1.91. kg m s. Solving for m s yields m s kg. Now we can use the fundamental definition of water content to get the mass of water. m w = 0.

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1 1 CE 46 - Homework #1 Solutions Problem #.1. Problem - A contractor will want to charge you more for coring through that "soft rock layer". However, your expertise tells you that that layer is not rock - it is hard soil. Who wins? Problem #.3. Sedimentary rocks contain fossils that accumulate along with sediments. (This is why petroleum is found in sedimentary rocks). Problem #4.. Given V total m 3 m total kg w ρ s.66. gm cm 3 We also know that Find... unit weight γ ρ w 1. gm cm 3 m. total g γ Nm. 3 V total density ρ m total ρ kg m 3 V total We know that w m w m s m total m s m s Thus, kg m s m s Solving for m s yields m s kg Now we can use the fundamental definition of water content to get the mass of water w m w so m w wm. s m w 0.4 kg m s The volume of water is V water m w ρ w V water 4.15 cm 3 volume of water S V water V voids V water V water V total V s m s V total ρ s S V water m s V total ρ s S Saturation is 76.3%

2 Problem #4.3. Given w 0.38 ρ s.73 gm Find.... S 1 We also know that ρ w 1. gm cm 3 cm 3 void ratio e We can use the magic formula porosity n n e 1 e ρ. w S. e w. ρ s so e n porosity w. ρ s ρ. w S e void ratio unit weight Think about the phase diagram: we can assume that this soil is only composed of solids and water (of course, it is fully saturated). Lets assume that the volume of solids Vs is 1 cubic meter. V s 1m. 3 The corresponding mass of solids is m s ρ. s V s m s kg mass of solids Now we can use the fundamental definition of water content to get the mass of water w m w so m w wm. s m w kg mass of water m s The corresponding volume of water is V water m w The unit weight is γ m. total g m s m. w g γ V total V s V w ρ V w water m 3 m s m. w g γ N V s V water m 3 Problem #4.5. Given (general) dia 60. mm length 15. mm V total m total dia π.. length V total cm 3 ρ s.70. gm cm gm We also know that ρ water 1. gm cm 3

3 3 Given (moisture content) mass can.01. gm mass canandmoistsoil gm mass cananddrysoil gm mass moist mass canandmoistsoil mass can mass moist kg mass dry mass cananddrysoil mass can mass dry kg mass water mass moist mass dry mass water 1.5 gm mass of water Now we can use fundamental definitions w mass water w water content mass dry unit weight γ m. total g γ Nm. 3 V total dry unit weight γ d γ 1 w γ d N m 3 We can use the following equation (it's in the book - keep it in mind) W s W total 1 w which translates to m s m. total g 1 w g m s kg mass of solids Now we can use fundamental definitions to get the mass of water and the saturation w m w so m w wm. s m w kg mass of water m s m w S V water V voids V water V water V total V s m s V total ρ s S ρ water m s V total ρ s S Saturation 59.4% void ratio e We can use the magic formula ρ. w S. e w. ρ s so e w. ρ s ρ. w S e void ratio

4 4 Problem #4.6. Assume ρ s.65. gm ρ w 1 gm cm 3 cm 3. S 1 below water table typically saturated e max e min 0.38 w 0.56 We need to get the in situ void ratio e insitu We can use the magic formula Dr e max e insitu Dr e max e min w. ρ ρ. w S. e insitu w. s ρ s so e insitu ρ. w S Dr 59.1 percent e insitu According to Table 4.4, this soil is Medium Dense. void ratio

5 CIVE 46 HW # SOLS 5.3. No, it is not necessary to perform hydrometer analyses to determine the USCS classification. A sieve analysis is sufficient to determine the percent fines and the distinction between clays and siltes is based on the Atterberg limits, not grain size. If the soil contains less than 1% fines, the USCS requires a values of D 10. Technically, a sieve analysis would be insufficient to obtain this value if the soil contains 10 to 1 percent fines, since the GSD curve would not extend to D 10. However, in this case, D 10 could be obtained with sufficient precision by extrapolating the GSD In the USCS, the noun represents the primary component, and the adjective represents the ondary component. Thus, a silty sand is primarily sand, whereas a sandy silt is primarily silt see table below.

6 1 CE 46 - Homework #3 Solutions kpa Pa Problem #7. kn N i ( 543 3) ( ) 50 i 0.14 Problem #7.3 γw kn h pl 74. ft h pk 3. ft m 3 u K γwh. pk u K kpa u L γwh. pl u L 1.67 kpa Problem #7.6 d 110. mm A π. d L 70. mm q 910. ml ( ). q cm3 h 010. mm mm i h L i 1.56 k q ia. k cm reasonable for fine sand. Problem #7.7 We may use Hazen's correlation, because the D10 of the soil, D mm, is between 0.1 and 3.0 mm, and Cu 1.6, which is less than 5. Coarse sand k k k 1. k 4 cm cm Fine sand k k 0.01 cm note the difference in permeability for fine vs. coarse sand (same soil type, but very different permeability values!)

7 Problem "#5". q k 1cm. 3 d 10. d cm A π. A cm 10. cm L 0. cm e 0.6 v D q A v D cm n v p e 1 e q An. n v p cm pore velocity is ALWAYS larger than Darcy velocity. Problem #7.0. k cm q k. i ( A) k..( 6cm km) q k..( 6cm km) 500 q m3 hr Problem "#7". k cm L 1000 ft h1 ( 1 30). ft head at first equipotential line h last 30. ft N D 7 # of drops N F # of flowpaths h h1 h last hlpd h hlpd m head lost per drop: "hlpd" N D h a h1 ( hlpd. 1) h a m h ea 4. ft h pa h a h ea h pa m u a γwh. pa u a kpa h b h1 ( hlpd. ) h b m h eb 4. ft h pb h b h eb h pb m u b γwh. pb u b 6.79 kpa etc... q N k. L. F h. N D q 0.01 m3 hr

8 3 Problem #8.8. γw lbf ft 3 z f 40. ft depth of foundation dimensions L 00. ft B 150. ft Area L. B Area grondwater tables: best case worst case for uplift (we use this one for design) z gwt 5. ft z gwt_f 13. ft Uplift pressure pore water pressure at foundation base u f γw. z f z gwt_f u f Pa Uplift force pore water pressure at foundation base times the area of the foundation F u u. f Area F u lbf F u N

9 46 HW Solution 7kp5 Problem 5 q : 1 cm3 e : 0.6 k 10 cm 10 : A : cm π A cm L : 0 cm q kia i : q ka i 1.73 vd : k i Darcy velocity vd cm e n : porosity n e ki vp : Pore velocity vp cm n vp vd.613 The pore velocity is.6 times larger than the Darcy velocity. vp > vd is always the case because the pore velocity incorporates a smaller area (the area of flow; not the total area, as incorporated in the vd equation). Problem 6 (7.0) L1 : 4.5 km L : 6 cm h : 187 m 11 m h 66 m L : 500 m k 910 cm : A : L1 L A 55 m i : q : ki A q m3 hr h L

10 Problem 4 (7.7) Yes we can, because the soil meets the requirements stated in p. 7. Hazen tested only these types of soil to obtain his empirical equation. D10 : 0.73 C : 1 k : C D10 k cm/ D10 : k : C D10 k 4 cm/ D10 :. k : C D10 k 0.04 cm/ Problem 3 (7.6) A : 110 mm π A cm L : 70 mm h : ( ) mm h i : i 1.56 L 910 ml q : q cm3 ( ) q k i A k : q ia k cm This is reasonable because this is a fine sand. (see typical k values in your notes... Problem 1 (7.) L : 50 ft he K : 543 ft he L : 461 ft hp K : 3 ft hp L : 74ft h K : he K + hp K h L : he L + hp L ( ) h : h K h L h K m h L m h m h i : i 0.14 L Problem (7.3) γw : 9.81 kn kpa : 1000 Pa m 3 u K : hp K γw u K kpa u L : hp L γw u L 1.67 kpa

11 46 HW solution 8fn7 1. Flownet problem Consider that the datum is on the interface between the soil and the impermeable layer. To compute the total head at all eqp lines, we need the total head loss through the whole system: h e1 h p1 : 30 ft elev head at first equipotential : 1 ft press head at first equipotential h t1 : h p1 + h e1 total head at first equipotential h t1 51 ft h e8 h p8 : 30 ft elev head at last (eigth) equipotential line : 0ft press head at last (eigth) equipotential line h t8 : h p8 + h e8 total head at last (eigth) equipotential line h t8 30 ft THE HEAD LOSS THROUGHTHE SYSTEM IS EQUAL TO THE DIFFERENCE IN TOTAL HEADS: h t : h t1 h t8 h t 1 ft N D : 7 this is the number of equipotential DROPS Now, the head loss per equipotential DROP is (see notes) h per_drop h t : h per_drop 3ft this is how much "energy" is lost per N D drop transversed. Now we can compute the total heads at all equipotential lines: h t1 51 ft total head at first eqp line h t : h t1 h per_drop h t 48 ft total head at ond eqp line h t3 : h t h per_drop h t3 45 ft total head at third eqp line h t4 : h t3 h per_drop h t4 4 ft total head at fourth eqp line h t5 : h t4 h per_drop h t5 39 ft total head at fifth eqp line h t6 : h t5 h per_drop h t6 36 ft total head at sisxth eqp line h t7 : h t6 h per_drop h t7 33 ft total head at seventh eqp line h t8 : h t7 h per_drop h t8 30 ft total head at eigth (last) eqp line 1

12 Now let's compute the flowrate through the system - Q we are given these parameters: L : 1000 ft k : 10 5 cm and we know from the flownet that N D : 7 this is the number of equipotential DROPS N F : this is the number of flow channels h t 1 ft this is the total head loss through the system So we use the equation given in class: N F Q: k L h t Q ft 3 or Q cm3 N D Now let's find the uplift pressures along the base of the dam. This means that we have to find the pressure heads along the base of the dam in order to use the equation u γ w h p We know the total head at points a,b,c,d,e, and f (calculated above), and we know that the elevation head at all these points is 30ft - 6 ft 4ft (from the drawing). The difference is the pressure head: h ta h ea : the total head at point a is equal to the total head at ond eqp line h ta 48 ft h t : 4ft elevation head at point a h pa : h ta h ea h pa 4 ft pressure head at point a now we repeat this for all points, b,c,d,e, and f: h tb h eb : the total head at point b is equal to the total head at third eqp line h tb 45 ft h t3 : 4ft elevation head at point b h pb : h tb h eb h pb 1 ft pressure head at point b h tc h ec : the total head at point c is equal to the total head at fourth eqp line h tc 4 ft h t4 : 4ft elevation head at point c h pc : h tc h ec h pc 18 ft pressure head at point c h td h ed : the total head at point d is equal to the total head at fifth eqp line h td 39 ft h t5 : 4ft elevation head at point d h pd : h td h ed h pd 15 ft pressure head at point d

13 h te h ee : the total head at point e is equal to the total head at sixth eqp line h te 36 ft h t6 : 4ft elevation head at point e h pe : h te h ee h pe 1 ft pressure head at point e h tf h ef : h t7 the total head at point f is equal to the total head at seventh eqp line h tf 33 ft : 4ft elevation head at point f h pf : h tf h ef h pf 9ft pressure head at point f OK, now we have the pressure head at points a,b,c,d,e, and f. To get the pore pressure (uplift pressure in this case) we just need to use the fundamental equation, cosidering that γ w : 6.4 lbf so here it is: ft 3 u a u b u c u d u e u f : γ w h pa uplift pressure at point a u a lbf (this unit is psf) : γ w h pb uplift pressure at point b u b : γ w h pc uplift pressure at point c u c : γ w h pd uplift pressure at point d u d 936 lbf : γ w h pe uplift pressure at point e u e lbf : γ w h pf uplift pressure at point f u f lbf lbf lbf so now, lets plot this (assume that points a,b,c,d,e, and f are equally spaced along the 4ft long dam) You can plot this easier by using excel - it is just an x-y plot. U : u a u b u c u d u e u f D : ft 3

14 uplift pressure (psf) U D distance along base (ft) u f + u b The "average" uplift pressure is u ave : u ave 936 lbf The preliminary factor of safety is FS : 1500 lbf u ave FS

15 HW #5 - Solutions Problem kpa : 1000 Pa σeff : 181 kpa σ : N m 3 (.5 m) σ 48.5 kpa Immediately after fill is placed: σeff i : 181 kpa After consolidation is completed: σeff f : 181 kpa + σ σeff f 9.5 kpa Problem kpa : 1000 Pa L : 1m Volume old : 1m 3 ε z dl L dl : L dl m Volume new : ( L dl) L L Volume new m 3 Volume squeezed : Volume old Volume new Problem Volume squeezed 85 liter σeff p : 850 lbf σeff i : 797 lbf σeff f : σeff i lbf ft 3 ft 3 ft 3 For this case, σeff i < σeff p < σeff f Equation 11.5 should be used. That is, both portions of the consolidation curve must be used. σeff f lbf ft 3 1

16 Problem #4. Lab data for specimens at A and B locations σeff pb : 195 kpa eo B : 1.0 Cc B : 0.9 Cr B : 0.06 At point B: σeff ib : [( ) ( ) ] kpa σeff ib kpa σeff fb : [( ) ( ) ] kpa σeff fb 1.95 kpa σeff B : σeff fb σeff ib σeff B 6.48 kpa σeff i < σeff p < σeff σeff f pb OCR B : σeff ib Use both portions of the curve for point B. OCR B 1.96 The overconsolidation ratio at point B at time 0 is > 1. Thus, as t 0, the soil is overconsolidated. δ c_ult : Cr B 1 + eo B ( 11 m) log σeff pb σeff ib + Cc B 1 + eo B ( 11 m) log σeff fb σeff pb δ c_ult m Problem 1.4 γdrysand : 11 lbf γsatsand : 13 lbf γsatclay : 14 lbf ft 3 ft 3 ft 3 γfill : 10 lbf ft 3 c v : 0.17 ft γw : 6.4 lbf day ft 3 Solution u H : ( 16 ft) γw u H lbf σeff i : [( 4ft ) γdrysand + ( 6ft ) γsatsand + ( 10 ft) γsatclay] u H σeff i lbf σeff f : [( 4ft ) γdrysand + ( 6ft ) γsatsand + ( 10 ft) γsatclay + ( 0 ft) γfill] u H σeff f lbf σeff : σeff f σeff i σeff lbf

17 H dr : 5 ft H dr 5 ft single drainage t1 : 1 yr t : yr t4 : 4 yr t8 : 8 yr t16 : 16 yr T v_1yr c v t1 c v t c v t4 c v t8 c v t16 : T v_yr : T v_4yr : T v_8yr : T v_16yr : H dr H dr H dr H dr H dr T v_1yr T v_yr T v_4yr T v_8yr T v_16yr 1.59 z dr : 10 ft z dr 0.4 H dr From chart 1.4: u e_1yr σeff 0.6 (interpolated) u e_yr σeff 0.46 u e_4yr σeff 0.31 u e_8yr σeff 0.11 u e_16yr σeff.0 excess pore pressures u e_1yr : 0.6 σeff u e_1yr lbf u e_yr u e_4yr u e_8yr u e_16yr : 0.46 σeff u e_yr lbf : 0.31 σeff u e_4yr 744 lbf : 0.11 σeff u e_8yr 64 lbf : 0.0 σeff u e_16yr 48 lbf total pore pressures u total_1yr : u H + u e_1yr u total_1yr lbf u total_yr : u H + u e_yr u total_yr lbf u total_4yr : u H + u e_4yr u total_4yr lbf u total_8yr : u H + u e_8yr u total_8yr lbf 3

18 u total_16yr : u H + u e_16yr u total_16yr lbf final total stress (t infinity) : ( 4ft ) γdrysand + ( 6ft ) γsatsand + ( 10 ft) γsatclay + ( 0 ft) γfill σ final lbf σeff 1yr : σ final u total_1yr σeff 1yr lbf σ final σeff yr : σ final u total_yr σeff yr lbf σeff 4yr : σ final u total_4yr σeff 4yr lbf σeff 8yr : σ final u total_8yr σeff 8yr lbf σeff 16yr : σ final u total_16yr σeff 16yr lbf Problem 6 Solution kpa : 1000 Pa Lab data for specimens at A σeff pa : 450 kpa eo A : 1.3 Cc A : 0.5 Cr A : 0.05 c v : 0.00 m day At point A: u H : kpa u H 9.43 kpa σeff ia : [( ) ( ) ] kpa σeff ia kpa σeff fa : [( ) ( ) ] kpa σeff fa kpa σeff A : σeff fa σeff ia σeff A 300 kpa load (a) What is the value of the total pore water pressure at point A before the fill is placed? u H 9.43 kpa (b) What is the value of the total pore water pressure at point A at time infinity? u H 9.43 kpa (c) What is the value of the excess pore water pressure at point A before the fill is placed? u e : 0 4

19 For t 5640 days, H dr z dr : : 7.5 m 3m z dr 0.4 H dr Tv : c v ( 5640 day) H dr Tv 0.01 From chart: u e σeff 0.46 Thus, u e_5640days : σeff A 0.46 u e_5640days 138 kpa Answer for (d) u total_5640days : u e_5640days + u H u total_5640days kpa Answer for (e) Ultimate settlement calculations At point A: σeff ia σeff fa kpa kpa σeff A 300 kpa σeff pa 450 kpa OCR A : σeff pa σeff ia OCR A 1.74 The OCR at t 0 is larger than 1 (i.e., the soil is overconsolidated). Therefore, we must use both portions of the consolidation curve! δ c_ult_a Cr A ( 15 m) log σeff pa Cc A ( 15 m) log σeff fa : + δ c_ult_a 0.3 m 1 + eo A σeff ia 1 + eo A σeff pa 5

20 Settlement calculations at t5640 days At point A: σeff ia kpa σeff at5640days σ f u total_5640days σ f : ( ) kpa This is the total stress at tinfinity, which is also the total stress at time anytime. 588 kpa σ f σeff at5640days : σ f u total_5640days σeff at5640days kpa σeff ia σeff pa σeff at5640days kpa 450 kpa kpa we must use the first portion of the curve because the preconsolidation stress is higher than the effective stress at the time in question. δ c_a_5640days : Cr A ( 15 m) log σeff pa 1 + eo A σeff ia δ c_a_5640days m Note that the settlement at this time is smaller than the ultimate settlement - this is expected! U at t5640 days U 5640days δ c_a_5640days : 100 U 5640days percent δ c_ult_a time required for U to be 90% U time 90 Solution δ c_a_t 100 δ c_ult_a δ c_a_t : 90 δ c_ult_a 100 δ c_a_t 0.08 m The layer must suffer this settlement in order for U to be 90% Since U90% means that the layer is "90 percent consolidated", it seems logical to assume that the effective stress that is necessary for this settlement is very close to the final effective stress at infinity (which is 558 kpa). Thus, we choose both portions of the consilidation curve for the settlement equation to solve for that effective stress, which we call σeff at_t : 6

21 0.08 m Cr A 1 + eo A ( 15 m) log σeff pa σeff ia + Cc A 1 + eo A ( 15 m) log σeff at_t σeff pa Solving the equation above for σeff at_t yields σeff at_t : kpa So, let's check if we made the right choice for the settlement equation σeff at_t 1.01 Yes we did, because our effective stress at time t (that is, the time for 90% consolidation) is σeff pa larger than the preconsolidation stress. Now, lets use our value of σeff at_t to get the total pore pressure at time t, u total_t u total_t : σ f σeff at_t u total_t kpa This means that u e_t : u total_t u H u e_t 18.4 kpa This is the excess pore pressure at time t What is u e_t σeff A? u e_t σeff A Now we are ready to use the chart to get the value of Tv that we need! Tv 0.9 Now we use Tv 0.9 c v t and solve for t H dr t : day Thus, 5310 days must go by for the layer to consolidate 90%. 7

22 46 Homework 6 solutions problem 1 kn d: 3 m u up : 1.6 m 9.81 m 3 A π d : u up kpa A m P : P σ : A 300 kn σ kpa σ.704 Tank will not float up. It would if the stress were equal or smaller u up than the uplift pressure. problem σeff v : ( ) kpa σeff v kpa K : 0.45 σeff h : K σeff v σeff h kpa problem 3 σeff v kpa σeff h kpa These stresses are principal because the ground is flat AND the stress as geostatic. The planes on which they act are principal planes, so there is no shear on those planes. The Mohr circle is shown below. Note that the shear stress on the plane in question is ~7 kpa, and acts clockwise.

24 1 46 HW 7 solutions Problem 1 kn N kpa Pa σ eff3 30. kpa σ effp σ effcs 98. kpa 76. kpa Draw Mohr circle at peak failure: The effective cohesion is zero because the sand is clean (free of fines). The sand was probably dense befopre shearing because the response exhibits a peak.

25 Problem T 8. kn A 00. cm Su T A Su 400 kpa The 100 kpa vertical load is not considered because the clay was loaded undrained (faster than the excess pore pressure dissipation rate. The clay was normally consolidated before shearing (before impsing the T force), because no peak is exhibited by the load deformation curve. Problem 3 part (a) σ effa_before ( 30. ) ( ) σ effa_before σ effa_after ( 1. 19) ( 3. 0) ( ) σ effa_after load σ effa_after σ effa_before load 8 all in kpa The strength of the soil at point A on the horizontal plane is S A σ. eff_onhorizplane tan φ eff c eff S A_before S A_after σ. effa_before tan( 0. deg) 5 S A_before kpa σ. effa_after tan( 0. deg) 5 S A_after kpa note that soil gets stronger on this plane when loaded in this manner. part (b) The strength of the soil at point B on the plane shown is S B σ. eff_onplaneshown tan φ eff c eff We must draw the Mohr Circle at point B to determine the value of effective stress that acts on the plane shown. For this purpose we need the vertical effective stress at point B and the horizontal effective stress at point B σ eff_v_b ( 15. 0) ( 1. 1) ( ) σ eff_v_b kpa K 0.5 given σ eff_h_b K. σ eff_v_b σ eff_h_b 8.0 kpa

26 3 STEPS: - draw the element - draw the plane shown next to the element - select the right side (the vertical side) of the element, and shade it - rotate the element 30 degrees clockwise, so that the shaded side aligns with the plane shown - say "i rotated the element 30 degrees clockwise" - the horizontal line that joins the two initial points in the MC must therefore be rotated x30 60 degrees clockwise, as shown below. - the 'new point' is the the one that was previsouly associated with the horizontal effective stress (because this stress acts on the plane we shaded). S B tan( 35. deg) 0 S B kpa remember, clean sand has no cohesion intercept, that is, c' 0

YOUR HW MUST BE STAPLED YOU MUST USE A PENCIL (no pens)

YOUR HW MUST BE STAPLED YOU MUST USE A PENCIL (no pens) Spring 2008 CIVE 462 HOMEWORK #1 1. Print out the syllabus. Read it. Write the grade percentages in the first page of your notes. 2. Go back to your 301 notes, internet, etc. and find the engineering definition