Introduction to geotechnical engineering 2nd edition das solutions manual 1

Introduction to Geotechnical Engineering 2nd Edition Das Solutions Manual

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Chapter 5

Therefore, M, = 2 2727 kg and M,,= 0 2273 kg

When raised to moisture content of 16.0%: M,= 0.16x2.2727 = 0.3636 kg

Therefore, quantity of water that has to be added:

0 3636 -0 2273= 0 1363 kg =0.136 kg

The plot of y vs w is shown. From the graph, the maximum dry unit weight = 109.5 I/ft' and the optimum moisture content= 16%.

The plot ofy%a vs w is shown

5.6 Refer to the following table.

w b y = ,w(%) 100

The plot of y vs w is shown. From the graph, the maximum dry unit weight = 100.8 Ib/ft and the optimum moisture content = 15%.

5.7 For w = 5.0%, mass ofthe moist specimen= 1850 g and the volume is 1000 ml.

Therefore, moist unit weight, _ I850,9 81 =18 15 kN/m° 1000

nm Ea. @.12 Dry nit weight a-, ',},-11.2av#a

From Eq. (3.20): Void ratio, % 2.68x9.8l 0.521

From Eq (3 21): Degree of saturat.on, S

Repeating these steps for all six moisture contents, the following table can be developed

a From the figure,

b. The variation ofvoid ratio and degree of saturation against the moisture content is shown in the following figure

c. From the figure in Problem 5.7b, at optimum moisture content, void ratio= 0.40

Degree of saturation= 0 66 or 66%

5 8 Eq (5 7): D = [ Yd(field) -yd(min) ][ Yd(max) ] ' las9) 7la(mnts)

D, 0.64, so

0.64 = [ Yd(field) - 97 ][ 116.5 ]; Yd(field) = 108.6 1b/f? 116.5- 97 yd(field)

From Eq (5 6): R(%)= 7a«ta) ,100 a(max) 1016 8 6 5,00 =93.2%

5.9 In the field, sand used to fill the cone = 16.733 - 10.538= 6.195 lb

Sand used to fill the hole = 6 195 - 1.202 = 4 993 lb

5 10 From Figure 5 23, the optimum moisture content= 16 0% and maximum dry density = 17 15 kN/m' Specifications require that:

a. 16.0% < wme1 < 19.0%, and

b Ye1) > 0 95 17 15 (=16 30) kN/m

Thus, the control test measurements must fall within the shaded zone in the figure to meet the specifications

Control Test 1: Lies to the left of the shaded zone. Dry unit weight is adequate, however, the moisture content is too low (i.e. too dry). Does not meet the specifications.

Control Test 2: Falls within the shaded zone and meets the specifications.

Control Test 3: The dry unit weight is adequate (in fact, well in excess ofthe required minimum value of 16.3 kN/m), but the water content is slightly high and falls outside the range. This does not meet the specifications. Another problem is that the control test point lies to the right ofthe zero air void curve, which is not possible theoretically Either the control test measurement is incorrect or the assumed G, value is incorrect This has to be checked A larger G, value would shift the zero air void curve upwards.

Control Test 4: Lies within the shaded zone and, hence, meets the specifications.

5 11 From Figure 5 24, the optimum moisture content= 10 0% and the maximum dry unit weight= 18.5 kN/m. Specifications require that:

a. 8 5% < we1< 11 5%

b Yaei)> 17 6 kN/m? (i e 0 95x 18 5 kN/m')

While the dry unit weight ofthe compacted soil meets the specification, the moisture content lies outside the specified range (i e the soil is too wet) It does meet the specification.