Birete Dam Project - Design of Diversion Structure

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/UNITED NATIONS DEVELOPMENT PROGRAMME DTCD PROJECT ETH-75/OO5
Ministry Of Mine?, Energy and Water Resources Ethiopian Water Resouroe  Authority
-
Addis Ababa
I
UN Engineering Adviser August 1980
r
■.'
■■
■ >3.        *             -      *
* *?<•*’*“- i -
ABSTRACT
Birete Dam diversion structure design report covers the scheme proposed for diversion, leaving a gap for the first rainy season thus making it posaiblo to have ar. economical size of diversion conduit. It elso includes the hydraulic
and structural design of oonduit and dotails the inlet and ' outlet structures.- 11 -
TABLE OF CONTENTS
PAGE
ABSTRACT .................................................................................... - i_
TABLE OF CONTENTS ................................................................  11
DESIGN CONSIDERATIONS ................................................  -1
2.   GAP .... ..... ............................. i. *........................................... * 2
3.   CONDUIT .................................... i..........................................  3
a)   Hydraulie Design .......... ..................................................... 3
b)   Structural Design............................. *......................... ..  4
4.   INTAKE STRUCTURE............................................................ 24
5.   OUTLET ......................................................... .........................  26
6.   ACKNOWLEDGEMENT................................. .. ..................... 27
7.   REFERENCES ..........................................................................  27
i
4.BIRETE DAM PROJECT DESIGN OF DIVERSION STRUCTURE
1.         DESIGN CONSIDERATIONS
The diversion of the river is planned in such a way that the construction of the dam is facilitated and is completed
in the stipulated 'iime  On the other hand it is kept in
<
view that the^sCnf the diversion remains low.
The"oinstruction" of the diversion structure is scheduled
to be finished in the first dry season. In the second dry
season, it is suggested to continue construction of the dam
fmm both "sides leaving a gap in the embankment in addition
the diver£i*n conduit. The dimensions of the gap are
decided hy the fnTlwwing considerations:
(a)          The gap should have an adequate capacity (area) to pass safely a flrwad discharge having a return period of 25
years.;
(b)       • The bettmm width should be such that the construction by machines is feasible.
(c)           Side slopes of gap should be such that a proper bond between - tJ^e earlier and later construction is ensured. Side
slop 6*4H is adopted. This side slope would also minimise
.the’'pessibilities'wf damage to the slope during floods and Avoid crocking* of the dam on account of differential settlement.
(d)          Closing'wf the gap machines        and labour available
should        be        possible        with        the in a reasonably short time.
The gap would facilitate passing of the flood in the rainy season fellowing the second dry season. In the third dry soason the river is diverted through the conduit and the gap in the embankment is closed.2
The ab^ve sequence of constru^*4 ori on one hand would enable the cons true t ion of the embankment in th^ p portion quick
enough to build up head of water sufficient for the diversion
©nnduit pass at least a discharge having a return period
of five ydars during the dry season avoiding any eventuality
ff damage to the dam and on the other hand would make it possible to keep the size of the conduit small thus affecting saving in
test. Drawing No. &W101 shows the alignment of the conduit.
2.         GAP
During the oonstruction of the dam to pass flood discharge at the end of first dry season, the gap ?-oposed to be left
in the embankment would have the following dimensions:
Discharge for once in 25 years frequency = -64 nr/sec. Bed elope nf the river at the dam axis is 1 ,ini60 metres
S = 1            = 0.00625 Considering a depth of 1*5 metres in the gap
A « (22+10)4x1* 5 
P = 12*36 +10 
R = A . llO73
R’ >        1.048
V = 1 x 1.048 x (0.00625)^
■ 24.0m2
= 22.36 m
5TC28
q ■ 2.96 x 24 
» ■y
. - 2.96 O.K.
m/sec- 3 -
Total       earthwork       and       rockfill       volume       in       the       gap       for
w
20.5 m height,
Area of gap =
Volume
2(20.5x4) * 20 x x 20.5 =      (1886x8) + (9^3 x 77.9)
=      1886 m
=      88,5^8 «3
say 90,000 m
A volume sf 90,000 m^ ran be easily done in one dry season.
3.        CONDUIT
(a) Hydraulic Design
At the beginning of the second dry season the river would be diverted through the conduit and the gap closed.
Discharge for once in 5 years frequency x 40 tn /sec.
A circular conduit having a diameter of 2.00 in is adopted and the head of water required to pass 40 iir/sec is
checked.
«
3
*3
Area of conduit,
A = ’jd  =
2
T“ 
Hydraulic mean Radius, R = A
Velocity,              V = 4o 
STTFl
3.141x4
Li
= d x
F
2
R* = 0.63
= 12.73 m/sec
= 3.141 m2
2 x 0.50
5“
Friction loss is calculated by Manning
V = 1 x R7 x S
n
T  £ 
4-
formula
12*7*3  = 1  x 0.63 x S*
0.014
S c 0.08length     of     conduit     =     (20.5     x     1.8)     +     8     +     \2O-5     x     2) Head loss due to friction = 85*9 x 0.08
Entrance loss » 0*10 V =
Exit loss = 0.20 V2
2g 
2
« 
0.1 (12.73,
19.8
02  (12.73)2
“T976
Total bead required = 6.87+0•83+1*66 Likely bed level of conduit at exit
Centre of conduit at exit = 108.80 + 1.00 Level of water on the upstream to pass the
=
discharge =
109180 +9.32 
=
Likely foundation level of the dam at the
deepest point
Dam height to be achieved in the gap 11? i 2-107.50 = Including a little margin the height to be achi^nred
in the gap may be taken Vqlume of earth and rockfill,
-
=
Bap area for 12.5 m height = Bioping length = (l2.5x1.8) + (12.5*2) Horizontal length « 85.9-47*5
2( 12..5x4) 4-20 xyxl2.5 =
=
85*9 m
6* 87 m
0.83 m
1.66 m
9.32 m
108.80
109.80
119-12
107.50
11.62
12.5 m
750 m2
U7.5 m
Volume of fill
Volume of fill
it can be completed
x
 4 . )
7 5   +f750x38.^)
=         38.4
46,688
say 
sx
in
in
the       gap      upto       12.5      metros a reasonably short time,
height is
therefore
47,000
such that
the size
of the conduit adopted would suffice.
(b) Structural Design
(i) Lighter loaded portions
For structural thickness conduit is divided into two, the heaviest loaded central portion and the lighter loaded portion of sides. Considering the lighter loaded po? tirns upto the fill
level of 120•BO,
Assume a thickness of conduit, t = 0.30 m which is^r«**
r
- 5 -
Foundation level at the load point = 1O9-OO-O.3O
Top level of conduit = 108.70 +2.60 
Height of fill over the conduit = 120.00-111.3°
For a general condition ofloading, (when the conduit projects above the natural ground or the .ditch in which it is cast is wide
enough not to influence loading) , the extreme cases to cater for
the initial and the long term effect of loading recommended in the Engineer Manual EM 1110-2-2902, Corps of Engineers, U.S.A., are
as below:
= 108.70
= 111-30
= 8.70 m
Case i)
* Case li)
We
Pe
we
1.5 yis,
0.5 « H
P
e
= XtfH
Where
w
G
-
Vertical load of earth per m
2
P
e
=
Horizontal load of earth per m
2
=s
Unit weight of soil
Since the
Height         of         fill         above         horizontal
■        diameter of conduit
H        s=      Height        of        fill        above        point
considered.
depth of the fill on the conduit is 16.70 m, the
effect of concentrated 11Ve load would be negligible
For worst condition the fill on top of the conduit is considered saturated and the conduit empty.- 6 -
Case (i),
Vertical load, wg
A uniform vertical
considered.
Horizontal pressure, Pg H at top of conduit
H at bottom of conduit
P at top of conduit
e
P at bottom of conduit
e
reaction of the
same intensity is
0. r X H
8.70 m
8.70+2.60                  =     11.30 m 0.5x2150x8.70 = 9,353 kg/m2
0.5x2150x11.00 = 12,110 kg/m2
Loading including the self wight of circular conduit is shown in the Figure.
=     12,148 - 9,353 
=     2,795 kg/m2P a6t bottom of conduit Loading      including      the shown in the Figure.
- 7 -
0.5 x 2150 x 11.30 = 12,148kg/nr self weight of circular conduit is
^’=12,148-9,353=2,795kg/m2
2
Unit weight of reinforced concrete,   c=2400 kg/m^
+ sign  corvention is shown  in  the Figure- 8 -
Mement, Thrust and Pheai* are calculated in the kllewing taking coefficients from Engineering Monograph J. - ’’ B&ggs Deformeter Stress Anaylsis of Single Barrel jnduits ” by U„S.B.R.
Coefficients
Values
Vert. Lead
1
•X Heriz, Unif. Lead
ioriz.' Triang.
Lead
Ta tai
Self
Load
Joeff.1
2
cw r
e
3oeff<2
2 
e
3oeff.3
Coefij x cr^
1
•
3
4 s
1
3
4
kgm
1
>
3
4
5
6
7
8
9
0
1
2
I3
*0.333 +0,289
+0,167
0
-0.16T -0,289
•0.333 -0.28* -0.167
0
*0,167 +0,28> +0,333
-0.333 -0.289 -0,167
•
+0,167 +0.28* +0.333 +0.289 +•*467
•
-0,167
•0,281
-0.333
-0.147 -0.132 -0.087
-0,020
*0,057
+0.126
+0.165
+0.151. +O,1A«
*0,016
•0.083
-0,161
-0.190
+0.111 +0.097 +0.057
+0.001
-0.058
-0.103
-0.117
-0.095 -0.045
+0.015 +0.064 +0.08E +0.093
+9,343 +8,109 +4»686
0
-4,686 -8,109 -9,343 -8,109 -4,686
0
+4,686 +8,109 +9,343
2
-3,115 -2,703 -1,562
0
+1,562 +2,703 +3,115 +2,703 +1,562
0
-1,562 -2,703
-3,115
-411
-369
-243
- 56
+159
+352
+461
+444
+296
+ 45
-232
-450
-531
+266 +233
► +137 + 2
-139
-247
-281
-228
-108
+ 36 +154 +2H +223
+6,083 +5,270
+3,018
-       54
-3,104
-5,301
-6,043
-5,190
-2,936
> SI
+3,046
+5,i67
+5,9209
Case (i)
Ceefficients
Values
Vert.
Load
Heriz.
Unif. 
Load
leriz. Triang. Load
Self
Load
Coeff.1
%1
Coeff.2
Coeff.3 A '■
Coeff.z
Total
«•
1
2
3
4
1
2
3
4
kg
0
+0.089
+0.333
+0.667
+1.000
+1.244
+1.333
+1.244 +1.000 +0.667
+0.333 +0.089
0
+4.333 +1,244 +1,000 +0.667 +0.333 +0.089
0
+0.089
+0.333
+0.667
+1.000
+1,244
+1*333
+0.395 +0.381 +0.336 +0.259 +0.156 +0.055
0
+0.035
+0.178
+0.408
+0.664
+0.863
+0.939
+0.024
+0.049
+0.122
+0,233
+0.364
+0.498
+0.6.11
+0.017
+0.474
+0.264
+0.08(
-0.00(
-0.02'
0
+2,497 +9,343
■08,715 +2B.O58
-04,904
■07,401 +34904
■28058 +18715
+9,343 +2,497
0
+12,468 +11,635 + 9,353 + 6,238 + 3,115 + 832
0
+ 832
+ 1,115
+'6;238
+ 9,353
+11,635
+12,468
+1,104 +1,065
+ 939 + 724 + 436
+ .154
0
+ 98 + 498 +1,140
+1,856 +2,412 +2,624
+ 58
+118
+293
+559
+874 +1195 +1466 +1481 +1138 + 634 +206
-14
- 58
+13,630 +15,315 +19,928 +26,236 +32,483 +37,085 +38,867 +37,315
+32,809 +26,727 +20,758 +16,530 +15,034
:t- 10 -
Case (i)
--------- ——|[
...... 1
]                    Values
— -
 . 1
Coefficients
Vert.
Load
1
Horiz. tfioriz. Unif, iTriang. Load jLoad
Total
Self
Load
L
Coeff.1 ;    Coeff.2 xw r              xP r
6e
Coeff.3
xP^r
e
Coeff.4
xC r^
i
1
2i3
4 I           1
2
3
4
kg
0
•0.333
•0.577
■0.667
■0.577 -0.333
0
-0.333
-0.577
K)> 667
-0.577 -0.333
0
0
-0.333 | -0.102
-0.577      -0.194
-0.667      -0.259
-0.577      -0.270
-0.333 -0.204
0        -0.061
+0.333    +0*129
+0.577    +6* 308
+0.667    +0.408
+0;577 +0.383 +0.333 +0.231
00
0
+0.092
+0.165
+0.199
+0.183 +0.109
-0.024
-0*190
-0.301
-0*298 -0.196 -0.069
0
0
+9,343 +16,189 +18,715 +16,189 + 9,343
0
- 9,343 -16,189 -18,715 -16,189
- 9,343
0
0
-3,115 -5,397 -6,238 -5,397 -3,115
0
+3,115 +5,397 +6,238 +5,397 +3,115
0
.0
-285
-542
-724
-755
-570
-171
+361
+861
+1140 +1070
+ 646
0
0
+221
+396
+478
+439
+262
- 58
-456
-722
-715
-470
-166
0
0
+6,164 +10,646
+12,231
+10,476
+ 5,920
- .. 229
- 6,323
-10,653
-'*2,052
-10,192
- 5,748
011
Case (ii)
Vertical load, w  <=>^
e
« 2150 x   ™ -
n
<o  '7"" vg/m2
A uniform vertical reaction of the raanr- intensity is
considered.
Horizontal pressure, P = i
P  at top of conduit = 2150 x 3,.70 « i8,705 kg/m2
g
P at bottom of conduit => 2150 x 11.30 = 24,295 kg/m2
e
P^ - 24,295-18,705 => 5590 kg/m2
Coefficients     remain     the     same     as     in     the     previous     case, values    of    Be    Mding    moment,    Shear    and    Thrust    for    case    (ii) loading are calculated below.
Valuers
Ceeff.1
x w
Coeff.2 x P
Coeff.3
Coeff
1   .2       x Or e ~
+6,229 
-6,229 
-822
+266 
Total
kgm
-556 
+5,406 
+3,124
-5,406 
-3,124
0 
+3,124 
+5,406 
+6,229 
+5,406 
+3,124
-738 
-486 
-112 
+319 
+704
;-922 
+889 
+593 
+233 
+137 
+    2 
-505 
-349 
-MO
-5,406 
-6,229 
-5,406 
-3,124
-139 
-247 
-:-231 
-228
-108 
+180
+457 
+641
+661 
+485 
0 
+3,124 
+5,406 
+6,229
0
-3,124 
-5,406 
-6,229
+  89 
+  36 
-464 
-900 
-1062 
I
I
f
+134
I
:  ii
+223 
+125 
-310
-689
-839Thrust
12 -
Valuer
No
Total
1
2
3 I4
5 6 7 8 9 10
Coeff.1 x vr t
e
1
0 +   1,665 
+   6,223 
+12,476 
+18,705 
+23,269 +24,934 +23,269 +18,705 +12,476 
Coeff.3
11      +   6,229 
Coeff.2
x P  r
e
2
+24,934 +23,269 +18,705 +12,476 +6,    229 + 1,665
0 
+ 1,665 
+    6,229 
+12,476 
+18,705 
+2,208 +2,130 +1,873 +1,448 -r  87  2 + 308 |
0 +    196 +    995 +2,281 
+3,712 
4
+ 58 + 118 
+    2.93
+    559
+     874 
+1,195 
+1,466 
+1,481 
+1,138 +    634. 
+ 206
leg
+27,200 +27,182 +27,105 +26,959 +26,680 +26,437 +26,400 +26,611 +27,067 
+27,867 
+28,852 
12
13
+ 1,665
0
+23,269 
+24,934 
+4,824 
+5,249
-
14
-
58
+29,744 
+30,125Values
No
Total
Coeff.2
x P_r
e
Coeff.3
x P e r
Coeff.4 
2
x C r
1
2
3
4
kg
h
cd <D
x:
co
1
2
3
4
5
6
7
8
9
10
11
12
13
0
+ 6,229
+10,793 +12,476 +10,793
+ 6,229
0
- 6,229
-10,793
-12,476
-10,793
- 6,229
0
0
- 6;229 -10,793 -12,476 -10,793
- 6,229
0
+ 6,229 +10,793 +12,476 +10,793 + 6,229
0
0
- 570
-1,085
-1,443
-1,509
-1,140
-     341
+ 721
+1,722
+2,280
+2,141
+1,291
0
0
+221
+396
+478
+439
+262
- 58
-456
-722
-715
t47O
'-166
G
0 .
—  349 .
-     689
-     970
-1,070
-     878
-     283 + • 265
+1,000
+■1,565
+-1,671
+1,125
0
Reinforcement required ie worked out in the following Loading Case (i)
- At crown (See.D
Maximum +ve moment at crown (snc.1), m e» 6,083 kgm ThruBt at crown (sec.1), N = 13,630 kg
a * so
<T> H>
•
—b  = 100cm
d  = 26cm
d' = d" = E =
NE ♦
Resisting moment, KF = K y^-g
= 15.65 x 
4cm
15-4   11cm
9
M  ll
IT a+ d
6 °8f^3U~ + 11 = 55.63cm 13,630 x 0.5563 = 7,58^ kgm
—        — bcl
100
= 1O,579kgm
Since Iff is more than NE, no compression reinforcement is needed.
Area of steel required for tension,
An
7,582x100 
(1-0.874x26)
1406x0.87x26 
55.63
= 23.73 (1-0.408)
= 14.1 cm
o
,tension inside
- At springing (Sec.7)
Maximum -ve moment at Sec.7, M=6,048 kgm
Thrust at seo.7 
e = 135^667^'
, N = 38,867 kg
»■
73  26.56cm a 0.2656m
NE = 38,867x0.2656 = 10,323
KF « 1O,579kgm
Since KF is higher than NE, no compression reinforcment is necessary.
f
Area of steel required or tension
=  10,323x100—             (1-0,874x26^ B  1406x0.874x26                        2K756
“     32.31     (1-0.855)     =     4.7     cm^     ,     tension outside.- 15 -
A.t . Invert (Sea.1>3)'1
Moment, M = 5,920kgm\
Thrust, N =>15,034kg
E = § + d"
° ~5~T5o34^■+ 11 = 50.38cm                  = 0.5038m.
* ■ ?}
NE » 15,084 x 0.5038 = 7,574kgm
KF is higher than NE, no compression reinforcement is needed.
7,574x100 m 0.874x26 1W>x.S74x2S u “30758“> 23.71(1-0.451) = l3.0cm2
Loading case(ii)
-At     Invert (Sec, 13)
Moment         ’,   M  = 839kgm
, tension inside
Thrust E   839x100 
o
,   N  = 30,125kg
+
0 -13.79
30,125
NE o 30,125x0.1379 ° 4,154kgm
A  --------- 4154x100— (1-0.874x26)
s  1406x0.874x26                     13.79
.. = 13.00 (1-1.648)
It shows that no structural steel is required. As the thrust is high, no tension is developed on the outside because of very little moment at invert in loading case(ii). Shear.
Loading Case(i)
Max Shear = 12,231kgShear Stress = —^j-231—   4.7kg/cm2
=
100x26
Allowable shear stress according to the code ■=> 4.24kg/cm Shear stress computed is slightly more. Since these porticu'1
of conduit on the sides would be under the rockfill where the longterm differential settlement to induce loading case(i) when the vertical load is multiplied by a factor of 1.5, may not occur, little extra shear computed may be tolerated and no reimforcement may be provided.
2 - 17 -
(ii) Heavier loaded central portion of conduit.
i'oundation of the conduit: 108.40
Top of the conduit: 108.40 + 2.80 = 111.20
So Case (i) Case (ii)
H  = 128 - 111.20 = 16.80m
h
w  = 1.5 H^
e
P  « 0.5 JH
e
p e = Zh
Case ( i ) Vertical load
w = 1.5 * 2150 x 16.80 = 54,180 kgf/m2 e
A uni, form vertical reaction of the same intensity is considered.
Horizontal pressure: P => 0.5 
H at top of conduit =• 16.80m
<|.H
H    at    bottom    of    confrait    16.80    ♦    2.80    =    19.60m
I*                —
P_ at top of the conduit =« 0.5 x 2150 y 16.80 = 18,06C£ jf/m2
P © at    bottom    of    the    conduit=>0.5*215CDE-1    9.60=21,07Gkigf/m2
p*e «= 21070 - 18,060 => 3,010 kgf/m2
Case (ii) Vertical load:
w  o 2150 x 16.80 = 36,120 kSf/m2
e
At top P  = 2150» 16.80 = 36,120 kgf/m2
oe
At bottom - P  => 2150 3* 19.60 = 42,140 kgf/m2
e
P*  =, 42,140 - 36,120 => 6,020 lbgf/m2
e
Looking      at      the      analysis      carried      out      for      lighter      portions of the conduit, it is evident that the loading case (ii)
will not gover n as such no computations have been carried out.Moment, Thrust and Shear are calculated, in the follorying taking the (linear interpolated) coefficients from Engineering monograph 14- "Beggs Deformeter Stress Analysis of Single Barrel Conduitfl" by U*S.B.R.Coefficients
Values
Vert.
Load
Horiz,
Unif. 
Load
Horiz. Triang
Load
Self
Load
4
Coeff.1 * 2 
*w r
e
Coeff.2 * Pe r2
Coeff.3 *pi r2 
e
Coeff.4
3 
*c r
4
Total
1
2
3
1
2
3
Kgm
J
)
)
2
3
0.354 0.307
0.178
0
-0.178
-0.307
-0.354
-0.307 -0.178
0
+0.178
+0.307
+0.354
+0.354 -0.307
-0.176
+0.176
+0.307
+0.354
+0.301 +0.17?
0 -0.171
-0.30
-0.35
-0.158 -0.141
-0.093 -0.020 +o.ofe +0.13? +0.176 +0.169 +0.112
+0.018
-0.086
I -0.168
/ -0.196
+0,146 +0.127 +0.075
+0..001
(£.076
-0.133
-0.146
-0.115 -0.052
+0.023
+0.084
+0.115
+0.121
19,180 16,663
9,644
0
-9,644 -*6,633
-4 9,180
-16,633
-3,644
0
+9,644 fl 6,633
-A 9,180
-6,393 -5,544 -3,215
0
+3,215 +5,544 +6,393
+5,544 +3,215
C -3,215
-5,5 4.4 -6,5 9;
-476
-424
-280
- 60
+187
+406
+530
+509
+337
+ 54
-259
-506
-596
+350
+305 +•>80
+ 2
-182
-319
-350
-276
-125
+ 55
+202
+276
+290
L.
+12,661 +10,970
+ 6,329
58
-6,424
-11,002
-12,607
-10,856
- 6,217
+         99
+ 6,372
+10,859
+12,48120 -
Coefficients
Values
r" I
V PT*‘t Load
’ Horiz.
Unif.
Load
Horiz. Triang. Load
Load
?oeff.1
*W  r
eo
Coeff.2
♦p r
e
Coeff.3
--
e
Codff.4
*C 2
r
1
0
0.095
0.354
0.709
1.063
1.322
1.417
1.322 1.063 0.709 0.354 0.095
0
2
3
4
1
2
3
Total
>
_i
i
kf
!
1.417
1.322
1.063
0.709
0.354
0.095
0
0.095
0.354
0.709
1.063
1.322
1.417
0.415
0.401
fl. 354
0.272
0.164
0.057
0
0.038
0.191
0.437
0.709
0.922
1.002
0.044
0.077
0.171
0.313
0.482
0.652
0.797
0.802
0.613 0.338 0.103
-0.020 -0.044
0
5,147
19,180
38,414
57,593
71,626
76,773
71,626 57,593 38,414 19,180
5,147
0
25,591 23,875 19,198 12,805
6,393
1,716
0
1,716
6,393 12,805 19,198 23,875 25,591
_____
1,249 ws
1,207           185
1,066           410
819          751
494       1157
172       1565
0         1913
114       1925
575       1471 1,315           811 2,134           247 2,775           -48 3,016        -106
i
26,946' 30,4141
39,854i 52,789 65,637 75,079 78,636 75,331
66,032
53,345
40,759
31,749 28,50121-
—
Coefficients
Values
I
Vert
Horiz. Unif.
Load
Soriz. Triang. Load
Coeff.1 *W  r
e
Coeff.2 C
r
e
oeff.3
*P
er
i Joeff.4
* 2
Total
1
2
3
Self
Load
4
1
2
3
4
kg
0
! 0.354
0.614
I- 0.709
5   0.614 > 0.354
ro
3 -0.354
9 -0.614
9 -0.709
1 -0.614 2 -0.354
3        0
0
-0.354' -0.614 -0.709 -0.614 -0.354
0
+0.354
+0.614
+0.709
+0.614 +0.354
0
0
-O.1’O7'
-0.204
-0.272
-0.282
-0.212
-0.065
+0.141
+0,331
+0.436
+0.409 +0*247
0
0
0.117
0.208
0.251
0.228 0.130
-0.044
-0.261
-0.405
-0.400
■0.266
■O.097
0
0
19,179
33,267
38,414
33,267 19,179
0
-.19,179
-33,267
-38,414
-33,267 -19,179
0
0
- 6,393 -11,089 -12,805 ^11,089
- 6,393
0 .
+ 6,393 +11,089 +12,805 +11,089
+ 6,393
0
0
- 322
- 614
- 819
- 849
- 638
- 184 +• 424. + 996 +1312 +1231 + 743
0
0
+281 +499 +602 +547 +312. -106
-626
-972
-960
-638
-233
0
0
12,745
22,063
25,392
21,876
12,478
- 290 -12,988
-22,154
-25,257
-21,585
-12,276
0
Reinforoement:
Thrust
d = 38om
d’ o 4 cm
d"«s - 4 t» 17cm Maximum moment at crown; (Seci 1)
M =» 12,661 logfm/m N «= 26,946 kgf/m3o NE s» 17,242 togm/m
Resisting moment kF o 15.65 * 10° + Q8)__  22,599kgsi/m 100
which is more than NE, so no compression steel will be needed.
Area of tension steel for the combined effect:
17,242 
0.874x38)
1406x0.874x0.38 
- 17.76 cm2/m—>0 25 5 27 (at inside)
- At springing (Seo.7)
M w 12,607 kgf/m Thrust: N e 78,686 kgf/m
U
E =, .l
2- 6o
78,686
e 0.330m
So: NE « 25,984lEgfm/m              , which is more than JjF , so compression steel will be needed.
Area of tension steel:
A„ "        88 ' Cl -
f .3.4
s
J.d o 0.87'. • ,i8 ■ 33.2om                   , which is more than E , %
so no tension steel will be needed*
i    2 i 100 + 17 = 33.0cm
1’
Area of compression steel: A « NE - k*?With: c = f  x
0
= 1406 x | x
(1
So: A’
=       489.5kgf/cm2 (25,984-22,599)
489.5 x 0.38
(1-0.3777)
= 18.20 cm2/m
0 25       27
Shear: Maximum
shear to be taken by
S = 38x100x4.24 =
the concrete:
16,1'!2 kgf/m
So in sections 3,4 and will be necessary
Take U-formed stirrups distances of          20 cm So;
5 and
around the main reinf crcanont
SB * * f s. d
in which: S
g
t 9,280 x
A s ’ 1,406 x 38
= part of the shear taken by the shear reinforcement
= 25,392 - 16,112 -- 9,280 kgf/m (See 4) = distance of stirrups
—        = 3.47cm2/m
Because per m’ there will be bars, the area per b 
will be :
x 3.47 = 0.94cm2 , provided d 16.4.         INTAKE STRUCTURE
Intake is provided with a stop log slot and upstreem of the stop log slot, the entrance is given. a. bell mouth shape with an elliptical curve.
Equation for the elliptical curue is adopted recommended by U.S.B.R,
' 1 'i
-^-2  +
(0.5D)2
» 1
Where D = diameter of conduit « 2.00m Therefor^equation of ellipse becomes
2 2 
z___ + y                  = 1
1             (0.30)2
The coordinates of ellipse ar^calculated below,
X
y
1
0
0.20
0.40
0.60
0.80
1.00
0.30^
0.29
0.27
0l24
0i18
0
r**"        •>
z
->■
•F
s,
Dimension of the slot and intake in .the figure on the next page.
walls are shown- 25
Drawing No. D.S. tot abowa the details of rexnforewae^t.5.   OUTLET
Outlet is provided with a diverging chf.c~‘'l 1 la P
for a length of 6 metres increasing the bed width of ch*rEt»A to 6 metres at the end. The side walls are flared out
ondir,g in a side slope of 1s1. This wtructuree i?* necessary to direct the diverted flow away from the to© o*
dam
Details of reinforcement are shown in drawing Xo6.   ACKNOWLEDGEMENT
The help of Mr. Simon de Haan is acknowledged with 
thanks for computing structural part of one portion of the
conduit.
7.   REFERENCES
A)  Design of small Dams
by United States Bureau of Reclamation, Denver, Colorado, U.S.A.
B)  Design of concrete structures by Winter, Urguhart, 
O’Rnurke, Nilson, McGraw Hill Book Company.
C)  Engineering Manual EM1110-2-2902- "Conduits Culverts and Pipes" by the Department of the Army, Office of the
Chief of Engineers, Washington D.C., U.S.A.
D) Engineering Monograph No.14 - " Beggs Deformeter Stress Analysis of Single Barrel Conduits" by United States Bureau of Reclamation, Denver, Colorado, U.S.A.

Summery

Project Overview

Project: Birete Dam Project

Document: Design of Diversion Structure

Author: Nafis Ahmad, UN Engineering Adviser

Date: August 1980

Location: Addis Ababa, Ethiopia

Project Code: ETH-75/005 (UNDP DTCD Project)

Abstract

The design report covers the diversion scheme proposed for the Birete Dam project, which includes leaving a gap for the first rainy season to enable an economical size of diversion conduit. The report details the hydraulic and structural design of the conduit, including inlet and outlet structures.

Key Design Considerations

Diversion planned to facilitate dam construction within stipulated time
Construction scheduled to be finished in first dry season
Gap left in embankment during second dry season to pass flood discharge
Gap dimensions determined by:
- Capacity to pass 25-year return period discharge
- Machine construction feasibility
- Proper bonding between construction phases
- Reasonable closure time

Gap Design

For passing flood discharge at end of first dry season:

25-year frequency discharge: 64 m³/sec
River bed slope: 1 in 160 meters
Depth in gap: 1.5 meters
Calculated discharge capacity: 77.7 m³/sec (with safety factor)
Total earthwork and rockfill volume: ~90,000 m³

Conduit Design

Hydraulic Design

5-year frequency discharge: 40 m³/sec
Circular conduit diameter: 2.00 m
Calculated velocity: 12.73 m/sec
Total head required: 9.32 m (including friction, entrance, and exit losses)
Conduit length: 85.9 m

Structural Design

The conduit is divided into lighter loaded portions and heavier loaded central portion:

Thickness: 0.30 m for lighter portions
Two loading cases considered for analysis
Reinforcement calculations performed for various sections
Shear stress analysis conducted
Heavier central portion has greater fill height (16.80 m)

Reinforcement Summary

Location	Loading Case	Required Reinforcement
Crown	Case (i)	14.1 cm² (tension inside)
Springing	Case (i)	4.7 cm² (tension outside)
Invert	Case (i)	13.0 cm² (tension inside)
Central Crown	Case (i)	17.76 cm²/m (Φ25@27cm inside)
Central Springing	Case (i)	18.20 cm²/m (Φ25@27cm compression)

Intake Structure

Features stop log slot
Upstream entrance has bell mouth shape with elliptical curve
Elliptical curve equation: x²/1 + y²/(0.30)² = 1
Detailed dimensions and reinforcement shown in drawings

Outlet Structure

Diverging channel with 1:3 slope for 6 meters length
Final bed width: 6 meters
Side walls flared to 1:1 slope
Designed to direct flow away from dam toe during construction

References

Design of Small Dams - USBR
Design of Concrete Structures - Winter et al.
Engineering Manual EM1110-2-2902 - US Army Corps of Engineers
Engineering Monograph No.14 - USBR