An Introduction to Geotechnical Engineering 2ED

AN INTRODUCTION·.TO .· GEOTECHNICAL ENGINE.ERING Second Edition

Robert D. Holtz, Ph.D., P.E., D.GE University of Washingt()n

William D. Kovacs, Ph.D., P.E., D.GE . University of Rhode Island ..

·

Thomas C. Sheahan~ Sc.D.,' P.E. Northeastern University ·

·

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Library of Congress Cataloging-in-Publication Data Holtz, R. D.. (Robert D.) An introduction to geotechnical engineering I Robert D. Holtz, William D. Kovacs, Thomas C. Sheahan.-- 2nd ed. p.cm. Includes bibliographical references and index. ISBN 978-0-13-031721-6 (alk. paper) 1. Soil mechanics--Textbooks. 2. Rock mechanics--Textbooks. I. Kovacs, William D. II. Sheahan, Thomas C. III. Title. TA710.H564 2011 624.1'51--dc22 2010028254

Prentice Hall is an imprint of 10

PEARSON

9 "8 7

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·rsB~-13: 978-d-13-249634-6 · www.pearsonhighered.com

ISBN~10:

0-13-249634-8

(p 7.-t.( •151 3

H7Lf5.i ~t'J

I),

Contents Preface Chapter 1

viii

Introd~ction to:,)

J.!

(2.13)

,.

p

Pd

=

1: w

1.88 = ~ 1.15

1:63- Mg/m3

(2.14)

You should verify that Pd = p/(1+'-·w), which is another very useful relationship to -· · ·' remember. •

I'

.

: •

~ ... •' ·, • :

c. Water content for S ='100%: . ·. . From Eq. (2.4), we know that Vw .::: Vv = 0.62 m3. From Eq. (2.8), Mw = VwPw = 0.62 3 (1 rvlgh:~3 ) -~• 0.62 Mg. Therefore w for s,=: ·lOOo/o_ must be ·· · .

mx

.· . Mw .0.62 W(s=lOO%) = Ms = . = 0.234 or 23.4% 2 65

2.3 d. Psat:

Solution of Phase Problems

21

From Eq. (2.10), we know Psat = (Ms + Mw)fVr, or ~ (2.65

Psat-

+ 0.62) Mg _ 1.62m3 .

-

2.019 or 2.02 Mg/m

3

Check, by Eq. (2.13): _ Ps(1 + w) _ 2.65(1 + 0.234) _ ·• ' 3 Psat + e 1.62 - 2.02 Mg/m 1

Example 2.5 Given: The definitions of the degree of saturation S, void ratio e, water content w, and the solid density Ps [Eqs. (2.4), (2.1), (2.5), and (2.7), respectively]. L

'',

Required: Derive a relationship between s; e, w, and Ps·

Solution: Look at the phase diagram with V.· = 1 (Fig. Ex. 2.5): Mass

Volume

v

t=

A e

!

f w t

V.

= Se

v. = 1

st

w

s

f

M =w

+ Ms

= PsV.s

i

FIGURE Ex. 2.5

From Eq. (2.4) and Fig. Ex. 2.5, we know that Vw = SVv = Se. From the definitions of water content [Eq. (2.5)] and Ps [Eq. (2.7)], we can place the equivalents forMs and Mw on the phase diagram. Since from Eq. (2.8), Mw = PwVw, we now can write the following equation:

Mw = PwVw = wMs

==:

WPsV.

or

PwSe = WPsV. Since V. = 1 m3, · (2.15) Equation (2.15) is among the ~ost useful of ~11 equations for phase pioblems. You can also verify · · its validity from the fundamental definitions of Pw• S, eiw, and Ps·

22

Chapte~2

Index and Classification Properties of Soils

Note that, using Eq. (2.15), we can write Eq. (2.13) another way: PwSe) 'Ps ( 1 + .· Ps P= · l+ e

Ps

+ PwSe +e

(2.16)

1

When S = 100%, Eq. (2.16) be~omes Psat =

Ps+'pwe. 1+e

(2.17)

Example 2.6

Given: A soil contaminated with gasolih~·(specific gravity = 0.9) with the follo~ing characteristics: Ps = 2.65 Mg/m3, w = 25%, volume of the gasoline is 20% of the volume of the water, and 85% of the void space is filled with gasoline and water (after T. F. Wolff). Required: Complete the phase diagram in Fig: Ex. 2.6a. Find the void ratio and porosity of the specimen. c. Find the total and dry density of the specimen. IL

b;

Volume (m 3)

Va=-.

Vv

= _

i.

r-

V9

= -·-·-.

v =

Vr=_·

w .. . -

Vs= - ~----------------J-----

Ps(Mg/m 3 )

Air

.Mass(Mg)

I

~~;:;·,·.~~~-.=~:~-,.:.~;,·.;;:;·,·.~~;~·.; ~M g = __

·:;·;~-:-:v:Gasoline';';'•.,:.;v':•,'

?~:;~~~:;~~~~~ ~

=-=-::water=-=-=-=

M

=

:~~~f,~~~;(.f~;~. ~......................,...... .· I .

/~~f:-~~;(~~:\~~~~;;~t;{\;~/K~~-

}~=r/tt:/~~~:;/W/K:~=:t?;:~

Ms

=l-

M,=_'

l

.

FIGURE Ex. 2.6a

Solution: IL

As we did with Examples 2.3 and 2.4, assume V5 = 1m3• Then; using the basic definitions for water content, deilsity, and degree of saturation, fill in the bla~ks as shown in Fig. Ex. 2.6b.

1

2.3 Solution of Phase Problems

23

Mass,(Mg)

Vv = 0.93

M1 = 3.43.

FIGURE Ex.. 2.6b ' 1 '

. (

'~

::

b. Again, use the basic definitions of e and n. We find that e = 0.93 and n = .48.2%. c. For both pi a~d Pd; ~imply take the values for M1 (3.43 Mg) and Ms (2.65 Mg) and divide each by 1.93 m3 to obtain p 1 = 1.78 Mg/m3 and Pd = 1.37 Mg/m3. Note that using Eqs. (2.13) and (2.14) will give you erroneous results. Here are the details: Calculate the mass of water by noting that Mw = wMs = 0.25 X 2.65 Mg of solids. (We assumed that Vs = 1m3, remember?) So, Mw = 0.66 Mg. Add that to the phase diagram. Also, the volume of water is MwiPw = 0.66 Mg divided by 1 Mg/m3, or V. = 0.66 m 3. Then the volume of gasoline = 20% of Vw ,;,; 0.2 X 0.66 m3 = 0.13 m3. Because the specific gravity (Sec. 2.3.2) of gasoline is 0.9, its density is 0.9 X Pw· So the mass, oLthe. gasoline Mg = 0.9 X Pw X Vg = 0.9 Mg/m3 x 0.13 m3 = 0.12 Mg. Add these items to the phase diagram (Fig. Ex. 2.6b). Because 85% of the voids are filled with water and gasoline, the ,total amount of voids is (Vw + Vg)/0.85 = (0.66 + 0.13)/0.85 = 0.93 m3. · Subtracting Vw + Vg from V,, we find that Va = 0.14 m3. Now all the "holes" on the phase diagram are filled. The rest is a piece of cold apple pie , , ' (parts b and c). 1

e

= Vv = 0.93m: = 0.93 .V.

LOOm 0.93 · n =- =X 100 = 48.2% v, '1.93 M1 3.43Mg Mg Pt = V, = 1.93m3 = 1.78 m3

Yv

Ms

,,

Pd =

V,

2.65Mg

=: 1.93 m3

;=:

Mg 1.37 m3 .

In summary, for the easy solution of phase problems, you don't have to memorize lots of complicated formulas. Most of the formulas you need can easily be derived from the phase diagram, as illustrated in the preceding exainples. Just remember the following simple rules: 1. Draw a phase diagram. 2.· Remember the basic definitions of w, e, p;, S, and so on.

24

Chapter 2 ·. · . Index and Classification Properties of Soils

3. Assume either V, = .1 or Vr = 1, if no masses or volumes are given. 4. Write out equations in symbol form. Then,place the numerical value along with its units in the same order, and solve. '· 5. Check units and reasonableness of your answer. · . ' : ·>: . ._,._ ' ... ·..-

2.3.1

Submerged or Buoyant Density In Eq. (2.11),we simply defined the submerged or buoyant density as p = 'Psat- Pw• without any explanation other than giving some typical values p in Table '2.l.Strictly speaking, the total p should be used instead of Psat in Eq.{2.11), but in mostcases' submerged soils are also completely saturated, or at least that is a reasonable assumption.::· ...·, • . , : • _So when ·a.· soil is submerged, the total density as expressed hy Eqs. (2.13) and (2.16) is partially balanced by the --buoyant effect 'of the water. You will recall from Archimedes' principle thai the buoyancy effect is equal to the weight of M,g water displaced by the solid particles in the soil mass. This is shown in Fig. 2.3(a), where the submerged (net) weight is 1

of

1

-¥-rh"' ~

v,TF:~v.M

.

·'

w' = Ws __:.· Fb .

FIGURE 2.3a Free-body diagram of a submerged soil particle. ·

'

'

· In terms of masses, W 1 = Msg- V,pwg ~ M 1 g ", i>

Thus, the submerged (net) mass

.

'Jc

;,•,·

·~ ~

:

';:.::....

M -

.....:,·

.

1\!s- V.pw

We obtain de'nsities by dividing by the total volume Yt: :·. " •

~

' • •

_! • •

M! · .Ms .. Vs -.=p = - - - p 1 '

, Vr

Becauseps=

· . ·· Vr

'Vr ~ 1

Ms!Vs[andu~i~~Eq.(2:i2)];weobtain;;" 1

P

Ms

•

· ·

Ms Pw

:.V~s~ (Vf.Plp:)P: 1:+ e

I';,

(

p,;;) 1-

p;.

Ps (2.18)

Ps- P:.V · 1+e There are several oth~r ways to get Eq: (2.18). One way is to use Eq. (2.13). '~

••

<

PI

=·

ps(1 + w) 1 + e ' - Pw \:

· Using Eq.'(2.15), PsW =: Pwe(S ~ 100% ), · ·

• 1

··'ps +·pwe .,-:_ Pw. -:-·Pwe ·.

p= p

1

';t

=

·

Ps- Pw 1 +e

1+e (2.19)

'

1

: 2.3

Solution of Phase Problems

25

Buoyant mass

Saturated mass

Marbles in bucket with holes in it, dipped · into water.

FIGURE 2.3b .. Schematic showing the relevant change in mass due to buoyancy.

Note carefully the different meanings of the densities described above. The saturated density is · · the total soil and water density whenS = 100%, while the submerged density is really abuoyant or an effective density. Note, too, that the difference between the saturated and: submerged densities is exactly the density of water [Eq. (2.11)]. Some physical examples will help you understand the concept of submerged or buoyant density. First, consider a bucket full of marbles; the relevant density is, of course, the dry density. Then fill the bucket with water, and the relevant density is Ps~t· If the marbles are now placed in a bucket that has numerous holes in it so that water can move freely into and out of the bucket submerged in a tank of water, then the correct density of the marbles is the submerged or buoyant density p' [Fig. 2.3(b )]. If we remove the bucketfrom the tank but keep it completely saturated, then the appropriate density is again Psat• A second, more realistic example is the case. ()f rapid drawdown, which occurs when the water level in a reservoir, canal, or river is quickly lowered. The result is that the density ofthe ~oils in the adjacent dam or slope increases from submerged or buoyant to saturated. This is a critical case for the stability of the dam or slope; because the gravitational forces acting on the embankment approxi.. ·mately double in magnitude. Therefore the factor ofsafety against slope instability is approximately cut in half. See Table 2.1 for typical values of Ps~t and p'; · d·

','

'

Example 2.7 Given:

. , A silty clay soil with Ps = 2700 kg/m3, S = 100%, and water content w = 46%. Required:

Compute the void ratio, the saturated density, and the' buoya~i

~r submerged density in kg/m3•

Solution: Place the given information on a phase diagram (Fig. Ex. 2.7).

26

Index and Classification Properties of Soils

Chapter 2

Mass (kg)·

Volume (m 3) ·

t

w

Vv =: fw= 1.24

~ ..

~

-

r

= 1242

.

~ -- - -

T

v. = 1.0

s

L

.S = 100%

FIGURE Ex. 2.7

t

Mw

M5

= 2700

_j_

'

Assume Vs = 1m3; therefo~e Ms = V.Ps =:2700 kg. From Eq. (2.15a),we can s?lve fore directly: WPs

e= -

=

PwS

0.46 X 2700 kg/ni.J . . = 1.242 1000 kg/m3 X 1.0

But e also equals Vv, since V,: .= 1.0; likewise Mw_:';. 1242kg,.since Mw is numerically equal to Vw, because Pw = 1000 kg/m3. Now t)lat all the unknowns have been .found, we may readily_ calculate the saturated density [Eq. (2.19)]: .

+ Ms _ (1242 + 2700)kg·_ ' · . _3 - 1758 kg/m ·v; · ., 1 + e . . . > , ( L + 1.24) m3 .

_ M1

Mw

_

Psat - - -

.

We c~uld. also use Eq.{2.i7) directly; •. . . ' - ,., ·. · -~ Ps Psat-

' ·· ·

+ Pwe ~ [2700 +1000(1.242)] kg _. • ,. '3 .+ e -:- . (1 + 1.242 )m3 -. -1758kg/m

1

The buoyant density p' from Eq. (Z.i1).is: -,; ' ; . _· ' ;

~'

'

~

;

,

· p' = Psat :.:. ·'

~ ' ; . ' ,.' '

Pw ,;; 175S kg/rri3 :_

·-

1000 kg/m3 ~ 758 kglm

3

In this example, p'•is less than the density of water. Go back and look at Table 2.1 for typical values of p'. The submerged or buoyant density of soil will be very important later on in our discussion of consolidation, settlement, and strength properties of soils.

2.3.2

Unit Weight and Specific Gravity· In geotechnical practice it is oft~n con~enient to use unit weight rather than density in engineering calculations. Unit weight 'Y is simply weight per unit volume; thus its SI units are N/m3, because the 3 Newton (N) is the SI unit of force, and in British units it is typically expressed in lb/ft , sometimes abbreviated pcf. Recall that the weight of an object is d~e to the force exerted by the earth's gravitational field, or W = mg, where g is the acceleration due to gravity. Then to get unit weight 'Y we simply divide the weight by the unit volume V, or: · '•

'

.

•

~ •

'

'

':0 •

•. j

C

•

'

'

W

'}' -~ v

L.

•• • C

-m ~ g

v

~

.=

pg

(2.20)

i

I

I

L

2.3

Solution of Phase Problems

27

As noted in Appendix A, the value of g varies slightly with latitude and elevation, but for ordinary engineering purposes we usually assume it is a constant (standard g = 9.807 m/s2) for most places on the Earth. The unit weights analogous to the ~ensities described earlier [Eqs. (2.6), (2.7), and (2.8)] are the total, wet, or moist unit weight y, the unit weight of solids y s, and the unit weight of water y w. In terms of the basic weights and volumes, they are'

Wr

y=-

(2.21)

v; w. Ys = V. ' s

(2.22) (2.23)

To convert between density and unit weight, use Eq. (2.20) and g ·= 9.81 m/s2• If you round off g to 10 m/s2, the error,is only 2%.; ·

-Example 2.8

Given:

The densities p., Psat• and p' in Example 2.7. ','

'Required:

,.

Compute the equivalent unit weights, using a. SI units, b. British engineering units.

Solution: a. SI units: From Example 2.7: Ps = 2700 kg/m3 = 2.7 Mg/m3 3 Psat = 1758 kg/m = 1.76 Mg/m3 3 p' = 758 kg/m = 0.76 Mg/m3

From Eq. (2.20), y = pg, we obtain: . Ys

= 2.7 Mgim3 X 9.81 m/s2 = 26.5 kN/m3 (Note: 1 kg

1.76 Mg/ni3 y' = 0.76 Mg!m3

Ysa; =

X X

9.81 m/s 2 = 17.2 kN/m3 9.81 m/s2 = 7.4 kN/m3

X m/s~

= 1 N)

'

·

Ify~~ use the rounded-off value of the values of Ys• Ysat• and y' are, respectively, 27, 17.6, and 7.6 kN/m3• As mentioned, the difference is only ·about 2% and normally negligible. - ., - . ' .' - ' ' . g = 10 m/s2/

28

Chapter 2

Index and Classification Properties of Soils

b. British engineering units: To convert the unit weight from kN/m3 to lbf/ft3 (pcf), merely multiply the value in 3 kN/m by 6.366. Or 'Ys = 26.5kN/m3 = 168.7 pcf 'Ysat = 17.2 kN/m3 = 109.5 pcf . 'Y' = 7.4 kN/m3 = 47.1 pcf .

.'

.

Note that the product of 6.366 and g (9.807) equals 62.43 or 62.4lbf/ft3, the unit weight of water in British engineering units. (If the rounded-off value of 10 m/s2 is used, the conversion factor is slightly more or 63,7.) So to convert directly from density in.Mg/m3 to unit .. weight in lbf/ft3, just multiply by 62.4! If you need some practice converting between density and unit weight and between the various systems of units, review the ·examples in Appendix A

In summary, to convert density in Mg/m3 to unit weight in kN/m3, multiply by 9.8lor 10. To convert density in Mg/m3 to unit weight in lbf/ft3 (pcf), multiply by 62.4. To help you get a feel for the magnitude of unit weights in both SI and British engineering units, we have converted the typical densities in Table 2.1 to unit weights in Table 2.2 in terms of kN/m3 and pcf. You may recall from physics that the specific gravity G of a substance is the ratio of its unit weight 'Y to the unit weight of water, usually pure water at 4°C (symbol: 'Yo), or ,G = _!_ 'Yo'

(2.24)

Although several different specific gravities can be defined, only the bulk specific gravity Gm, the specific gravity of solids G., and the specific gravity of water Gw are of interest in geotechnical engineering. These are defined as ' ·' ' · ' ' Gm = _!_ 'Yo

(2.25)

TABLE 2.2 . Some 1)rpical Values for Different Unit Weights of Common Soil Materials in Units of kN/ni3 and pcf Unit Weight

Soil'JYpe Sands and gravels Silts and clays Glacial tills Crushed rock Peats ' ·Organic silts an~. clays

y'

'Yd

'Ysat

kN/m3

pcf

kN/m3

pcf

kN/m3

pcf

19-24 14-21 : 2f-24 19.:.22 10-11 .. ,13..:.18

119.:_150 .. 87-131 131-150 119-137. 60-69 81:_112

15.,-23 6-18 17-23 15-20 1.C.3 5-15

94-144' 37-112 106-144 94-125 6-19 31-94

9-14 4-11 11-14 9-12 0-1 3-8

62-:-81 25-69 69-87 56-75

Note: Values are rounded to the nearest 1 kN/m3 and 1 pcf.

0-6

19-50

2.3

Solution of Phase Problems

G = 'Ys s : 'Yo '

':'

.

Gw

29

(2.26)

')'i/;

(2.27)

=----::

'Yo

Because the density and therefore the unit weight of water are a maximum at 4°C, the specific gravity of water is exactly 1.0000 at that temperature. Because the value of Gw ranges between 0.9999 at ooc and 0.9922 at 40°C, it is sufficiently accurate for most geotechnical work to assume Gw = 1.00 and 'Yw ;::j 'Yo = constant. Note that specific gravity is a dimensionless quantity and its numerical values are similar to what we used for d~nsities in Mg/m3• For example; the specific gravity of solid quartz (that is, if we could ~reate a piece that had no void space) is 2:65, and typical values for most soils range from 2.60 to 2.80. Organic soils will have lower specific gravities, while heavy metallic minerals may occasionally have,higher:values. , · ' If you need to determine the specificpayity of a soil, use ASTM (2010) standard D 854.

Example 2.9 Given:

A sample of soil has a bulk specific gravity of 1.91 and · water content of29% (after Taylor, 1948).

aspecific gravity of solids of2.69, and a ··

Required:

Determine the void ratio, porosity, degree of saturation, and the dry density of the sample in a . . British engineering units and b.SI units. ,

Solution: As before, when the size of the sample is not given, assume any convenient weight or volume. , For the SI case, let's assume the total volume V =. 1 ft 3, and for part b, assume V = 1m3• Draw the ·, : · , · phase diagram for each' case. a. British engineering units: FromEqs. (2.25) and (2.21); Gm = Wrl~yw; so Wr = Gm~'Yw = (1.91)(1ft3 )(62.4lbf/ft3 ) = 119lbf. From the definition of water content, we know that Ws + 0.29Ws = 119lbf. There. .. ; , , fore, Ws = 92 lbf and W ~ = 27 lbf. FromEqs. (2.26) and (2.22), V, = WsfGs'Yw; so 92lbf/(2.69)(6Z.4lbf/ft3 ) = 0.55 ft 3 . From Eqs. (2.27) and (2.23), Vw = WsfGw'Yw; so 27lbf/(1.0)(62.4lbf/ft3 ) = 0.43 ft 3 • Figure Ex. 2.9a is the completed phase diagram for·part a of this example. Therefore, the answers are e =. 0.82; n = 45%; S = 96%; and I'd = 92 pcf. Weight (lbf)

I

= 1 ft3

.

+

. V8 =!0.02

.

t

Vw = 0.43

j v,~

A

w s

~

t27

+ w,~o2

1

'Ym = 11 9 pet

j FIGURE Ex. 2.9a

30

Chapter 2 ·

Index and Classification Properties of Soils

b. S.l. units: . . . 3 The solution for part b is basically the same as for part a, except you use Vr = 1 m3 and 'Yw = 10.0 kN/m3 • The 'a~wers for e, n, and S are identical, and Pd = 1.48 Mg/m • See Fig. Ex. 2.9b for the completed . phase diagram..

r· =r02, ··

-Mass(Mg)·

,; Volume (m 3 )

_'" .f _. . -: ,1·. · ·

A.

_ ., :: Va

. . .... . .

1\Tm'v·~ •• v:__;;,o.5_5·

w

~f

...

· M.·~1_.:4a ..

s

<

'·''

/

,.

'f

·,·,_. · '·

•'

·,"

'

''

G5 ~

2.69

FIGURE Ex. 2.9b

Example ·· · '

2~10

· Given; ·

' i

~

' '

Equations (2.12), (2.15), and (2.16) . •Required:

·;··

,;!

Develop the corresponding relationship for these equations in terms of unit weights and specific gravity of solids. · ·. · · ·

· ··::

Solution: Use Eq. (2.20) ( 'Y = pg) and the appropriate definitions for specific gravity of solids'and water. From Eq. (2.12): Ps Pd = 1 ~Fe ';:

'{

"

i

·,:

' / ·'

Substituting, ,·

. 'Yd'

~- 'Ys g

-=--

·:g . J+e or

'

y; .. -

.

.

'Yd = l+ e = Gsywf(1 +.e)

From Eq. (2.15):

pwSe =.wp, . ;J.

'Yw Se·= W 'Ys = g .. 'g .

WGs'Yw ·g~·

(2.28)

2.4

Soil Texture

31

Therefore, (2.29) For Eq. (2.16),

~-:u;s~

.,. ,,_. 'Ys

!

= g

. g

1

g

+e

· · Therefore, (2.30)

. " ... u~i~g J,ro.cedures simila~·t~t·li~s~ shown in :Examplez.io, you can.·~e~dily develop the corresponding relationships for Eqs. (2:13), (2.14),(2.17);and (2:18) in terms of unitweights and specific . griwity of solids. nie corresponding equations, Eqs. (2.31) through (2.34), are giveninTable 2.3, as . .. . .. '. . : . . ' . . . are those developed in the above example. '

,

,

,

'

'

-

'.

. ..

J

,.

'

.I

', ,_;,- ~.

,,

\ ..-;_: .-· ,t'

TABLE 2.3

Corresponding Equatio!lsShowing Density and UnityWeig~~~elatio11ships

Equation (2.13) . (2.14). (2.12)

-I.

(2.17) (2.18)

2.4

Equati~n

Density

,•

· ps(1 + w) p= 1+e p .. Pa= 1+ w Ps Pa = 1 +e. p, + Pwe Psat = 1+e ·ps-Pw 1 p=~

(231)

.

(2.32) ,, . (2.28)' (2.33) ..

. Gs(1 +w) y= 'Yw 1+e " 'Y :.. ·- .: 'Yd = 1 + w 'Ys 'Ya = 1 + e G, + e 'Ysat=~,"Yw

~.

(2.34)

'Y

'

G :_·1 s

=~'Yw

SOIL TEXTURE So far we haven't said much about what makes up the "solids" part of the soil mass~ In' Chapter 1 we gave the usual definition of soil from an engineering point of view: the relatively loose agglomeration of "mineral and organic materialii found above the bedrocK. briefly described how weathering ·and other geologic processes act on the rocks at or near the earth's surface to form soil. Thus the solid part of the soil mass consists primarily of particles of mineral and organic matter in various sizes and amounts... The texture of a soil is its appearance or "feel," and it" depends on the relative 'sizes and shapes of the particles as well as the range or distribution of those sizes. Thus coarse-grained soils such as sands . or gravels obviously appear coarse textured, while a fine-textured soil is composed mainly ofvery small mine.ral grains invisible to the naked eye. Silts and clay soils are good examples of fine-textured soils.

We

32

Chapter 2

Index and Classification Properties of Soils

The texture of soils, especially of coarse-grained soils, has some relation to their engineering behavior. In fact, soil texture has been the basis for certain soil classification schemes, although these are more common in agronomy than' in geotechnical engineering. Still, textural classification terms (gravels, sands, silts, and clays) are useful in a general sense in geotechnical practice. A convenient dividing line is the smallest grain that is visible to the naked eye. Soils with particles larger than this size (about 0.075 mm) are called coarse grained, while soils finer than the size are (obviously) called fine grained. Sands and gravels are coarse grained while silts and clays are fine grained. For fine-grained soils; the presence of water greatly affects their engineering response-much more so than grain size or texture alone. Water affects the interaction between the mineral grains, and this may affect their plasticity (roughly defined as the soil's ability to be molded) and their cohesiveness (its ability to stick together). While sands are tionplastic and noncohesive (cohesionless), clays are both plastic and cohesive. Silts fall between clays and sands: they are fine grained yet nonplastic and cohesionless. These relationships as well as some ge11eral engineering characteristics are presented in Table 2.4. , Note that the term clay refers both to specific minerals called clay minerals (discussed in Chapter 4) and to soils which contain clay l:ninerllis. The behavior of some soils is strongly affected by the presence of clay minerals. In geotechnical engineering, for simplicity, such soils are usually called clays, but we really mean soils that contain enough clay minerals to affect their engineering behavior. ' ' 'n is a good idea to get some pnictice'identifying soils according to tex'ture and other general , characteristics, such as plasticity and cohesiveness. This process is best done i~ the laboratory, and in fact ASTM standard D 2488 provides an excellent guide for 'describing and identifying soils visually and manually. Visual-manual description of soil is also mentioned ~hen we· discuss soil classification later in this chapter.

TABLE2.4

Soil Name

Grain size

Characteristics

Effect of water (m . · engineering behavior Effect of grain-size distribution on

2.5

Gravels, Sands

Silts

Clays

Coarse grained Can see individual grains by eye,

Fine grained Cannot see individual· grains Cohesionless Nonplastic Granular Impor.tant

Fine grained Cannot see individual grains Cohesive Plastic

Relatively · , , unimportant . ,,; ,,

Relatively unimportant

Cohesionless Nonplastic Granular Relatively unimportant (exception: loose, saturated granular materials and dynamic loadings) · Important

Very important

' GRAIN SIZE ANb GRAINSIZE DISTRIBUJI