Solution Manual for Foundation Design: Principles and Practices, 3rd Edition

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Chap. 2Uncertainty and Risk in Foundation Design2.1Classify the uncertainty associated with following items as either aleatory or epistemic andexplain your reason for your classification: average wind speed over a 30 day period, location ofa certain applied load, change in strength of a soil caused by sampling method, capacitydetermined by a certain analysis method, magnitude of live load caused by vehicles travelling ona bridge, soil shear strength as measured by a certain method.Solution•Uncertainty of the average wind speed is aleatory.This is a random process that wecannot affect.•Uncertainty of location of an applied load is mostly aleatory. There is a certain accuracywith which a structure can be built and the designer had little or no control over thisaccuracy.In theory there is some epistemic uncertainty in that could be reduced withbetter construction techniques, but from a practical standpoint this uncertainty is aleatory.•Uncertainty in the change in strength of a soil caused by sampling method is an epistemicuncertainty. Improved sampling techniques can reduce this uncertainty.•Uncertainty in the capacity determined by a certain design method is generally epistemic.With improved analytical tools we can reduce this uncertainty.•Uncertainty in magnitude of live load caused by vehicles travelling on a bridge isinherently aleatory. This is a random process which we cannot affect.•The uncertainty in the soil shear strength asmeasured by a certain method is acombination of epistemic and aleatory uncertainty. The uncertainty caused by the qualityof the equipment used and the care of the technician making the measurement isepistemic and can be reduced by the use of more precise equipment and better training ofthe technician. However, there is aleatory uncertainty in the soil strength inherent in thenatural processes that created the soil.Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-1

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Chap. 2Uncertainty and Risk in Foundation Design2.2Figure 2.1 shows the PDF for a normal distribution determined from the unconfined compressiontests shown in the histogram.Does the mean and standard deviation of this PDF representaleatory or epistemic uncertainty? Explain.SolutionThe mean and standard deviation of this PDF contain both aleatory and epistemic uncertainty.The mean of 20.8 and standard deviation of 7.30 are estimate valued of the true mean andstandard deviation of the unconfined compressive strength of this sandstone. The epistemicuncertainty is associated with the number of samples used to estimate the parameters. If we hadtaken more samples, we would have better estimates. However, this particular sample obviouslycontains a large number of measurements.Therefore the estimated standard deviation isprobably very close to the aleatory uncertainty and testing more specimens is unlike to reducethe uncertainty significantly.Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-2

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Chap. 2Uncertainty and Risk in Foundation Design2.3List three sources of epistemic uncertainty associated with determining the soil strength at agiven site and describe how you might reduce these uncertainties.SolutionSourceHow do reduceSmall sample sizeTake and test more samplesSloppy laboratory techniquesImprove laboratory methodsOldor poor quality testing equipmentAcquire improved testing equipmentDisturbance of soil samples before orduring testingUse better sampling and testing methodsMixing up results from different samplesImprovedocumentationmethodstoeliminate mixingup samplesSolutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-3

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Chap. 2Uncertainty and Risk in Foundation Design2.4Using a random number generator create a sample of 4 relative densities using the PDFpresented in Figure 2.2. Repeat the exercise to create 3 different sample sets. Compute the meanand standard deviation of your sample.Compute the mean and standard deviation of eachsample set.Compare the means and standard deviations of your samples with each other andwith the mean and standard deviation of the original distribution. Discuss the differences amongthe sample sets and the original distribution, including the type of uncertainties you are dealingwith.How many samples do you think are needed to reliably determine the mean and standarddeviation of the relative density of this particular soil?SolutionThere are an infinite number of solutions to this problem.The table below shows Excelspreadsheet formula that can be used to generate the random sample sets.ABCD1μσNZ294.95.7=NORM.S.INV(RAND())=$A$2+$B$2*C23=NORM.S.INV(RAND())=$A$2+$B$2*C34=NORM.S.INV(RAND())=$A$2+$B$2*C45=NORM.S.INV(RAND())=$A$2+$B$2*C56μ=AVERAGE(D2:D5)σ=STDEV.S(D2:D5)The table below shows three sample sets generated with the Excel spreadsheet shown above.Note that the average of the samples ranges from 6.5 below the distribution mean to 6.8 above it.Also one estimate of the standard deviation is nearly twice that of the original distribution.It ispossible, using sampling theory, to determine the number of sample required to have a certainconfidence level in the estimated parameters. However, this is well beyond the scope of this text.Students should note that increasing the sample size to 3 to 7, significantly reduces the variabilityof the estimated mean and standard deviation.Sample #Trial 1Trial 2Trial 31107.3492.4495.79298.7578.4783.103101.50100.5583.95499.02102.0790.80Sample mean101.6593.3888.41Sample Standard Deviation3.9910.806.01Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-4

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Chap. 2Uncertainty and Risk in Foundation Design2.5A certain column will carry a dead load estimated to be 400 k with a COV of 0.1 and a live loadof 200 k with a COV of 0.25. What is the mean and standard deviation of the total column load?What is the probability that this load will exceed 750 k?SolutionFirst we must compute the standard deviation of each random variable from their mean and COVusing Equation 2.10.0.1(400)400.25(200)50DLσσ====Then we compute the mean and standard deviation of the total column load using Equations 2.17and 2.1822400200600405064TotalTotalμσ=+==+=Then using Equation 2.15 we compute the probability that the load exceeds 750 and 1 minus theprobability that it is less than 750()()3750600Load >750112.349.51064P−−=− Φ=− Φ=×Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-5

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Chap. 2Uncertainty and Risk in Foundation Design2.6A simply supported beam has a length of 3 m and carries a distributed load with a mean of 5kN/m and a COV of 0.2.What is the mean and standard deviation of the maximum moment inthe beam? What is the probability the maximum moment will exceed 7 kN-m?SolutionThe equation for the maximum moment in a simply supported beam subject to a distributed loadis22max31.12588wlMww===Using Equations 2.10, 2.17 and 2.18 the mean and standard deviation ofMmaxis()()maxmaxmax1.1255.621.1251.125 COV1.125 0.2 51.125MMMwwwμσσμ====⋅=⋅=Then using Equation 2.15 we compute the probability that the load exceeds 750 and 1 minus theprobability that it is less than 750()()max75.62>7111.230.111.125P M−=− Φ=− Φ=Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-6

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Chap. 2Uncertainty and Risk in Foundation Design2.7Using the data shown in Figure 2.5, determine the probability that tangent of the friction anglefor the mudstone at the Confederation Bridge site is less than 0.25.SolutionThe data in Figure 2.5 is lognormally distributed withμ= -1.09 andσ= 0.270. Using Equation2.16()()ln 0.25( 1.09)tan0.251.0970.1360.270Pφ− −<= Φ= Φ −=Or there is a 13.6% chance that tanφwill be less than 0.25.Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-7

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Chap. 2Uncertainty and Risk in Foundation Design2.8The capacity for a certain foundation system is estimated to be 620 kN with a COV of 0.3. Thedemand on the foundation is estimated to be 150 kN with a COV of 0.15.Compute the meanfactor of safety of this foundation and its probability of failure.SolutionThe mean factor of safety is6204.1150Fμ==The standard deviation of demand and capacity are computed using Equation 2.10()()0.15 15022.50.3 620186DCσσ====The mean and standard deviation of the safety margin,m, are computed using Equations 2.17and 2.18620150470mμ=−=2218622.5187mσ=+=And the probability thatm< 0 is computed using Equation 2.15()()3047002.51610187P m−−≤= Φ= Φ −=×Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-8

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Chap. 2Uncertainty and Risk in Foundation Design2.9We wish to design a shallow foundation with a probability of failure of 10-3. The footingsupports a column carrying a dead load with a mean of 30 k and COV of 0.05 and a live loadwith a mean of 10 k and COV of 0.15.Based on the uncertainty of soil properties and ouranalysis method, we estimate the COV of the foundation capacity to be 0.2.For what meancapacity does the foundation need to be designed?SolutionWe want to select a value ofμCsuch thatPf= 10-3or310Cmmμμσ−−Φ=. From Equations 2.10,2.17, and 2.18()()()222222COV+ COVCOVmCDLCCDLCCDDDLμμμμσσσσμμμ=−−=++=+And()()()()322210COV+ COVCOVCDLCCDDDLμμμμμμ−−+Φ=+Substituting know values of for the COVs and means()()()()32223010100.2+ 0.05 300.15 10CCμμ−−+Φ=⋅+⋅Solving this equation iteratively using Excel we getμC= 106 kSolutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-9

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Chap. 2Uncertainty and Risk in Foundation Design2.10Assume the foundation in Problem 2.9 was to support a high voltage transmission line near theDanish city of Århus.If the transmission line fails it will potentially kill 50 people.If thecomputed probability of failure is for a design life of 100 years, is risk associated with the failureof design acceptable based on the Danish guidance in Figure 2.8? Explain.SolutionThe probability of failure in Problem 2.9 was set to 10-3. If this is the total probability of failureover 100 years, then the annual probability of failure is approximately 10-3/100 = 10-5. The pointwith 50 deaths and an probability of 10-5is plotted on Figure 2.8 below. This point lies betweenthe negligible line and limit of tolerability for the Danish code.In this zone the project mustinclude mitigations to make the risk “as low as reasonably practicable” or the probability offailure must be reduced to an annual probability of 10-6.1.E-081.E-071.E-061.E-051.E-041.E-031.E+001.E+011.E+021.E+031.E+041.E+05Annual ProbabilityTotal DeathsArea ofIntenseScrutiny (2)Limit ofTolerability (2)Limit ofTolerability (1)10-310-410-510-610-710-8101100102103105104Negligible (1)Negligible (2)(50,10-3)Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-10

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Chap. 2Uncertainty and Risk in Foundation Design2.11For the footing in Example 2.2, compute factor of safety required for a probability of failure of5×10-4assuming the COV of the demand is 0.15SolutionFrom Example 2.2 we know that the mean capacity is 11,910 lb/ft2with a standard deviation of2,280 lb/ft2. The question is what is the greatest mean demand that will give us a probability offailure of 5×10-4.To compute this we must compute the mean and standard deviation of thesafety margin,m, as a function of the mean and standard deviation of the demand,D.2211,9102, 280COV2, 2800.15mCDDmCDDDDμμμμσσσμμ=−=−=+=++And()4411,9105102, 2800.15211,9105102, 2800.15DDDmmDDDμμμμσμμμ−−−−−×= Φ= Φ +−×= Φ +Solving the above equation iteratively using Excel, we computeμD= 4135And the mean factor of safety,F, is then11,9102.94,135F==Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-11

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Chap. 2Uncertainty and Risk in Foundation Design2.12If the ASD design method has work satisfactorily for over 50 years, what’s the value in changingto LRFD method?SolutionThere are two major advantages to LRFD when compared to ASD.First, since LRFD usesmultiple partial factors of safety, it is more flexible and produces designs with more consistentprobabilities of failure for different load combinations, and different material property variability.Second, the partial safety factors in LRFD are selected based on an optimization process thatuses probability theory explicitly include the variability of the loads, material properties, andanalysis methods.Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd2-12

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Chap. 3Soil Mechanics3.1Explain the difference between moisture content and degree of saturation.SolutionMoisture content of a soil is the ratio of the weight of its water to weight of its solids, whereasthe degree of saturation is the ratio of the volume of water to volume of voids.The degree ofsaturation can range from 0 to 100%, whereas the moisture content can be larger than 100%.Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd3-1

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Chap. 3Soil Mechanics3.2A certain saturated sand (S= 100%) has a moisture content of 25.1% and a specific gravity ofsolids of 2.68.It also has a maximum index void ratio of 0.84 and a minimum index void ratioof 0.33. Compute its relative density and classify its consistency.SolutionFirst we must compute the void ratioeusing the given specific gravity and moisture content. Thevoid ratio calculated using specific gravity and moisture content is25.1%2.680.67100%swGeS×===Using Equation 3.1, the relative density ismaxmaxmin0.840.67100%100%33%0.840.33reeDee−−=×=×=−−Look up the consistency in Table 3.3 based on the calculated relative density of 33%: the soil isclassified as loose.Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd3-2

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Chap. 3Soil Mechanics3.3Consider a soil that is being placed as a fill and compacted using a sheepsfoot roller (a piece ofconstruction equipment). Will the action of the roller change the void ratio of the soil? Explain.SolutionYes, whenever a soil is compacted, the volume of the soil decreases.Since the volume of thesolids does not change with compaction, the volume of the voids decreases; and since the voidratio is the ratio of the volume of voids to the volume of solids, the void ratio decreases as aresult.Solutions ManualFoundation Engineering: Principles and Practices, 3rdEd3-3

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