Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition

Preview (16 of 297 Pages)

100%

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 1 preview image

Loading page ...

CHAPTER 2INDEX AND CLASSIFICATION PROPERTIES OF SOILS2-1.From memory, draw a phase diagram (like Fig. 2.2, but don’t look first!). The “phases” havea Volume side and Mass side. Label all the parts.SOLUTION:Refer to Figure 2.2.2-2.From memory, write out the definitions for water content, void ratio, dry density, wet or moistdensity, and saturated density.SOLUTION:Refer to Section 2.2.2-3.Assuming a value ofρs= 2.7 Mg/m3, take the range of saturated density in Table 2.1 for thesix soil types and calculate/estimate the range in void ratios that one might expect for these soils.SOLUTION:Create a spreadsheet using input values from Table 2.1 and Eq. 2.18.ρ' - minρ' - maxemaxemin(Mg/m3)(Mg/m3)0.91.40.890.210.41.13.250.551.11.40.550.210.91.20.890.420.00.1∞16.000.30.84.671.13(Given)(see Eq. 2.18)

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 2 preview image

Loading page ...

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 3 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22-4.Prepare a spreadsheet plot of dry density in Mg/m3as the ordinate versus water content inpercent as the abscissa. Assumeρs= 2.65 Mg/m3and vary the degree of saturation, S, from100% to 40% in 10% increments. A maximum of 50% water content should be adequate.SOLUTION:Solve Eq. 2.12 and Eq. 2.15 forρd= f(ρs, w, S, Gs), or use Eq. 5.1.ρSwρ=dryρww+SρsS =100908070605040302010wρdryρdryρdryρdryρdryρdryρdryρdryρdryρdry(%)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)0.02.652.652.652.652.652.652.652.652.652.655.02.342.312.272.232.172.091.991.841.591.1410.02.092.051.991.921.841.731.591.411.140.7315.01.901.841.771.691.591.481.331.140.890.5320.01.731.671.591.511.411.291.140.960.730.4225.01.591.531.451.361.261.141.000.830.610.3530.01.481.411.331.241.141.020.890.730.530.3035.01.371.311.231.141.040.930.800.650.470.2640.01.291.221.141.050.960.850.730.580.420.2345.01.211.141.060.980.890.780.670.530.380.2150.01.141.071.000.920.830.730.610.490.350.19Problem 2-40.000.501.001.502.002.503.000510152025303540455055w (%)ρdry(Mg/m3)S=100%S=90%S=80%S=70%S=60%S=50%S=40%S=30%S=20%S=10%

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 4 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22-5aPrepare a graph like that in Problem 2.4, only use dry density units of kN/m3and poundsper cubic feet.SOLUTION:ρSwFrom Eq. 2.12 and Eq. 2.15:ρ=dryρww+SρsS =100908070605040302010wγdryγdryγdryγdryγdryγdryγdryγdryγdryγdry(%)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)0.026.0026.0026.0026.0026.0026.0026.0026.0026.0026.005.022.9522.6622.3021.8621.2920.5519.5318.0315.6411.1810.020.5520.0819.5318.8618.0316.9915.6413.8011.187.1215.018.6018.0317.3716.5815.6414.4813.0411.188.705.2320.016.9916.3615.6414.7913.8012.6211.189.407.124.1325.015.6414.9714.2213.3612.3511.189.798.106.033.4130.014.4813.8013.0412.1711.1810.048.707.125.232.9035.013.4912.8012.0411.1810.219.117.836.354.612.5340.012.6211.9411.1810.349.408.337.125.734.132.2445.011.8611.1810.449.628.707.686.535.233.732.0150.011.1810.529.798.998.107.126.034.803.411.82Problem 2-5a0.005.0010.0015.0020.0025.0030.000510152025303540455055w (%)γdry (kN/m3)S=100%S=90%S=80%S=70%S=60%S=50%S=40%S=30%S=20%S=10%

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 5 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22-5bPrepare a graph like that in Problem 2.4, only use dry density units of pounds per cubicfeet.SOLUTION:swdrysGSSG wγγ=+S =100908070605040302010wγdryγdryγdryγdryγdryγdryγdryγdryγdryγdry(%)(pcf)(pcf)(pcf)(pcf)(pcf)(pcf)(pcf)(pcf)(pcf)(pcf)0.0165.36165.36165.36165.36165.36165.36165.36165.36165.36165.365.0146.01144.14141.86139.04135.45130.72124.21114.7099.4671.1210.0130.72127.75124.21119.95114.70108.0899.4687.8071.1245.3015.0118.33114.70110.47105.4799.4692.1282.9471.1255.3533.2420.0108.08104.0799.4694.1187.8080.2771.1259.7745.3026.2525.099.4695.2590.4584.9678.5971.1262.2551.5438.3421.6930.092.1287.8082.9477.4371.1263.8555.3545.3033.2418.4835.085.7981.4476.5871.1264.9557.9249.8340.4129.3316.0940.080.2775.9371.1265.7759.7753.0045.3036.4826.2514.2645.075.4271.1266.3961.1655.3548.8541.5333.2423.7512.7950.071.1266.8962.2557.1651.5445.3038.3430.5321.6911.60Problem 2-5b0.0020.0040.0060.0080.00100.00120.00140.00160.00180.000510152025303540455055w (%)γdry(pcf)S=100%S=90%S=80%S=70%S=60%S=50%S=40%S=30%S=20%S=10%

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 6 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.6.Prepare a graph like that in Problem 2.4, only for S = 100% and vary the density of solidsfrom 2.60 to 2.80 Mg/m3. You decide the size of the increments you need to “satisfactorily”evaluate the relationship asρsvaries. Prepare a concluding statement of your observations.SOLUTION:ρSwFrom Eq. 2.12 and Eq. 2.15:ρ=dryρww+SρsNote: The relationship betweenρdryand w is not overly sensitive toρs.ρs=2.62.652.72.752.8wρdryρdryρdryρdryρdry(%)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)0.02.602.652.702.752.805.02.302.342.382.422.4610.02.062.092.132.162.1915.01.871.901.921.951.9720.01.711.731.751.771.7925.01.581.591.611.631.6530.01.461.481.491.511.5235.01.361.371.391.401.4140.01.271.291.301.311.3245.01.201.211.221.231.2450.01.131.141.151.161.170.000.501.001.502.002.503.000510152025303540455055w (%)ρs(Mg/m3)

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 7 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.7.The dry density of a compacted sand is 1.87 Mg/m3and the density of the solids is 2.67Mg/m3. What is the water content of the material when saturated?SOLUTION:wdryws3w33drysSFrom Eq. 2-12 and Eq. 2-15:;Note: S = 100%wS1111Solving for w: wS(1Mg / m )(100%)16.0%1.87 Mg / m2.67Mg / mρρ=ρ+ ρ⎛⎞⎛⎞= ρ−=−=⎜⎟⎜⎟⎜⎟ρρ⎝⎠⎝⎠2.8.A soil that is completely saturated has a total density of 2045 kg/m3and a water content of24%. What is the density of the solids? What is the dry density of the soil?SOLUTION:a) Solve using equations or phase diagrams:3tdrywsatdrywswdry3sdrywsat20451649.2 kg / m(1w)(10.24)1(1000)(1649.2)2729.6 kg / m(1649.210002045)ρρ===++⎛⎞ρρ=−ρ+ ρ⎜⎟ρ⎝⎠ρ ρρ===ρ+ ρ− ρ+−b) Solve using phase diagram relationships:ttwwsstwssss3swwww3sss3sdrytassume V1.0M2045 kgM0.24M0.24MMMMM20450.24MMM1649.19 kg0.24MM1000V0.3958 mV1000M1649.192729.6 kg / mV10.3958M1649.191649.2 kg / mV1===→==+→=+→=ρ==→==ρ===−ρ===

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 8 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.9What is the water content of a fully saturated soil with a dry density of 1.72 Mg/m3? Assumeρs= 2.72 Mg/m3.SOLUTION:wdryws3w33drysSFrom Eq. 2-12 and Eq. 2-15:;Note: S = 100%wS1111Solving for w: wS(1Mg / m )(100%)21.4%1.72 Mg / m2.72 Mg / mρρ=ρ+ ρ⎛⎞⎛⎞= ρ−=−=⎜⎟⎜⎟⎜⎟ρρ⎝⎠⎝⎠2.10.A dry quartz sand has a density of 1.68 Mg/m3. Determine its density when the degree ofsaturation is 75%. The density of solids for quartz is 2.65 Mg/m3.SOLUTION:drydry(final)wdryws3w33drystdryRecognize that(initial)(final);S= 75%SFrom Eq. 2-12 and Eq. 2-15:wS1111Solving for w: wS(1Mg / m )(75%)16.34%1.68 Mg / m2.65 Mg / mfinal(1ρ= ρρρ=ρ+ ρ⎛⎞⎛⎞= ρ−=−=⎜⎟⎜⎟⎜⎟ρρ⎝⎠⎝⎠ρ= ρ+3w)1.68(10.1634)1.95 Mg / m=+=2.11.The dry density of a soil is 1.60 Mg/m3and the solids have a density of 2.65 Mg/m3. Findthe (a) water content, (b) void ratio, and (c) total density when the soil is saturated.SOLUTION:wdryws3w33drysswGiven: S = 100%SFrom Eq. 2-12 and Eq. 2-15:wS1111(a) Solving for w: wS(1Mg / m )(100%)24.76%1.60 Mg / m2.65 Mg / mw(24.76)(2.65)(b) From Eq. 2.15: eS(100)(1ρρ=ρ+ ρ⎛⎞⎛⎞= ρ−=−=⎜⎟⎜⎟⎜⎟ρρ⎝⎠⎝⎠ρ==ρ3tdry0.656.0)(c)(1w)1.60(10.2476)1.9962.00 Mg / m=ρ= ρ+=+==

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 9 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.12.A natural deposit of soil is found to have a water content of 20% and to be 90% saturated.What is the void ratio of this soil?SOLUTION:ssww = 20% and S = 90%; assume G2.70w(20.0)(2.70)From Eq. 2.15: e0.60S(90)(1.0)=ρ===ρ2.13.A chunk of soil has a wet weight of 62 lb and a volume of 0.56 ft3. When dried in an oven,the soil weighs 50 lb. If the specific gravity of solids Gs= 2.64, determine the water content, wetunit weight, dry unit weight, and void ratio of the soil.SOLUTION:Solve using phase diagram relationships.wtswstttsdryt3www3ssswvtv(a) WWW625012 lbW12(100)w100%24.0%W50W62(b)110.7 pcfV0.56W50(c)89.29 pcfV0.56W12(d) V0.1923 ft62.4W50V0.3035 ftG(2.64)(62.4)VVV0.560.30350.2565e=−=−==×==γ===γ======γ===γ=−=−=vsV0.25650.84510.84V0.3035====2.14.In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in theoven and 100.06 g after the soil had dried. The container alone had a mass of 49.31 g. Thespecific gravity of solids is 2.80. Determine the void ratio and water content of the original soilsample.SOLUTION:Solve using phase diagram relationships.swwsswM100.0649.3150.75 gM113.27100.0613.21 gM13.21(100)(a)w100%26.0326.0%M50.75w2.80(26.03)(b) From Eq. 2.15: e=0.72880.73S(1)(100)=−==−==×===ρ===ρ

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 10 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.15.The natural water content of a sample taken from a soil deposit was found to be 12.0%. Ithas been calculated that the maximum density for the soil will be obtained when the watercontent reaches 22.0%. Compute how many grams of water must be added to each 1000 g ofsoil (in its natural state) in order to increase the water content to 22.0%.SOLUTION:wwsstswsssswswsNatural stateMw0.12M(0.12)MMMMMM0.12M1.12M1000 gM892.857 gM(0.12)(892.857)107.143 gTarget state(Note: Mdoes not change between natural state and target state)MwM(0.22)(89==→==+=+======×=2.857)196.429 gadditional water necessary = 196.429107.14389.28689.29 g=−==2.16.A cubic meter of dry quartz sand (Gs= 2.65) with a porosity of 60% is immersed in an oilbath having a density of 0.92 g/cm3. If the sand contains 0.27 m3of entrapped air, how muchforce is required to prevent it from sinking? Assume that a weightless membrane surrounds thespecimen. (Prof. C. C. Ladd.)SOLUTION:33333t3vt3stv3kgsssstmtkgtmtkgbuoytoilmbuoybuoyV1m1,000,000 cmVnV(0.6)(1.0)0.60 mVVV1.00.600.40 mMGV(2.65)(1000)(0.40 m )1060 kgMM10601060V1.01060920140g(140)(9.81)1373.4===×===−=−==× ρ ×===ρ===ρ= ρ − ρ=−=γ= ρ×==33Nm3NbuoymForce(entrapped air)1373.40.27 m370.8 N= γ×=×=

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 11 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.17.A soil sample taken from a borrow pit has a natural void ratio of 1.15. The soil will be usedfor a highway project where a total of 100,000 m3of soil is needed in its compacted state; itscompacted void ratio is 0.73. How much volume has to be excavated from the borrow pit to meetthe job requirements?SOLUTION:tstvss3t3s(emb)ts(borr )s(emb)3t(borr )sborrVVVVVVeee1EmbankmentV100,000 m100,000V57,803.47 m0.731Borrow PitVVVe1VV(e1)(57,803.47)(1.151)124,277 m−==→=+===+==+=×+=+=2.18.A sample of moist soil was found to have the following characteristics:Total volume: 0.01456 m3Total mass: 25.74 kgMass after oven drying: 22.10 kgSpecific gravity of solids: 2.69Find the density, unit weight, void ratio, porosity, and degree of saturation for the moist soil.SOLUTION:333kgtmNkNttmm3sssw3v25.74(a)1767.85717680.01456(b)g(1767.857)(9.81)17,342.6817.34M22.10(c)V0.00822 mG(2.69)(1000)V0.014560.008220.006344 m0.006344e0.77180.770.008220.7718(d) n10.7ρ===γ= ρ ×======ρ=−=====+ww10043.5643.6%718(e) M25.7422.103.643.64V0.0036410000.00364S10057.37757.4%0.006344×===−====×==

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 12 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.19.A gray silty clay (CL) is sampled from a depth of 12.5 feet. The “moist” soil was extrudedfrom a 6-inch-high brass liner with an inside diameter of 2.83 inches and weighed 777 grams. (a)Calculate the wet density in pounds per cubic feet. (b) A small chunk of the original sample had awet weight of 140.9 grams and weighed 85.2 grams after drying. Compute the water content,using the correct number of significant figures. (c) Compute the dry density in Mg/m3and the dryunit weight in kN/m3.SOLUTION:()()()()()323ttt33twwtsstdrylbdryft1lb777 g453.6 gM(2.83)(a)V637.741in ,78.42978.4 pcf4V1 ft37.741in12 inM55.7(b) MMM140.985.255.7 g,w100%65.3865.4%M85.278.4(c)47.40 pcf(1w)(10.654)1 ft47.40.3π=×=γ=====−=−==×===γγ===++ρ=() ()33333kgMgmmNkNdrymm0.4536 kg759.2880.759048 m1lb(759.288)(9.81)7448.67.45==γ===2.20.A cylindrical soil specimen is tested in the laboratory. The following properties wereobtained:Sample diameter3 inchesSample length6 inchesWt. before drying in oven2.95 lbWt. after drying in oven2.54 lbOven temperature110°CDrying time24 hoursSpecific gravity of solids2.65What is the degree of saturation of this specimen?SOLUTION:233t33sssw33vtswts33wwwwv(3)V642.4115 in0.02454 ft4W2.54V0.01536 ft26.542 inG(2.65)(62.4)VVV42.411526.54215.869 in0.009184 ftWWW2.952.540.41lbW0.41V0.00657 ft11.354 in62.4V1S100%Vπ=×======γ=−=−===−=−=====γ=×=1.35410071.5%15.869×=

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 13 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.21A sample of saturated silt is 10 cm in diameter and 2.5 cm thick. Its void ratio in this state is1.35, and the specific gravity of solids is 2.70. The sample is compressed to a 2-cm thicknesswithout a change in diameter.(a) Find the density of the silt sample, in prior to being compressed.(b) Find the void ratio after compression and the change in water content that occurredfrom initial to final state.SOLUTION:3323twwvvvss3tvsssss3vwgwwwcmgssswcm(10)(a)V2.5196.350 cm4VS1VVVVeV1.35VVVV1.35VV2.35V196.350V83.553 cmV(1.35)(83.553)112.797 cmVMV(1)(112.797)112.797 gMGV(2.70)(83.553)(1)π=×===→==×==+=+==→===== ρ×===×× ρ=33ttgkgtcmmt23t23s3vtsfinalinitial225.594 gM112.797225.594338.391 gM338.3911.7231723V196.35(10)(b)V2.0157.08 cm4V83.553 cm(no change)VVV157.0883.55373.527 cm73.527e0.8883.553112.(c)w−==+=ρ====π=×===−=−====3wvwsfinal797100%50.0%225.594final conditionsVV73.527 cm ; M73.527 g; M225.594 g (no change)73.527w100%32.6%225.594w50.032.617.4%×======×=Δ=−=

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 14 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.22.A sample of sand has the following properties: total mass Mt= 160 g; total volume Vt=80cm3; water content w = 20%; specific gravity of solids Gs=2.70. How much would the samplevolume have to change to get 100% saturation, assuming the sample mass Mtstayed the same?SOLUTION:3wsstsss3gwwwcm33ssvswitMwM(0.20)MMM0.20M160 gM133.33 gM(0.20)(133.33)26.667 g;V(1)M26.667 cmM133.33V49.383 cm ;V8049.38330.617 cmG2.7026.667S100%87.10%30.617Desired condition: S100%V ch=×==+==========−=ρ=×==ss3wwvw3tsv3anges, but Vand Mremain the sameM26.667 g;V26.667 cmVV(when S = 100%)VVV49.38326.66776.053 cmV8076.0533.95 cm====+=+=Δ=−=

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 15 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.23.Draw a phase diagram and begin to fill in the blanks: A soil specimen has total volume of80,000 mm3and weighs 145 g. The dry weight of the specimen is 128 g, and the density of thesoil solids is 2.68 Mg/m3. Find the: (a) water content, (b) void ratio, (c) porosity, (d) degree ofsaturation, (e) wet density, and (f) dry density. Give the answers to parts (e) and (f) in both SI andBritish engineering units.SOLUTION:wws33sssw3vwww(a) M 14512817 gM17w100%10013.28113.3%M128M128(b)V47.7612 cm47,761.2 mmG(2.68)(1)V80,00047,761.232,238.8 mm32,238.8e0.6747,761.232,238.8(c) n100%40.3%80,000(d)VM(17)(1)−==×=×======ρ=−====×==× ρ=33333333gkgtcmmkglbmtmftkgdrymkgldrym17 cm17,000 mm17,000S100%52.7%32,238.8145(e)1.81251812.58011812.5113.2(see Appendix A)16.0181812.5(f )1600.0(10.13281)11600.099.916.018===×=ρ===⎛⎞ρ=×=⎜⎟⎝⎠ρ==+⎛⎞ρ=×=⎜⎟⎝⎠3bm ft(see Appendix A)

Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition - Page 16 preview image

Loading page ...

Index and Classification Properties of SoilsChapter 22.24.A sample of soil plus container weighs 397.6 g when the initial water content is 6.3%. Thecontainer weighs 258.7 g. How much water needs to be added to the original specimen if thewater content is to be increased by 3.4%? After U.S. Dept. of Interior (1990).SOLUTION:twstwssssswswsM397.6258.7138.9 gM0.063MM138.9MM0.063MM1.063MM130.668 gMw0.034MMwM(0.034)(130.668)4.44 g=−====+=+==ΔΔ==Δ= Δ×==2.25.A water-content test was made on a sample of clayey silt. The weight of the wet soil pluscontainer was 18.46 g, and the weight of the dry soil plus container was 15.03 g. Weight of theempty container was 7.63 g. Calculate the water content of the sample.SOLUTION:swwsM15.037.637.40 gM18.4615.033.43 gM3.43(100)(a)w100%46.35146.3%M7.40=−==−==×===

Preview Mode

This document has 297 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

Solution Manual For Elementary Surveying: An Introduction To Geomatics, 13th Edition

Solution Manual for Reliability Engineering, 1st Edition

Elementary Surveying: An Introduction to Geomatics, 15th Edition Solution Manual

Solution Manual For Steel Structures: Design And Behavior, 5th Edition

Dynamics of Structures, 5th Edition Solution Manual

Solution Manual for Design Concepts for Engineers, 5th Edition

Solution Manual For Structural Analysis, 8th Edition

Solution Manual For Civil And Environmental Systems Engineering, 2nd Edition

Structural Analysis, 10th Edition Solution Manual

Solution Manual For Dynamics of Structures, 4th Edition

Company

Explore

Study Tools