## Sunday, September 23, 2012

### Design Of Small Dam

My boss asked me to redesign this small Gabion Dam because it has some scouring problems in the downstream., taking account in the design are seepage , hydraulic's fall and foundation design.

Reference books are :-

1. “GEOTECHNICAL GUIDELINES FOR DID WORKS”
2. “DESIGN OF SMALL DAMS
3. “BS8110 - 1985”

Detail Calculations are as followed :-

SEEPAGE DESIGN

Checking Existing Design below.

Using Lane’s Weighted Creep Theory,
i.       Length of Seepage Path,  = 0.75 + 0.3 +1/3(2.2)+(0.3+0.15)
= 2.23
ii.     Difference Head / Gabion Height / Retain water Height = 1 meter
iii.    Lane’s Weighted Creep Ratio = Length of Seepage Path / Difference Head
= 2.23/1
= 2.23
From Table 11, Page 82 of “GEOTECHNICAL GUIDELINES FOR DID WORK"

Lane’s Weighted Creep Ratio = 2.23 is between Medium Clay (2.0) and Soft Clay (3.0), which means the structure is safe on Medium Clay and Not safe on Soft Clay.

iv.     Assume The Structure is on Soft Clay, therefore the existing design is not safe
v.       Redesign Length of Seepage Path,

Extend base another 2 meter length to fulfill Soft Clay criteri
Length of Seepage Path,
= 0.75 + 0.3 +1/3(2.2)+(0.3+0.15)+1/3(2)
= 2.9 , still < 3.0 , Soft Clay , Not Safe

Adopt extended base to 2.5 meter length ;
Length of Seepage Path,
= 0.75 + 0.3+1/3(2.2)+0.3+0.15+1/3(2.5)
= 3.063  > 3.0
Therefore safe for Soft Clay

Table 11, Page 82 of “GEOTECHNICAL GUIDELINES FOR DID
WORKS”

CONTROL STRUCTURES HYDRAULIC DESIGN

From Figure 9.22, Page 368 of “DESIGN OF SMALL DAMS
Drop Length =  2.758 Ho

Assume Design Head, Ho = 0.5 meter ( which is over ) , Ho can be
determined from formula q=CHo3/2  ( Figure 9.21, Page 366 )

Drop Length = 2.758 * 0.5 = 1.379 m ( Water Drop on base slab since extended base slab is 2.5 meter length)

Therefore protection such as rip rap for scouring is not needed.

Figure 9.22, Page 368 of “DESIGN OF SMALL DAMS

CONRETE DESIGN TO BS8110 - 1985

Design base as 1  X (2.2 + 2.5 ) meter base pad footing.

Load = Stone (Gabion ) + Water
Gabion Weight, m  =  Density / Volume
= 2000 / ( 1 X 1 X 1)  Kg
= 20 Kn
Water Weight , m = Density / Volume
= 1000 / ( 1 X 1 X 1)  Kg
= 10 Kn

Point Load = 20 + 10 = 30 Kn

Ultimate axial force in column
N=30 kN
Ultimate moment in column
M=0 kNm
Ultimate shear force top of base
V=0 kN
Column dimension
cx=1000 mm
Column dimension
cy=1000 mm
Length of base (bending dirn)
lx=4.7 m
Width of base
ly=1 m
Left distance to edge of column
la=2.5 m
Right distance to edge of column
lb=lx-la-cx/1000
=4.7-2.5-1000/1000
=1.2 m
Char strength of concrete
fcu=30 N/mm2
Depth of base
h=150 mm
Cover to reinforcement
c=40 mm

Ground pressures                                                          ----------------
Mlh=(N*(la+cx/2000)+V*h/1000+M)
=(30*(2.5+1000/2000)+0*150/1000+0)
=90 kNm

Distance of line of action
from LH edge
y=Mlh/N
=90/30
=3 m

N
|   |
|e |
|  v
A---------------Å---------------B
|
|------- y
-------

Eccentricity
e=y-lx/2
=3-4.7/2
=0.65 m

Centroid of load lies within middle third.  Pressure varies linearly
from pa at A to pb at B.

|-------------------------------|
pa |                                               |
|----------                                | pb
------------          |
----------|

Elastic modulus of base in plan
z=ly*lx^2/6
=1*4.7^2/6
=3.6817 m3

Pressure at end A
pa=N/(lx*ly)-N*e/z
=30/(4.7*1)-30*0.65/3.6817
=1.0865 kN/m2

Pressure at end B
pb=N/(lx*ly)+N*e/z
=30/(4.7*1)+30*0.65/3.6817
=11.679 kN/m2