BASIC DEFINITIONS AND TERMINOLOGY OF SOILS

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Gwendolyn Pierce
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1 1 BASIC DEFINITIONS AND TERMINOLOGY OF SOILS Soil i a thr pha atrial hich coit of olid particl hich ak up th oil klto ad void hich ay b full of atr if th oil i aturatd, ay b full of air if th oil i dry, or ay b partially aturatd a ho i Figur 1. Solid atr Air Figur 1: Air, atr ad Solid pha i a typical oil It i uful to coidr ach pha idividually a ho i Tabl 1. Pha olu Ma ight Air A 0 0 atr M Solid S M S S Tabl 1 Ditributio by olu, Ma, ad ight 2.1 Uit For ot girig applicatio th folloig uit ar ud: Lgth tr Ma to (1 to kg) Dity (a/uit volu) t/ ight kiloto (kn) Str kilopacal (kpa) 1 kpa 1 kn/ 2 Uit ight kn/ To ufficit accuracy th dity of atr ρ i giv by ρ 1 to/ 1 g/c I ot applicatio it i ot th a that i iportat, but th forc du to th a, ad th ight,, i rlatd to th a, M, by th rlatio M g hr g i th acclratio du to gravity. If M i aurd i to ad i kn, g 9.8 / 2

Pha olu Ma ight Air A 0 0 atr M Solid S M S S Tabl 1 Ditributio by olu, Ma, ad ight 2.

2 2 Bcau th forc i uually rquird it i oft covit i calculatio to u th uit ight, (ight pr uit volu). M g ρ g Hc th uit ight of atr, 9.8 kn/ 2.2 Spcific Gravity Aothr frqutly ud quatity i th Spcific Gravity, G, hich i dfid by G Dity of Matrial Dity of atr ρ ρ G Uit ight of Matrial Uit ight of atr It i oft foud that th pcific gravity of th atrial akig up th oil particl ar clo to th valu for quartz, that i G 2.65 For all th coo oil forig iral 2.5 < G < 2.8 ca u G to calculat th dity or uit ight of th olid particl ρ G ρ G ad hc th volu of th olid particl if th a or ight i ko M G ρ G 2. oid Ratio ad Poroity

atr It i oft foud that th pcific gravity of th atrial akig up th oil particl ar clo to th valu for quartz, that i G 2.

3 Uig volu i ot vry covit i ot calculatio. A altrativ aur that i ud i th void ratio,. Thi i dfid a th ratio of th volu of void, v to th volu of olid,, that i v hr v + a a + + A rlatd quatity i th poroity,, hich i dfid a ratio of th volu of void to th total volu. v Th rlatio bt ad ca b dtrid by otig that No - v (1 - ) v v ( 1 ) 1 ad hc Dgr of Saturatio Th dgr of aturatio, S, ha a iportat ifluc o th oil bhaviour. It i dfid a th ratio of th volu of atr to th volu of void S + a Th ditributio of th volu pha ay b xprd i tr of ad S, ad by koig th uit ight of atr ad th pcific gravity of th particl th ditributio by ight ay alo b dtrid a idicatd i Tabl. S v S

total volu. v Th rlatio bt ad ca b dtrid by otig that No - v (1 - ) v v ( 1 ) 1 ad hc 1 + 2.

4 4 a v - (1 - S) Pha olu Ma ight Air (1 - S) 0 0 atr S S ρ S Solid 1 G ρ G Tabl 2 Ditributio by olu, Ma ad ight i Soil Not that Tabl 2 au a olid volu 1, All tr i th tabl hould b ultiplid by if thi i ot th ca. 2.5 Uit ight Svral uit ight ar ud i Soil Mchaic. Th ar th bulk, aturatd, dry, ad ubrgd uit ight. Th bulk uit ight i iply dfid a th ight pr uit volu bulk h all th void ar filld ith atr th bulk uit ight i idtical to th aturatd uit ight, at, ad h all th void ar filld ith air th bulk uit ight i idtical ith th dry uit ight, dry. Fro Tabl 2 it follo that bulk G + S ( G + S) at ( G + ) 1 + S 1 dry G 1 + S 0 Not that i dicuig oil that ar aturatd it i coo to dicu thir dry uit ight. Thi i do bcau th dry uit ight i iply rlatd to th void ratio, it i a ay of dcribig th aout of void. Th ubrgd uit ight,, i oti uful h th oil i aturatd, ad i giv by 2.6 Moitur cott at -

Th bulk uit ight i iply dfid a th ight pr uit volu bulk h all th void ar filld ith atr th bulk uit ight i idtical to th aturatd uit ight, at, ad h all th void ar filld ith air th bulk uit ight i

5 5 Th oitur cott,, i a vry uful quatity bcau it i ipl to aur. It i dfid a th ratio of th ight of atr to th ight of olid atrial If xpr th ight i tr of, S, G ad a bfor obtai S ad hc G S G Not that if th oil i aturatd (S1) th void ratio ca b iply dtrid fro th oitur cott. Exapl Ma ad olu fractio A apl of oil i tak uig a thi alld aplig tub ito a oil dpoit. Aftr th oil i xtrudd fro th aplig tub a apl of diatr 50 ad lgth 80 i cut ad i foud to hav a a of 290 g. Soil triig cratd durig th cuttig proc ar ighd ad foud to hav a a of 55 g. Th triig ar th ov drid ad foud to hav a a of 45 g. Dtri th pha ditributio, void ratio, dgr of aturatio ad rlvat uit ight. 1. Ditributio by a ad ight Pha Triig Ma (g) Sapl Ma, M (g) Sapl ight, Mg (kn) Total Solid atr Ditributio by olu Sapl olu, π (0.025) 2 (0.08) atrolu, Solid olu, G

$oitur cott. Exapl Ma ad olu fractio A apl of oil i tak uig a thi alld aplig tub ito a oil dpoit. Aftr th oil i xtrudd fro th aplig tub a apl of diatr 50 ad lgth 80 i cut ad i foud to hav a a of 290 g.$

6 6 Air olu, a Moitur cott oid ratio v Dgr of Saturatio S v Uit ight bulk kn 181. kn / dry kn / If th apl r aturatd thr ould d to b a additioal of atr. Thi ould igh kn ad thu th aturatd uit ight of th oil ould b ( ) at kn / Exapl Calculatio of Uit ight A oil ha a void ratio of 0.7. Calculat th dry ad aturatd uit ight of th atrial. Au that th olid atrial occupi 1, th auig G 2.65 th ditributio by volu ad ight i a follo. Pha olu ( ) Dry ight (kn) Saturatd ight (kn) oid Solid

2-6 kn ad thu th aturatd uit ight of th oil ould b ( 2845 + 146. 2 ) at 19. 04 kn / 157. 1 Exapl Calculatio of Uit ight A oil ha a void ratio of 0.7. Calculat th dry ad aturatd uit ight of th atrial.

7 kn Dry uit ight dry kn / ( ) Saturatd uit ight at 19. kn / 17. If th oil r fully aturatd th oitur cott ould b 687. Moitur cott % Altrativly th uit ight ay b calculatd fro th xprio giv arlir hich ar o p. 5 of th Data Sht dry at G 1 + ( G + ) 1 + S alo pag 4 ad 5 of th Data Sht for ot of th dfiitio ad quatio giv i thi ctio.

0 Altrativly th uit ight ay b calculatd fro th xprio giv arlir hich ar o p.

Problem Set 6 Solutions

Problem Set 6 Solutions 6.04/18.06J Mathmatics for Computr Scic March 15, 005 Srii Dvadas ad Eric Lhma Problm St 6 Solutios Du: Moday, March 8 at 9 PM Problm 1. Sammy th Shar is a fiacial srvic providr who offrs loas o th followig