullet) indicating that both POOR EMBEDMENT and EXCESSIVE LOAD are necessary to cause the leak The probabilities, Pe, of the faults must be multiplied together to get through the and gate POOR EMBEDMENT is an event (rectangle) which could be the result of a HARD SPOT or LOOSE SOIL in the embedment Because either loose soil or a hard spot is sufficient to cause a leak, they are connected to POOR EMBEDMENT through an or gate (arrowhead) The probabilities, Pe , of the faults must be added together to get through the or gate Probability of Failure by Fault Tree A justifiable R, either for insurance coverage or elimination of the probability of failure, is the cost of failure times the probability that it will occur in any pay period A model for estimating probability of failure is the fault tree — a logic diagram that starts with an undesired event (failure) and traces back through all of the causes (faults) assessing and relating the probability of each fault to the probability of failure An example illustrates the fault tree Example A buried fuel tank costs $50,000 installed A leak could result in liabilities up to $100,000 What is the probability of a leak? ©2000 CRC Press LLC HARD SPOT is a diamond, a basic fault undeveloped If the constructor had a history of problems with hard spots, a probability might be developed Without such a history, probability can only be assumed and varied to determine the effect of hard spots on the probability of a leak This could serve to evaluate encroachment of the constructor’s hard spots into the safety zone (safety factor zone) Or from the history (track record) of the constructor, suppose that repair of a hard spot shows up in the bedding of one out of every 20 tanks Probability is 0.05 LOOSE SOIL is a circle because it is a basic fault, but developed From geotechnical tests, the soil was found to be loose in one out of every 40 tanks, (0.025) Figure 30-4 Example of a Fault Tree for finding the probability of a leak in a tank that is buried in poor embedment, and is subjected to excessive loads ©2000 CRC Press LLC With loose soil in one out of every 40 tanks (0.025), and hard spots in one out of every 20 tanks (0.050), the Pe 's are shown in Figure 30-4 for HARD SPOT and LOOSE SOIL They are added together through the or gate yielding a value of Pe = 0.075 = probability of a POOR EMBEDMENT EXCESSIVE LOAD is an event (rectangle) that might be caused by e ither a 50-YEAR FLOOD or LIVE LOAD If either one or the other is sufficient, the two are connected to EXCESSIVE LOAD through an or gate The 50-YEAR FLOOD is a triangle because it is developed elsewhere See Example The probability is shown on 50-YEAR FLOOD; Pe = 0.02 in any payment period (year) The LIVE LOAD is a diamond (undeveloped) Why not assume that a ready-mix concrete truck could pass over the tank inadvertently — say once every 40 years? Then the probability in any payment period (year) is Pe = 0.025 as shown on LIVE LOAD The probability of EXCESSIVE LOAD is the sum of the Pe's for the 50-YEAR FLOOD and LIVE LOAD For EXCESSIVE LOAD, P e = 0.02 + 0.025 = 0.045 The probability of a LEAK in any pay period is the product of the Pe's for POOR EMBEDMENT and EXCESSIVE LOAD Pe = 0.075(0.045) The probability of a leak is, Pe = 0.003375 = 1/296 or approximately one in every 300 tanks per year This 1-in-300 probability of a leak per year might justify either a periodic payment to reduce the possibility of a leak, or insurance premiums that cover the leak It is noteworthy that the above simple example does not include such effects as contents of the tank, time related consolidation of sidefill soil, vacuum in the tank, and liquefaction of the embedment A more complex fault tree might be justified ©2000 CRC Press LLC Example What beginning-of-period payment, R, can be justified to cover the cost of a leak in the tank of Example if the cost of the leak is $150,000 The probability of a leak is 1/296 = 0.003375 in any one year R = $150,000(3.375)10-3 R = $506.25 per year SAFETY FACTORS Simply defined, a safety factor is the ratio of performance limit to performance Not so simple is a number for that safety factor For above-ground structures, classical performance is measured by stress in the material Performance limit is strength For buried structures, performance is, usually, deformation — including leaks and excessive movement of the soil or structure Failure is usually reduced to a monetary equivalent What is the monetary equivalent of failure? Are there any mitigations for failure? What are the encroachments into the safety zone (demilitarized zone)? And, in the event of legal action, who is responsible for the encroachments? What are the monetary equivalents of encroachments? Whose are the responsibilities? Legal counsel usually becomes involved in these questions However, information upon which counsel directs legal proceedings must come from the engineer Reduction to monetary equivalence often comes from the engineer Comparative or contributory encroachment into the safety factor zone requires engineering knowledge Example A 12,000-gallon tank for storing gasoline, leaked just after installation The leak was a circumferential crack on the bottom at midspan The tank was buried in dry soil (no water table) What are the relative responsibilities for the leak (in percent) of the engineer, manufacturer, and installer? Neglect contributory negligence of other parties in this hypothetical case It is known that an HS-20 axle load of 32 kips passed over the tank at midspan See Figure 30-2 The dual wheels were separated by ft D = 84 inches, L = 42 ft, t = 0.6 inch, H = ft, g = 120 pcf = unit weight of dry soil, g G = 42 pcf = unit weight of gasoline, sf = ksi = yield strength of the tank wall, w T = 200 lb/ft = weight per unit length The safety factor is two based on yield strength of ksi Design stress is 2.5 ksi Therefore the safety zone is from 2.5 to ksi Weight per unit length of tank is the tank full of gasoline, plus prismatic soil load on top If the tank is a simply supported beam, including live load, the longitudinal stress in the bottom at midspan is 3.83 ksi ENGINEER specified compacted soil with no organics or large rocks The tank was designed for 40% of simply supported beam stresses — assuming that the bedding would provide some support But the beam was simply supported Therefore, design encroachment is (3.83-2.50)/(5-2.5) = 0.53 MANUFACTURER supplied a tank with wall strength of ksi according to post-leak tests — not ksi as advertised The encroachment is (5-4)/(52.5) = 0.4 INSTALLER leveled the tank on timbers at the ends thereby forcing the tank to act as a simply supported beam with no bedding There was no compaction under the haunches Had a bedding reduced stresses to 40% of the simply supported beam stresses, the maximum stress in the tank would be (0.4)(3.83) = 1.53 ksi The encroachment is (3.83-1.53)/(5-1.53) = 0.66 ©2000 CRC Press LLC Percent Encroachment into the Safety Zone (Safety zone = 0.53+0.40+0.66=1.59): ENGINEER MANUFACTURER INSTALLER = 33% = 25% = 42% PROBLEMS 30-1 What is the probability of a leak in a farm of 10 buried tanks if the probability of a leak in any one tank is 1/300 during any pay period? (1/30) 30-2 Two tanks are under consideration for a project Tank A costs $10,000 plus installation cost of $5,000 Service life is 30 years if an insurance premium of $400 is paid at the beginning of each year Tank B costs $15,000 plus installation cost of $7,000 and service life of 50 years What are the equal series payments, R, at year end for the two alternatives? Interest rate is 8% (A, $1764; B, $1798) 30-3 If the tanks of Problem 30-2 are to be replaced by identical tanks at the end of each service life, what are the capitalized costs, P , of tanks A and B? (A, $22,055; B, $22,479) 30-4 The head of a tank leaks near the bottom H = ft, D = 84 inches, t = 0.187 inch Find contributory negligence of: Manufacturer — insertion is only 0.3 inch Constructor — a timber is left under the end of the tank for vertical alignment Owner — a 16-kip wheel load passed over the tank the evening before the leak was detected  ... gasoline, plus prismatic soil load on top If the tank is a simply supported beam, including live load, the longitudinal stress in the bottom at midspan is 3.83 ksi ENGINEER specified compacted soil. .. assume that insurance covering leaks can be obtained for $2000 per year The tank is designed for 50 years of service If the interest rate is 8%, what is the equivalent annual R at beginningof-pay... could cause flotation in any one time period is only one in 50 years; i.e., 1/ncr = 0.02 R = ($150,000)(0.02) = $3000 Including a 30% profit for the insurance carrier, the insurance premium is