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248 SOIL MECHANICS AND FOUNDATION ENowegg.
__ Hence, the settlement (AH) can be easily determined by the above equation, afte =
ing C, from the consolidation test data. Compression index (C,) can also be roqa¢ ut.
mated more quickly and easily without a consolidation test, and on the knowledge ., ano co 249
5510"
pnt 0.45x5
oo" = p16 S014 m=1.522m—- 15.22 em
eee =15.22em Ans,
“ Ca 2 9.22 A ie aioe
2 ap ta dept [ he present
eit ly ‘8% and the specific gravity of eo" surface, The natural water
oil of! rh
gt nd clay, tM
met omer table is located at 4.4. m below the ground surface. The average submerged unit
ye we
is 10.6 KN/m? and the unit weigh; of sand k :
fH ee rolagloal evidence if le ‘ocated above the water-table is 17.2
3, Fro!
1 is normally consoli-
tt terete Building that will
we} on sand above the clay, is
ject ase the effective over-burden
yon the clay by 42 kN/m?. Esti-
pene iement ofthe building. Fine sand
elt s, We will first estimate the
yressure (Jp’ OF Pg) coming on
aie yer before the building is con-
oscil "ie will consider the centre of
“iy layer depth (A-A), to estimate
on ditions, as shown in Fig, 9.8,
wipe effective vertical pressure caused
gaAis due to sand and clay both. It is
egal to
12m
Fig. 9.8
Py
Now You for sand
Yor sand =
ths lor day is given as :
G-1
Yous = | Fae | ho
where G = 2.75, w = 0.38
But e = w . G for saturated soils = 0.38 x 2.75 = 1.045
eub for clay * 3-9 + Yous for sang % 6.8 ™ + Yor sang * 4.4m Ai)
0.6 kN/m! (given) —
7.2 kN/m3 (given)
141.045 2.045
Substituting the values in (i), we get
i. Po = 8.39 x 3.9 + 10.6 x 6.8 +17.2 x 4.4 = 180.48 kN/m?
"ng settlement Hq. (9.16), we have
2.75-1 15
i (73) x981= 23 «39 kim?
Ay = CeHo log, Pot?
1teg Po
where p, = 180.48 kN/m?, Ap = 42 kN/m? (given)
Hy = clay depth = 7.8 m ee ee
EAI
G
of clay in natural condition
= e, = void ratio
of, ga 10d talent areas
he s
Also, C, is given y ee — sg) = 0.009 16-101 = ae
032478 jog, BOMEES 21.129 m = 11.23 em
Hence, OH = 341.045 10 180.48
=11.23em Ans.
Hence, settlement expected = 1 De
Example 9.3. The following observation are recor’ on lest ona fay
saturated specimen an
Initial height of specimen =20m
Diameter of specimen. = 75mm
Specific gravity of soil grains =2.77
Water content (final) = 39%
Applied pressure 0 50 100 200 400 800 ;
in kN
Final dial gauge 100 359 499 632 768 39 oa
reading mm x 1
(a) Calculate the void ratio corresponding to each pressure increment,
() Plot e-log p curve and find the compression index.
ute the values of coefficient of compressibility and coefficient of volume change, for
() Compt
200 kN/m? ; and 200 to 400 kN/m?.
the increment of pressure from 100 to
Solution. Final m.c. 0.29
Final void ratio 4p = WG = 0.89 x 2.77 = 1.08
Since consolidometer gives AH, and we have to work out ¢ (or Ae first), we use Eq. (9.10), as
‘ete AH
ay ee or de= (1 +09)
Initial height of specimen = 20 mm :
Final height of specimen = 20 ~ 6.66* *[i.e. (766 - 100 1041
= 13.34 mm. a
Usi .
cued Speen nu and (i.e. at the end of the test), we can find a general relation
: , which will hold good for all incremental pressure intervals.
H,=13.34mm and e,= 1.08
AH
= i334 (1 + 1.08) mm
= 0.156 AH (where AH is in mm)
as shown in column (7) of Table 9.2. Ans.
e values are then coniputed oo Table 9.2 251
ee
Dial gauge, wt) Wena
reading mm * 10" inmm HWin'rmm!! |" becoasean Ta
| $0186 xc01.(4) | (Sraih Trebt beitomy
(3) (4) se |
a ©
259 20.00 |
9 OW
955 Oia 41 ()0.404
499 oe 16.01, eee
2 0:
. 136 14.68 ee
8 o21
76 ous. 13.32 i ce a
9 (-)0.204
89s (4) 1.33 12.91
766 13.34 (0.207
table 9.2 are plotted o fgg eee
ofe and p from table 9. na Semi-log pay :
(QyValues ore gp, 2s shown in Fig. 99, PaPer, So as to obtain a graph
yeene an
peo
T 0
22 For, p= 100, e= 1.496
1 For. p= 200, e 289
24 + Ae = 1.496 — 1.289
| rm S109 p= & log 200— 10g 100
19) : =—0207 0207
logjo-200. tog,.2
18) MAU 0910 OD ho
t 17] = 0.688 a
g 16
245
B14
3
743
1.2)
fiber ila
1 —
03] ts
08) N.
0 10 20 50, 100 200 400 800 1000
Pin kN/m? (log)
Fig. 9.9, Plot of e and log p tor example 9.3 (Table 9.2).
Compression index C,, from this plot is obtained, as the slope of the straight line portion of
curve,
: : 1496-1289 _ ee
Choosing any two points on this straight line, we get this slope = sr apo" = 5:
10810 F500
100
Hence, co
pression index = C, = 0.688 Ans.
© Coe,
Vm? :
f compressibility (a,). a, for increment of pressure between 100 to 200 kN/m’
Gaede. a
wy, = = 20741
AP Py ~ Po SOIL MECHANICS AND Fou
ATION
252
1,496 -1.289 _ 0.207 _ 9 97x 10-9 mean, i
=g00-100 ~ 100 s.
or
low ; _.
7 where é, is the first void ratio betwe
increment stages i.e. void ratio gs” b¥o
of 100 kN/m? = 1.496, Tes
modulus of volume change for pressure increm
200 kN/m?
Mt etree
-3
‘ = 207TX10™ _ 9,83 x 10-8 mY/KN = 0.88 x 10 as
= 741496 AN. An,
«a, for increment of pressure between 200 to 400 kN/m?
Similarly,
£9 — 41
“ Pi- Po
where py = 200 kN/m? and p, = 400 kN/m2
07
200
Similarly, the modulus of volume change for pressure increment between 209 to 400 kgs
4%, _ 1.06x10% :
% - Thep 141.289
(9 = void ratio corresponding to py of 200 kNin? = 1
1.46 x 10% mY/KN. Ans.
Example 9.4.4 uniform load of 145 kN /m? is distributed over a large area ofthe sufae
of the ground. The sub-soil consists of a bed of dense sand, containing two strata of day, xh
3 m thick. The top of the upper stratum is 6 m below ground and. of the lower 21 m bets
ground. The compression index for both layers is 0.35, the water content is 34%, and spetie
gravity of soil grains 2.75. The sand weighs 20 kN/m* and is completely submerged. Conpit
the total settlement under the load.
Solution. The given soil conditions are shown in Fig. 9.10.
= 1.06 x 10% m%KN. Ans,
m,
1.45 kglom?
ttt
Sand
Y= Yoat = 20 KNim>
Completely
submerged 21m 4) 270
Using Eqn: (9.32), we have
‘To determine typ above, we should first determine C, a
or
‘SOIL MECHANICS AND FOUNDATION ENGINe
ia
y
Cy
or 0.196 = GG «too
s follows : o
AH=m,.dp Hy la. (9.1)
1.73 em = m, x 62.5 “Xx (6 x 100 em)
i m’
1B» 4 m2
2 28 yan = 2.47 104 mi
™,* §2.5x500 ™
Further using Eq. (9.30), K=C, .m, «jy we have
or
K=8.5 x 10% ems; giv,
2 kN
10 ML 4) By 9.81 >
3.5 x 1010 > = C, x (2.47 x 10 ) in” mn
10
£__8.6x10 msec = 1.44 x 10-7 m%see
¥" 2.47% 10"! x 9.81
Substituting this value of C, in (i) above, we get
or
144x107? m*/sec ,
.196 =
ons (6m)? °
gg = ES soe = 3.403 x 107 see
1.44 x10
_ 3.403 x 107
= Soxamny days = 398.84 days. Ans.
mple 9.6. Find the time required for 50% consolidation in a soil stratum, 9 m thick with
ious strata on top and bottom. Also determine the coefficient of consolidation given that
K = 10 ml sec ;e9 = 1.5, a, = 0.003 m?/kN, Time factor = 0.2.
. H_9
Solution. d=5e aaa as it is draining at both ends = 4.5 m
C,=?, T,=0.2, K= 10 m/sec.
5, a,=8 x 10% m”kN
Using Eq. (9.32), wehave T,=—¢t
0.2
or t= too = G (4.5 my? oli)
To determine C,, we have
K=C,.m,.1,. (Bq. 9.30)
ay
+e
where m, = (from Eq. 9.2)
=3x193 ™
kN
1
ee 3
ie “3 1.2 x 10% m/kN
2
10 m/sec = C, x (22% 19-3 B kN
a ay) ©9815 GOMPRESSION AND CONSOLIDATION SETTLEMENTS 27
107°
Hence, c, Taxi0-Pxogi MUsee = 8.49 x 10 msec
., 2
Now from(@, tsp = G45 mj? = 02 ___ 00" yoy a
‘0 m’
8.49 x 10-5
— 0.2.x 4.5? x 108
= 49 Secs = 552 days. Ans.
Example 9.7. A soil of specific gravity 2.65 has a moisture content of 18% when fully
saturated. 1.9 cm thick sample of this soil tested in a consolidometer shows a compression of
0,050 em when the load is increased from 40 kN /sq m to 80 kN/sq m. Compute the compres-
sion index of the soil.
Solution. G=265, w=0.18
S = 1 (fully saturated)
w.G
n e= S = 0.18 x 2.65 bs e¢ = 0.477.
Now, compression AH = 0.050 cm.
Hy = Thickness of sample = 1.9 cm.
. 4H be
Now, using i a Tra’ we get
0.050 _ Ae 1.477 x 0.050
“To *Tsoai7 6 MRT 70.0389
Now, use Bq.: Ae = C, logy, 20*4?
Po
where p, = 40 kN/m?
Ap = 80 — 40 = 40 kN/m?
40440
40
3
2.0389 _ 9 199
log 19 2
0.0389 = C, logy,
C,logy2 or C,
Henge, C, = 0.129. Ans.
\ pam 9.8. The loading period for a building extended from July 1987 to July 1989. In
uly 1992, the average measured settlement was found to be 113 mm. It is known that the
ultimate settlement will be about 360 mm. Estimate the settlement in July 1997. Assume
double drainage to occur.
Solution. When loading is applied over a certain period of time, then all time calculations
are done by assuming the middle of the construction period as datum. Thus, in this problem,
the building was constructed between July 1987 to July 1989. Hence, the datum would be
July 1988, All further timings would be calculated from this datum.
Hence, 113 mm settlement observed in July 1992 would be after a period of 4 years from
this datum (July 1988).
Settlement S, after ¢, (4 yrs) = 113 mm = 0.113 m
Total settlement (S i.e. AH) = 360 mm = 0.360 m
Settlement in July 1997, i.e. after 9 yrs from datum = S, at t, (9 yrs) =?
, O13 _ 9 arg
Now, U, = degree of settlement after t, (4 yrs) time = 9 569m, = 9 27:
or
i
‘SOIL MECHANICS AND FOUNDATION ENGINEER
9
. S
U, = degree of settlement after t, (9 yrs) time = 0360 = 2-778 8,
Now, from Eq. (9.33),
And
r, =¢ U2 (for U < 0.60)
T= Fos 14)? ff)
x
Ty, = G2.778 8.)
i)
(assuming U, to be $0.6; to be checked after computing 8.) _ 4,
C, Cc,
it . Ty = Geety Ta = ote
k 2
Tt 7 gO _Ayrs
yt Fert S,)? 9yts
(0.314) _ f - 0814 01696 m
2.7788, \9 3 2.778
Let us check back that U, < 0.6
So
Hence, the expected settlement after 9 yrs, ie. in July 1997
KY = 0.1696 m = 169.6 mm. Ans.
ample 9.9.4 3 m thick clay
and is underlain by an imp
(which is less than 0.6, and hence
our assumption at (A) is OK)
layer beneath a building is overlain by @ permeable stratum
ervious rock. The-coefficient of consolidation of the clay was found
to be 0.025 cm?/ min. The final expected settlement for the layer is 8 cm.
(© How much time will it take for 80% of total settlement to take place ?
(i) Determine the time required for a settlement of 2.5 cm.
(ii) What will be the settlement in 6 months ?
Solution. 100% settlement = AH = 8 em
1,
0
0.008
0.031
0.071
0.126
0.196
0.287
0.403
0.567
0.848
(Engineering Services, 1984) Fr
cowPRESSION AND CONSOLIDATION SETTLEMENTS, 278
80% settlement = 80% x 8 em = 6.4 em
beon, = ?
tise the general Bgn. (9.82) as :
Cc,
T= Geek
where 7, for 80% settlement or consolidation = 0.567
(+ From the given values of U and T,, for U = 0.8
(80%)]
max. length of drainage path
full depth of clay, because it will drain only on one
side, since impervious rock exists below the clay
layer, and pervious soil exists above it
=3m
C, = 0.025 em*/min (given)
Substituting the values, we got for 80% consolidation
G, 0.025 em?/min
Tygo)= Se (tg) or = 0.567 = IID, 5
00) = ge «eo (800cmy>*
= tao = Time for 80% settlement
(300)? x 0.567
= mi
"ae = 1417.5 days = 3.88 yrs. Ans.
(ii) Time required for 2.5 cm settlement (¢,) =?
4H, _ fl _ fa | nw.
AH = T,, is a T, is time factor
ort, = 216.3 days
‘Time required for 2.5 cm settlement = 216.3 days. Ans.
(iii) The settlement in 6 months (i.e. 365/2 days or 182.5 days (say)
Using ft, _ AM, we have 1417.5 days _ 64cm
Sey 182.5days — Settlement (AH,)
Settlement in 6 months (182.5 days) = AH, = 2.3m. Ans.
Example 9.10. A building was to be constructed on a clay stratum. Preliminary analysis
indicated a settlement of 50 mm in 6 years and an ultimate settlement of 250 mm. The aver-
{ge increase of pressure in the clay stratum was 24 kN/m?,
The following variations occurred from the assumptions used in the preliminary analysis :
® The loading period was 3 years, which was not considered in the preliminary analysis.
i Borings indicated 20% more thickness for the clay stratum than originally assumed.
(G) During construction, the water-table got lowered permanently by 1 metre.
Estimate
(©) The ultimate settlement
a The settlement at the end of the loading period
The settlement 2 years after completion of building (Civil Services, 2001)
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