3.) In its natural state, a moist soil has a volume of 0.00935m3 and weighs 178 N. The oven dried weight of the soil is 154 N. If Gs = 2.67, determine the following: a. Moisture Content b. Moist Unit Weight c. Dry Unit Weight d. Void Ratio e. Porosity f. Degree of Saturation
Given: V = 0.00935 m3 W = 178 N W S = 154 N GS = 2.67
Solut So lutii on: a.) MOISTURE CONTENT
w=
WW WS
=
e.) POROSITY
178 NN-154 N
n=
154 N
e e+1
=
0.59 0.59 + 1
w = 0.1558/15.58%
n = 0.371
b.) MOIST UNIT WEIGHT
f.) DEGREE OF SATURATION
γwet =
W V
=
0.178 kN 0.00935
GSw = eS Gsw 2.67 0.1558 S= = e 0.590
m3
γwet = = 19.037 kN/ kN /m 3 c.) DRY UNIT WEIGHT
γdry =
γwet 1+w
=
19.037 kN/m3 1+0.1558
γdry = = 16.471 kN /m 3 d.) VOID RATIO
γdry =
GS (γw ) 1+e
16.47 kN/m3 =
e = 0.590 0.590
2.67 (9.81 kN/m3 ) 1+e
S = 0.705 0.705//70.5% 70.5%
4.) The dry density of a sand with a porosity of 0.387 is 1600 kg/m3. Determine the following: a. Void Ratio b. Specific Gravity of Soil c. Saturated Unit Weight of Soil in kN/m3
Given: n = 0.387 γdry = 1600 kg/m3
Solution: a.) VOID RATIO
e=
n 1-n
=
0.387 1-0.387
e = 0.631 b.) SPECIFIC GRAVITY OF SOIL
γdry =
GS (γw ) 1+e 3
1600 kg/m =
GS (1000 kg/m3 ) 1+0.631
G S = 2.61 c.) SATURATED UNIT WEIGHT OF SOIL IN kN/m 3
γsat = γsat =
GS +e γw e+1 (2.61+0.631)(9.81 kN/m3 ) 0.31+1
γ sat = 19.49 kN/m 3
5.) The moist unit weight of is 16.5 kN/m3. If water content is 15% and sp.gr. of soil is 2.7, determine the following: a. Dry Unit Weight b. Void Ratio c. Porosity d. Degree of Saturation e. Mass of water in kg/m3 to be added to reach full saturation
Given: γwet = 16.5 kN/m3 w = 15% GS = 2.7
Solution: e.) MASS OF WATER IN kg/m3 TO BE ADDED TO REACH FULL SATURATION
8.) The moist unit weight of 0.00283 m3 of soil is 54.4 N. If the moisture content is 12% and Gs = 2.72, determine the following:
a. b. c. d. e. f.
Moist Unit Weight Dry Unit Weight Void Ratio Porosity Degree of Saturation Volume Occupied by Water
Given: V S = 0.00283 m3 W S = 54.4 N w = 12% GS = 2.72
Solution: a.) MOIST UNIT WEIGHT
γwet =
W V
=
0.0544 kN 0.00283
GSw = eS (2.72)(0.12) = 0.555S
m3
γwet = 19.223 kN/m 3
S = 0.5881/58.81%
b.) DRY UNIT WEIGHT
γdry =
γwet 1+w
=
19.223 kN/m3 1+0.12
γdry = 17.163 kN/m 3 c.) VOID RATIO
γdry =
GS (γw ) 1+e 3
17.163 kN/m =
2.72 (9.81 kN/m3 )
e = 0.555 d.) POROSITY
n=
e e+1
=
n = 0.357
0.555 0.555 + 1
e.) DEGREE OF SATURATION
1+e
f.) VOLUME OCCUPIED BY WATER VW wGs VW=(0.12)(2.72) V W =0.326=32.6% =
9.) A granular Soil located in a borrow pit is found to have an in place dry density of 1895 kg/m3. In the laboratory, values of dry maximum density and minimum density are determined as 2100 kg/m3 and 1440 kg/m3, respectively. Compute for the following: a. Relative Density in the Borrow Pit b. Void Ratio of the soil in the natural condition if the maximum and minimum densities indicate void ratios of 2.8 at the maximum density and 0.92 at minimum density. c. Specific Gravity of the Soil
⁄⁄ ⁄
Given:
ρ Dry=1895 kg m3
ρ DryMax=2100kg m3 ρ DryMin=1440 kg m3 e Max =0.28 e Min =0.92
Solution: a. RELATIVE DENSITY IN THE BORROW PIT γDry-γDryMin γDryMax Dr= γDryMax -γDryMin γDry
Dr=
⁄⁄ ⁄⁄ ⁄⁄ 1895kg m3 -1440 kg m3
2100kg m3
2100kg m3 -1440 kg m3
1895kg m3
Dr = 0.764 kg m 3 /76.40%
Dr = eMaxeMax − −eMie n ⁄
b. VOID RATIO
0.764 kg m3 =
0.28-e
0.28-0.92
e = 0.769 c. SPECIFIC GRAVITY
γDry= 1895
GsγWater
( )
e+1 kg m3
Gs
=
Gs = 3.352
kg 1000 3 m 0.769+1
10.) The moist unit weight and degrees of saturation of a soil are given in the following table.
⁄
16.64 17.73 Determine the following:
Degree of Sat (S)% 50 75
a. Void Ratio b. Specific Gravity c. Saturated Unit Weight
11.) A sample of sand has relative density of 40% with a solid specific gravity of 2.65. The minimum void ratio is 0.45 and maximum void ratio is 0.97.
a. Void ratio of the sand having a relative density of 40% b. Density of the sand in saturated condition in pcf c. If the sand is compacted to a relative density of 65%, what is the decrease in the thickness of the 4 ft thick layer?
Given: Dr=40%=0.40 Gs=2.65 emin =0.45 emax =0.97
Solution: a. VOID RATIO OF THE SAND HAVING A RELATIVE DENSITY OF 40% eMax-e Dr= eMax-eMin 0.97-e 0.40= 0.97-0.45 e=0.762
⁄ ( )( ) ( ) ( ) ⁄
b. DENSITY OF THE SAND IN SATURATED CONDITION pcf Gs+e γWater 2.65+0.762 9.81kN m3 γSat /ρSat = = e+1 0.762+1
γSat /ρSat = 18.996
kN
1000N
m3
1kN
1
9.81
3
γ Sat /ρ Sat =120.885 lb ft
c. DECREASE IN THE THICKNESS eMax-e Dr= eMax-eMin 0.97-e 0.65= 0.97-0.45 e=0.632 ∆eH ∆H= eO +1 (0.762-0.632)(4ft) ∆H= 0.762+1 ∆H=0.295
1m
m s2
3.28ft
3
2.20lb 1kg
12.) A sample of soil weighing 2.62 kN is removed from a test pit. 1.33 kN of water will fill up the pit. A sample of the soil weighing 112.4 grams is dried in the oven and its weight after is 103.6 grams. Assume the specific gravity of th e solids is 2.67. Max attainable dry unit weight of soil is 18.47 kN/ m 3 and minimum attainable unit weight is 16.35 kN/m3. Determine the following:
a. Wet unit weight of the soil b. Dry unit weight of the soil c. Relative density of soil sample
Given: W S =2.62kN W W =1.33kN m MoistSoil =112.4g m DrySoil =103.6g γ MaxDry=18.47 kN m3
⁄⁄
γ MinDry=16.35kN m3
Solution:
a. WET UNIT WEIGHT WWater γWater = V kN 1.33kN 9.81 3 = m V V=0.136m3 WMoistSoil 2.62kN γMoist = = V 0.136m3 kN γ Moist =19.265 3 m b. DRY UNIT WEIGHT WW 112.4g-103.6g w= = WS 103.6g w=0.085 kN γMoist 19.265 m3 γDry= = 1+w 1+0.085 kN γ Dry =17.756 3 m
c. RELATIVE DENSITY OF SOIL SAMPLE γDry-γDryMin γDryMax Dr= γDryMax -γDryMin γDry
kN kN -16.35 3 3 m m kN kN 18.47 3 -16.35 3 m m
17.756
=
kN m3 kN 17.756 3 m 18.47
Dr=0.6899 or 68.99%
13.) Following are the results of a field unit weight determination test using the sand cone method. Dry unit weight of sand = 16.36 kN/m3 Wt. of sand to fill the cone = 11.15 N Wt. of jar + cone + sand (before use) = 58.9 N Wt. of jar + cone + sand (after use) = 27.65 N Wt. of moist soil from the hole = 32.55 N Moisture content of moist soil = 11.6% Compute for the following:
a. Weight of sand to fill the hole b. Moist unit weight of the soil c. Dry unit weight of the soil
Given: kN γ Dry=16.36 3 m W Sand to fill the cone =11.15N=0.01115kN W jar+cone+sand
before use
W jar + cone + sand
after use
=58.9N=0.0589kN =27.65N=0.02765kN
W moist soil from the hole =32.55N=0.03255kN w=11.6%=0.116
Solution: a. WEIGHT OF SAND TO FILL A HOLE WS =W jar+cone+sand before use -W jar + cone + sand after use -WSand to fill the cone
WS =0.0589kN-0.02765kN-0.01115kN W S =0.0201kN b. MOIST UNIT WEIGHT OF SOIL WS γDry= V kN 0.0201kN 16.36 3 = m V -3 3 V=1.229x10 m WMoist 0.03255kN γMoist = = V 1.229x10-3 m3 kN γ Moist =26.485 3 m c. DRY UNIT WEIGHT OF SOIL kN γMoist 26.485 m3 γDry= = 1+w 1+0.116 kN γ Dry =23.732 3 m
14.) A soil sample has a bulk unit weight of 19.6 kN/m3 at a water content of 10%. Determine the following if it has sp. Gr. of 2.7.
a. Void Ratio b. Percent air in void (air void ratio) c. Degree of Saturation
Given: γwet = 19.6 kN/m3 w = 10% GS = 2.7
Solution: a.) VOID RATIO
γdry =
γwet
=
1
1+w
9.6
kN/m3
1+0.10
γdry = 17.818 kN/m3 γdry =
GS (γw ) 1+e
17.818 kN/m3 =
2.7 (9.81 kN/m3 )
e = 0.487
1+e
b.) PERCENT AIR IN VOIDS
Air Void Ratio = Air Void Ratio =
e-wGS 1+e 0.487-(0.10)(2.7) 1+0.487
Ai r Void Ratio = 0.146/14.6% c.) DEGREE OF SATURATION
GSw = eS (2.7)(0.10) = 0.487S
S = 0.554/55.4%
15.) The hydraulic gradient for quicksand condition is equal to 1.13 with a void ratio of 0.50. The moist unit weight of soil is 19.04kN/m3.
a. Compute the sp. gr. of the soil b. Compute the dry unit weight of soil c. Compute the water content of the soil
Given: i = 1.13 e = 0.50 γwet = 19.04 kN/m3
Solution: a.) SPECIFIC GRAVITY OF SOIL
i=
GS -1
G0.5S011
e+1
1.13 =
-
G S = 2.695
b.) DRY UNIT WEIGHT OF SOIL
γdry =
GS (γw ) 1+e
=
2..+.8/
γdry = 17.625 kN/m 3 c.) WATER CONTENT OF SOIL
γdry =
γwet 1+w
17.625 kN/m3 =
.+/
w = 0.0802/8.02%
COLLEGE OF ENGINEERING AND ARCHITECTURE DEPARTMENT OF CIVIL ENGINEERING
CE 423
ASSIGNMENT IN SOIL MECHANICS (PROBLEM SET NO. 2)
SUBMITTED BY: ANGELICA E. CABRERA 4CE-D/T 10:00 AM 1:00 PM –
ENGR. RANDY R. AUREADA Instructor
COLLEGE OF ENGINEERING AND ARCHITECTURE DEPARTMENT OF CIVIL ENGINEERING
CE 425
ASSIGNMENT IN TRANSPORTATION ENGINEERING
SUB MI TTE D BY: ANGELICA E. CABRERA
SUB MI TTE D TO: ENGR. ANDILON ALBERT D. LODRIGA 4CE-A/T 6:00 PM 9:00 PM –
