No_10_October_1968 — Page 43

ples, compacted at the optimum mois- ture content and maximum density as obtained from the compaction tests.

Design of earth dam

In designing low dams, a stable em- bankment up to 30 ft. height can usually be secured with an upstream slope of 2.75 to 1 and a downstream slope of 2 to 1, provided there is no possibility of submergence by tail water. A minimum slope of 21⁄21⁄2 to 1 should be used for any part of the embankment subject to submergence.

The results of the tests indicate that

the dry density, when the soils are properly selected and sufficiently com- pacted, will range from 100 to 112 lb. per cu. ft. and are impervious. The maximum height of the dam is 26 ft. and hence a section 21⁄2 to 1 on the upstream and 2 to 1 on the down- stream, with a top width of 15 ft. was considered sufficiently stable.

There will be an inside core of well compacted impervious material, desig- nated A. The outside shell will be filled relatively pervious material, designated B. This material B will also be well compacted.

The foundation is not very good and hence certain precautionary meas- ures need be taken. The poor foundat- ion material will be removed by cutting a core trench to a depth of 1/3 rd. the

Sample No. Location

Trial Pit No. Depth

SEMIPERVIOUS

IMPERVIOUS

THE

ON

FT. WELL GRADED SAND FILTER,

AT LEAST 5 FEET INTO IMPERVIOUS MATERIAL

height of dam. After de-watering and cleaning the core trench, it will be filled with well compacted soil made up of the material A.

The core trench will have a bottom width of 15 ft. and side slopes of 2 to 1. In addition a line of sheet piles will be driven to good soil below the core trench. The sheet piling will be 3 ft. above the bottom of the core trench

Table 1

Results of Tests on Samples of Soils. Ulu Sembrong Dam, Johore, Malaysia.

2

* I * *

3

4

2

3

4

Description:

5 ft. 6 in.

Red Laterite

Classification:

Clay

Sandy Silty

Specific Gravity.

2.73.

5 ft. 6 in.

Red & Grey Laterite Clay

Sandy Silty Clay

2.75

5 ft. 6 in.

Red Laterite

5 ft. 6 in.

Clay

Sandy Clay

Grey Clay with

red Spots

5 ft. 6 in. Grey Laterite

Sandy Clay

Clay

Gravelly

& Shale

Clay

Mechanical Clay %:

26

30

2.72

28

2.70

2.55

19

12

Analysis:% Pass. 200:

59

62

64

48

28

%Pass. 4 in.

100

95

96

100

50

% Pass. 6 in.

100

100

100

Dry Density lbs./c. ft.: Optimum Moisture %: Pen. Resis. Ibs/sq. ins.

112.7

101

93

101.5

99.8

16

20

26

15

19

255

675

700

260

Undisturbed Percolation

13.5 x 103

Rate, ft./year.

6.4 x 103

18.1 x 103

15.5 x 103

1.75 x 103

Shear Cohesion Remoulded: lbs./ft."

Tan

650

0.1228

2160

0.4877

6350

0.4040

3600

0.4663

4320

0.2126

Table 2

Dry Density Cohesion

Material:

lb./cu. ft.

lb./sq. ft.

Tano

95

500

0.3

1.25 H

105

300

0.15

H

C

95

100

0.5

0.25 H

ABO

Pore Water Pressure

Ft. Head of Water

ON LE

•

Fig. 3

Rock

SLOPE

SLOPE

I ON I

SAND FILTER.

1.6"GRAVEL FILTER.

to prevent any seepage due to creep between the dam and the foundation. The section of the dam is given in Fig. 1 (a).

In order to check the stability of the earth fill structure, the following conservative values for the properties of the soils (shown in table 2) were used, based on the results of tests in table 1.

The pore water pressure is an as- sumed value, measured in feet head of water, H being the depth above the point of consideration.

The preliminary checking of the stability of the earth dam was based on the circle method illustrated in Fig. 2 (a) and modified by Taylor. According to this the Taylor Number T is given by:

C

T =

UWH

where (1) W is the dry density of the

soil, which is 100lb./cu. ft.

(2) C is the cohesion, which is

300 lb. per sq. ft.

(3) H is the height of the dam,

which is 26 ft.

(4) U is the factor of safety which is assumed as 1.5. (5) is the angle of friction

which is 15 deg. To simplify the problem only one set of very conservative figures for the properties of soils is used. The Taylor Number T is given in Fig. 2 (b), but instead of the actual value of 0, a value Oc will be used when using the table, where,

Oc=0/U = 15/1.5 = 100

T= (300)/(1.5) x 100 x 26 0.077

=

Far East BUILDER, October 1968.

41