PHILIPPINE NATIONAL

STANDARD PNS/BAFS/PAES 229:2017

ICS 65.060.35

Design of a Diversion Dam

BUREAU OF AGRICULTURE AND FISHERIES STANDARDS

BPI Compound Visayas Avenue, Diliman, Quezon City 1101 Philippines
Phone (632) 920-6131; (632) 455-2856; (632) 467-9039; Telefax (632) 455-2858
E-mail: bafpsda@yahoo.com.ph
DEPARTMENT OF Website: www.bafps.da.gov.ph
AGRICULTURE
PHILIPPINES
PHILIPPINE NATIONAL STANDARD PNS/BAFS/PAES 229:2017
Design of a Diversion Dam

Foreword

The formulation of this national standard was initiated by the Agricultural

Machinery Testing and Evaluation Center (AMTEC) under the project entitled
“Enhancement of Nutrient and Water Use Efficiency Through Standardization of
Engineering Support Systems for Precision Farming” funded by the Philippine
Council for Agriculture, Aquaculture and Forestry and Natural Resources
Research and Development - Department of Science and Technology (PCAARRD -
DOST).

As provided by the Republic Act 10601 also known as the Agricultural and
Fisheries Mechanization Law (AFMech Law of 2013), the Bureau of Agriculture
and Fisheries Standards (BAFS) is mandated to develop standard specifications
and test procedures for agricultural and fisheries machinery and equipment.
Consistent with its standards development process, BAFS has endorsed this
standard for the approval of the DA Secretary through the Bureau of Agricultural
and Fisheries Engineering (BAFE) and to the Bureau of Philippine Standards
(BPS) for appropriate numbering and inclusion to the Philippine National
Standard (PNS) repository.

This standard has been technically prepared in accordance with BPS Directives
Part 3:2003 – Rules for the Structure and Drafting of International Standards.

The word “shall” is used to indicate mandatory requirements to conform to the

standard.

The word “should” is used to indicate that among several possibilities one is
recommended as particularly suitable without mentioning or excluding others.

iii
PHILIPPINE NATIONAL STANDARD PNS/BAFS/PAES 229:2017
Design of a Diversion Dam

CONTENTS Page

1 Scope 1
2 Definitions 1
3 Types of Diversion Dam 1
4 Selection of Diversion Site 3
5 Design Procedure 9
6 Bibliography 26

ANNEXES

A Flood Discharge Analysis 27

B Determination of Crest Shape 35
C Structural Stability Analysis 39
D Sample Computation 43

ii
PHILIPPINE NATIONAL STANDARD PNS/BAFS/PAES 229:2017

Design of a Diversion Dam

1 Scope

This standard specifies the minimum design requirements of a diversion dam.

This type of dam shall be provided across the water source in cases where water
is too low to divert water in order to raise its water level to facilitate irrigation
by gravity. The height of this type of dam ranges from 3 m to 5 m.

2 Definition

For the purpose of this standard, the following definitions shall apply:

2.1
afflux elevation
rise in maximum flood level from the original unobstructed flood level which
result after an obstruction to the flow such as a dam, has been introduced

2.2
diversion dam
structure or weir provided across the river or creek to raise its water level and
divert the water into the main canal to facilitate irrigation by gravity.

2.3
hydraulic jump
occurs when a thin sheet of incoming flow moving at high velocity strikes water
of sufficient depth

3 Types of Diversion Dams

The different types of diversion dams and suitability in site conditions are shown
in Table 1.

Table 1. Types of Diversion Dams

Type Description Site Conditions

Ogee - a weir wherein the upper curve of - for most sites under
the ogee is made to conform to the normal conditions
shape of the lower nappe of a
ventilated sheet of water falling from
a sharp-crested weir
- has a high discharge efficiency

1
Vertical Drop - a weir which produces free- - for mountain streams
discharging flows and dissipates with very steep slopes
overflowing water jet with the and a hydraulic jump
impact in the downstream apron cannot form, the drop
- not adaptable for high drops on height (from the weir
yielding foundation crest to the downstream
apron) should not exceed
1.50 m and the
foundation is firm and
unyielding
Glacis - a weir with a surface that slopes - for weirs not more than
gently downward from the crest to 1 m high located on
the downstream apron where only rivers with large, rolling
the horizontal component of the boulders and other
overflow jet takes part in the impact debris during flood
with the tailwater while the vertical condition
component is unaffected
-has stable and predictable hydraulic
jump
- most adoptable for rivers that have
heavy sediment loads
Gated - a weir where the larger part of the - for use in rivers or
ponding is accomplished by the solid creeks where the afflux
obstruction or the main body of the level would affect
weir populated or cropped
- additional head can be achieved by areas on the upstream
installing gates on the crest of the side of the weir
weir which can be collapsed or - for sites where the river
raised during floods has heavy sediment loads
during floods which
could be allowed to pass
through the gate
openings
Trapezoidal - weir with sloping upstream and - for weirs more than 1 m
downstream slopes which allow in height but not
boulders and debris roll over and hot exceeding 4 m, located
the downstream apron with less on rivers with large,
impact rolling boulders and
other debris during flood
condition
Corewall - used to stabilize the river bed for - to be used to stabilize
intake type diversion structures or to the river bed for sites of
gain a limited amount of diversion intake structures
head requiring only a minimal
- the external part of the weir additional diversion head
exposed to water flow is made of and where there is a
pure concrete while the inside part is need to maintain a

2
 filled with stones and cobbles which
smooth flow into the
provides a more economical section
intake
-the maximum height
from the crest to the
existing river bed is 0.50
m
ADOPTED FROM: Design of Concrete Gravity Dams on Pervious Foundation

4 Diversion Site Requirements

4.1 The site shall have a stable and firm foundation. An impervious
foundation is preferred. Otherwise, proper safeguards shall be included in the
design.

4.2 Adequate water supply of good quality shall be available to provide the
irrigation needs of the service area. There shall be no potential problems of
pollution or saline intrusion.

4.3 The diversion structure shall be located on a straight river channel reach
and shall be located at a certain distance before the next river curvature to avoid
scouring of its downstream banks.

4.4 The site should be selected such that only a short diversion canal will be
required. If a long diversion canal is inevitable, there shall be a low diversion
dam or weir.

4.5 The site shall have an adequate waterway width to allow the passage of
the maximum design flood without overtopping its banks.

4.6 A suitable high ground shall be present nearby such that guide banks and
protection dikes can be anchored.

4.7 Construction materials should be readily available at the site.

4.8 The site shall be accessible to transportation and with no right-of-way

problems.

4.9 There shall be minimal works required for diversion, coffer damming,
dewatering or other special works during construction.

4.10 There should be no adverse effects on the environment. If inevitable,

provisions in the design shall be made to eliminate or mitigate them.

4.11 Should there be several potential sites for diversion, the most economical
site that will provide a hydraulically efficient structure and structurally safe shall
be selected.

3
Figure 1. Plan View of a Diversion Dam Structure
SOURCE: NIA Design Manual for Diversion Dams

4
Figure 2. Section View of an Ogee Diversion Dam Structure
SOURCE: NIA Design Manual for Diversion Dams

5
Figure 3. Vertical Drop Diversion Dam Structure
SOURCE: NIA Design Manual for Diversion Dams

6
 Figure 4. Glacis Diversion Dam Structure
SOURCE: NIA Design Manual for Diversion Dams

7
Figure 5. Trapezoidal Diversion Dam Structure
SOURCE: NIA Design Manual for Diversion Dams

8
5 Design Procedure

Figure 6. Design Procedure for an Ogee Type Diversion Dam

5.1 Gather required design data.

5.1.1 Topographic map of the site covering a radius of at least two (2)
kilometers, with 1-meter contour interval and a scale of 1:1000 and location of
the boreholes

5.1.2 Rectified aerial photographs of the area.

5.1.3 Cross-section of the proposed dam axis and at least four (4) cross-
sections: two to be taken upstream at points along the river spaced 200 meters
apart and the other two at the downstream side of the dam line similarly spaced.
Each cross-section shall have the following details:
• Character of the river bed, the nature and kind of vegetation on the banks
and flood plains
• Water surface elevation at the time the survey was made
• Maximum flood level elevation as obtained by repeated inquiries from old
folks residing in the vicinity
• Ordinary water surface drawn at a scale of 1:100

9
5.1.4 Profile of the river bed following the center of the waterway extending at
least one kilometer both upstream and downstream of the dam axis with the
following details:
• Water surface line at the time of the survey
• Maximum flood line
• Scale of 1:1000 Horizontal and 1:100 Vertical

5.1.5 Photographs to show the kind of vegetation along the river banks and
flood plains for determining the coefficient of roughness

5.1.6 Boring logs of subsurface explorations shown with the cross-section of

the dam axis as well as other logs not taken along the dam axis

5.1.7 Cores of the borings for further evaluation and interpretation by the
designing engineer and also for use as information to bidders

5.1.8 Stream flow measurements and more comprehensive study of hydrologic

data.

5.2 Determine design flood discharge.

5.2.1 The following methods may be used if streamflow records are available:
• Slope-Area method
• Gumbel Method and other probability concepts of estimating frequency of
occurrence of floods

5.2.2 The following methods may be used if streamflow records are not
available:
• Correlation Method using Creager’s Formula
• Flood Formulas derived from Envelope Curve for the region
• Drainage Area versus Discharge Frequency Curves
• Rational formula
• Modified rational formula

5.3 Generate a plot of the tailwater rating curve.

5.3.1 Select the river cross-section 50 meters away from downstream.

5.3.2 Determine the corresponding discharges at different water levels using

the Slope-Area Method discussed in Annex A.

5.3.3 Plot the values of discharge and elevation on the x- and y-axis,
respectively to generate the tailwater rating curve.

10
 Figure 7. Tailwater Rating Curve

5.4 Determine the length of the diversion dam based on the type of
foundation material and using the formula below. Table 2 shows the
corresponding allowable maximum flood concentration for each type of
foundation material.

Table 2. Allowable Maximum Flood Concentration for Various Foundation

Material

Character of Foundation Allowable Maximum Flood Concentration

Material (m3/s/m)
Fine sand 5
Coarse sand 10
Sand and gravel 15
Sandy clay 20
Clay 25
Rock 50
ADOPTED FROM: Design of Concrete Gravity Dams on Pervious Foundation

𝑄
𝐿𝑚𝑖𝑛 =
𝑞 𝑎𝑙𝑙𝑜𝑤
where:

Lmin is the minimum required length of the dam (m)

Q is the maximum flood discharge (m3/s)
qallow is the allowable maximum flood concentration (m3/s/m)

5.4.1 The minimum stable river width shall be checked with the computed
minimum dam length. It is preferred to take the average of these values for the
length of the dam.

11
 𝑃𝑤 = 4.825𝑄 1⁄2
where:

Pw is the minimum required length of the dam, m

Q is the maximum flood discharge, m3/s

5.4.2 For upper flood plains, with sandy-loam as the dominant material, the
allowable maximum flood concentration shall not be greater than 5 m 3/s/m with
velocity not exceeding 1 m/s to avoid scouring.

5.5 Determine afflux elevation using a trial-and-error method based on the

figure below.

Figure 8. Afflux Elevation in an Ogee Dam

SOURCE: Design of Concrete Gravity Dams on Pervious Foundation

5.5.1 Compute for the unit discharge.

𝑄
𝑞𝑟 =
𝐿
where:

qr is the discharge per meter run (m3/s/m)

Q is the maximum flood discharge (m3/s)
L is the dam length (m)

5.5.2 Assume a first trial value of afflux elevation and compute for the
following:
𝑑𝑎 = 𝐸𝐿𝑎𝑓𝑓 − 𝐸𝑙𝐷⁄𝑆
𝑞𝑟
𝑉𝑎 =
𝑑𝑎
𝑉𝑎2
ℎ𝑎 =
2𝑔
where:

da is the depth of approach (m)

ELaff is the afflux elevation (m)
ElD/S is the elevation of the downstream floor (m)
Va is the velocity of approach (m3/s)

12
 qr is the discharge per meter run (m3/s/m)
ha is the head due to velocity of approach (m)
g is the gravitational acceleration (m/s2)

5.5.3 Determine the energy elevation and the head above the dam.

𝐸𝐿𝑒𝑛𝑒𝑟𝑔𝑦 = 𝐸𝐿𝑎𝑓𝑓 + ℎ 𝑎

𝐻 = 𝐸𝐿𝑒𝑛𝑒𝑟𝑔𝑦 − 𝐸𝐿𝑑𝑎𝑚
where:

ELenergy is the energy elevation (m)

ELdam is the dam elevation (m)
ELaff is the afflux elevation (m)
ha is the head due to velocity of approach (m)
H is the head above the dam

5.5.4 Determine the coefficient of discharge for free flow condition, C o, using
Figure 9.

5.5.5 Calculate for the coefficient of discharge for flow over submerged dam.

100 − % 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒
𝐶𝑠 = × 𝐶𝑜
100
where:

Cs is the coefficient of discharge for flow over

submerged dam
Co is the coefficient of discharge for free flow condition
% Decrease is the decrease in coefficient of discharge in
Figure 10

5.5.6 Solve for the supplied discharge per meter run, q s. The obtained value
shall be equal to the previously computed qr, otherwise, assume another value
for the afflux elevation and repeat the procedure above.

𝐶𝑠
𝑞𝑠 = × 𝐻3⁄2
1.811

13
Figure 9. Coefficient of Discharge for Ogee Crest (a) Vertical-Faced (b)
Sloping Upstream Face
SOURCE: Design of Concrete Gravity Dams on Pervious Foundation

14
 Figure 10. Characteristics of Flow Over Submerged Dams
SOURCE: Design of Concrete Gravity Dams on Pervious Foundation

15
5.6 Perform hydraulic jump analysis based on the figure below.

Figure 11. Hydraulic Jump in an Ogee Dam

SOURCE: Design of Concrete Gravity Dams on Pervious Foundation

5.6.1 Assume a value of d1 less than the height of the dam and compute for the
head loss due to velocity.
𝑞
𝑉1=
𝑑 1

𝑣12
ℎ𝑣1 =
2𝑔
where:

v1 is the velocity of water just upstream before formation of

the jump (m)
q is the discharge per meter run (m3/s/m)
d1 is the assumed depth of water just upstream before
formation of the jump (m)
hv1 is the head loss due to velocity (m)
g is the gravitational acceleration (m/s2)

5.6.2 The sum of d1 and hv1 shall be almost equal to the difference between the
energy elevation above the dam and the downstream apron elevation. Otherwise,
assume another value for d1 and repeat the procedure above.

5.6.3 Calculate the jump height, d2, or use the nomograph in Figure 12.

𝑑1 𝑑1 2 2𝑣12𝑑1
𝑑=− +√ +
2
2 4 𝑔

5.6.4 It must be noted that the position of the hydraulic jump on a horizontal
and smooth can hardly be predicted.

5.6.5 The length of the jump should approximately be five times the jump
height.

16
 Figure 12. Nomograph for Hydraulic Jump
SOURCE: Iglesia, Design of Concrete Gravity Dams on Pervious Foundation

17
5.7 Compute for the length of downstream apron or follow the recommended
length based on the Froude number as shown in Table 3.

𝐿𝑎 = 5(𝑑2 − 𝑑1)
where:

La is the length of the downstream apron (m)

d1 is the depth of water just upstream before formation of the
jump (m)
d2 is the depth of water just downstream of the formation of
the jump (m)

𝑣2
𝐹=
√𝑔𝑑
where:

v is the water velocity

d is the hydraulic depth
g is the gravitational acceleration

Table 3. Recommended Length of Downstream Apron

Froude Type of Basin Formula for Length

Number
< 4.5 Type I basin with dentated end sills La = 5(d2 − d1)
> 4.5 Type II basin with dentated end La = 3.5 d2
sills
> 4.5 Type III basin with dentated end La = 4 d 2
sills
ADOPTED FROM: Iglesia, Design of Concrete Gravity Dams on Pervious
Foundation

5.8 Determine the size of chute blocks, baffle blocks and end sill using the
recommended values in Table 4.

Table 4. Recommended Sizes of Chute Blocks, Baffle Blocks and End Sill

Size (cm)
Froude Number
Chute Blocks Baffle Blocks End Sill
< 2.85 30 60 30
> 2.85 40 to 60 80 to 120 30 to 40
ADOPTED FROM: Iglesia, Design of Concrete Gravity Dams on Pervious
Foundation

5.9 Determine the extent of riprap. The minimum length of riprap shall not be
less than 10 meters.

18
 𝐿 = 𝑐 × 𝑑2

if F < 4.5, 𝐿𝑅𝑎 = 1.5(𝐿 − 𝐿𝑎 )

if F > 4.5, 𝐿𝑅𝑎 = (𝐿 − 𝐿𝑎)

𝑞
𝑣2 = × 3.28
𝑑 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑

0.65𝐻𝑜 3
2
𝐿𝑅𝑏 = ( )2 × 𝑣 2
𝑑𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝐿
𝐿𝑅𝑎 + ( 𝑅𝑏 )
3. 28
𝐿𝑅 =
2
where:

L is the length of natural jump, m

c is the value from Figure 10
d2 is the depth of water just downstream of the formation of
the jump, m
LRa is the first value for the extent of riprap
La is the length of the downstream apron, m
LRb is the second value for the extent of riprap, ft
Ho is ELaff – ElD/S
dsupplied is the supplied tailwater depth, m
V2 is the average tailwater velocity, ft/s
LR is the extent of riprap, m

5.10 The size of riprap can be determined using two different methods: based
on bottom velocity and required stone diameter. For a well-graded riprap, it is
recommended to contain about 40% of the size smaller than the required.

5.10.1 Use Figure 13 for the corresponding size of riprap using the bottom
velocity. The bottom velocity shall be computed using the formula below.
However, if bottom velocity cannot be determined, the average tailwater velocity
is acceptable.

𝑣𝑏 = 2.57√𝐷
𝑞
𝑣2 =
𝑑𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
where:

vb is the bottom velocity (ft/s)

D is the weighted mean diameter of river bed materials (in)

5.10.2 Use the required stone diameter for determining the size of riprap.

19
 4 3
𝑊𝑅 = 𝜋𝑟 × 165
3
where:

WR = weight of riprap, lbs

r = required stone radius, ft

5.10.3 The riprap shall be have a thickness 1.5 times greater than the stone
diameter.

5.10.4 The riprap shall be provided with gravel blanket with a thickness half the
thickness of the riprap but not less than 12 inches.

5.11 Determine the depth of downstream cut-off wall.

𝑑𝑐𝑜 = 𝑅 − 𝑑𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
where:

dco = depth of downstream cut-off wall, m

R = depth of scour, m (Figure 14)
dsupplied = depth of tailwater supplied

5.12 Determine the crest shape detailed in Annex B.

5.13 Analyze structural stability detailed in Annex C.

5.14 A sample computation is shown in Annex D.

20
 Figure 13 . Tentative Curve in Determining Riprap Sizes
SOURCE: United States Bureau of Reclamation, Design of Small Dams, 1967

21
 Figure 14. Curve for Determining Depth of Scour
SOURCE: United States Bureau of Reclamation, Design of Small Dams, 1967

22
6 Bibliography

Chow, V.T. 1959. Open-Channel Hydraulics. New York: McGraw-Hill Book

Company, Inc.

Iglesia, G.N. n.d. Design of Concrete Gravity Dams on Pervious Foundation. n.p.

National Irrigation Administration. n.d. Design Manual on Diversion Dams. n.p.

National Resources Conservation Service. 2011. Conservation Practice Standard:

Dam Diversion

Stephens, T. 2010. FAO Irrigation and Drainage Paper 64: Manual on Small Earth
Dams. Rome: Food and Agriculture Organization of the United Nations

United States Bureau of Reclamation. 1967. Design of Small Dams.

23
 ANNEX A
(informative)

Determination of Flood Discharge

A.1 Slope-Area Method

A.1.1 Determine the required parameters:

Table A1 – List of Required parameters

Parameter Symbol Unit
Slope of the river bed Srb
Slope of flood water Sws
surface
Water cross-sectional area A m2
Wetted perimeter P m
Hydraulic radius R m
Roughness coefficient n
NOTE: If the value of Sws can’t be determined, use Srb as “S” in substituting with
Manning’s Formula

A.1.2 Calculate for the average velocity, V (m/s)

1 2 1
𝑉= 𝑅3 𝑆 2
𝑛

A.1.3 Determine the discharge, Q (m3/s)

𝑄=𝐴×𝑉

Table A2. Roughness coefficient for various channel conditions

Values of n Channel Condition

0.020 Smooth natural earth channels, free from growth, little
curvature
0.0225 Average, well-constructed, moderate-sized earth channels in
good condition
0.025 Small earth channels in good condition, or large earth
channels with some growth on banks or scattered cobbles in
bed
0.030 Earth channels with considerable growth; natural steams with
good alignment; fairly constant section; large floodway
channels, well maintained
0.035 Earth channels considerably covered with small growth;
cleared but not continuously maintained flood ways
0.040 – 0.050 Mountain streams in clean loose cobbles; rivers with variable
section and some vegetation growing in banks; earth channels
with thick aquatic growths
0.060 – 0.075 Rivers with fairly straight alignment and cross section, badly

24
 obstructed by small trees, very little underbrush or aquatic
growth
0.100 Rivers with irregular alignment and cross section, moderately
obstructed by small tees and underbrush; rivers with fairly
regular alignment and cross section, heavily obstructed by
small trees and underbrush
0.125 Rivers with irregular alignment and cross section, with growth
of virgin timber and occasional dense patches of bushes and
small trees, some logs and dead fallen trees
0.150 – 0.200 Rivers with very irregular alignment and cross section, many
roots, trees, bushes, large logs, and other drifts on bottom,
trees continually
falling into channel due to bank caving
0.035 Natural (wide) channel, somewhat irregular side slopes; fairly
even, clean and regular bottom; in light gray silty clay to light
tan silt loam; very little variation in cross section
0.040 Rock channel, excavated by explosives
0.045 Dredge channel, irregular side slopes and bottom sides
covered with small saplings and brush, slight and gradual
variations in cross sections
0.080 Dredge (narrow) channel, in block and slippery clay and gray
silty clay clay loam, irregular side slopes and bottom, covered
with dense growth
of bushes, some in bottom

A.2 Gumbel Method

A.2.1 For 20-year floods, the graphical linearization approach shall be used.

A.2.1.1 Select the the highest flood discharge, Q, for each year of the 20-year
record.

A.2.1.2 Arrange the annual peak discharges in descending order and rank them
from 1 to N where N is the number of years of record.

A.2.1.3 Compute for the probability that an event will be exceeded or equalled
and the probability that an event will not occur.

𝑚
𝑃=
𝑁+1

𝑃𝑟 = 1 − 𝑃
where:

P is the probability that the event will be exceeded

Pr is the probability that the event will not occur
m is the rank
N is the number of records

25
A.2.1.4 Plot the values of Pr and Q on the x- and y-axis of the Gumbel probability
paper, respectively.

A.2.1.5 Determine the best-fit line through the plotted points.

Q m P Pr
highest 1

lowest N

A.2.2 For 25-year, 50-year and 100-year floods, mathematical approach using
statistical principles shall be used.

A.2.2.1 Determine the standard deviation for the values of the annual peak
discharges.
∑(𝑄 − 𝑄̅)2
𝐷𝑠 = √
𝑁−1
∑𝑄
𝑄̅=
𝑁
where:

Ds is the standard deviation

̅Q is the mean flood discharge
Q is the annual peak discharge
N is the number of records

A.2.2.2 Compute for the reduced variate for the required return periods.

𝑇𝑟
𝑦 = −𝐿𝑛(𝐿𝑛 )
𝑇𝑟 − 1
where:

y is the reduced variate

Tr is the return period

A.2.2.3 Compute for the corresponding discharge using the following relation.

𝑦 = 𝑎′ (𝑄 − 𝑄̅) + 𝐶
𝑟
𝐷𝑠
where:

y is the reduced variate

Ds is the standard deviation
Qr is the corresponding discharge
a’,C is the factors (function of N)
26
 Table A3. Values of a’ and C

N a’ C
10 0.970 0.500
15 1.021 0.513
20 1.063 0.524
25 1.092 0.531
30 1.112 0.536
35 1.129 0.540
40 1.141 0.544
50 1.161 0.549
100 1.206 0.560
1000 1.269 0.574

A.2.2.4 Plot the values of Tr and Qr on the x- and y-axis of the Gumbel
probability paper, respectively.

A.3 Runoff Formula

Determine the flood discharge using the formula below.

𝑄 = 0.028 × 𝑃 × 𝑓 × 𝐴 × 𝐼𝑐
where:

Q is the flood discharge (m3/s)

P is the percentage coefficient for catchment characteristics
(Table A4)
f is the coefficient for storm spread (Figure A1)
A is the catchment area (ha)
Ic is the rainfall intensity (cm/h)

Table A4 – Percentage coefficient for various catchment characteristics

Type of Catchment Maximum Value of P

Steep bare rock 0.90
Rock, steep but wooded 0.80
Plateaus lightly covered 0.70
Clayey soils, stiff and bare 0.60
Clayey soils, lightly covered 0.50
Loam, lightly cultivated or covered 0.40
Loam, largely cultivated 0.30
Sandy soil, light growth 0.20
Sandy soil, covered, heavy brush 0.10

27
 Figure A1. Coefficient of Storm Spread based on Catchment Area

A.4 Rational Formula

Determine the flood discharge using the formula below.

𝑄 = 𝐹 × 𝐶 × 𝐼𝑀 × 𝐴
where:

Q is the flood discharge (m3/s)

F is the conversion factor
C is the runoff coefficient of the catchment (Table A5)
IM is the rainfall intensity (mm/h)
A is the catchment area (km2)

Table A5. Runoff coefficient for various catchment characteristics

Type of Catchment Recommended Values of C

Parks, lawns and gardens 0.05 – 0.25
Open or Unpaved areas 0.20 – 0.30
Light residential areas 0.25 – 0.35
Moderate residential areas 0.30 – 0.55
Dense residential areas 0.50 – 0.75
Suburban areas 0.45 – 0.55
Agricultural lands 0.15 – 0.25
Steep sloped watershed 0.55 – 0.70
Moderately sloped watershed 0.45 – 0.55
NOTE: If the catchment area is of more than one type, use the weighted average value of C.

28
A.5 Modified Rational Formula

Determine the flood discharge using the formula below.

𝑄 = 𝐾 × 𝐶 × 𝐼𝑐 × 𝐴
where:

Q is the flood discharge, m3/s

K is the empirical time correction factor to account for
decrease of infiltration with time
= 0.943Tc0.1044 for Tc < 1.75; Tc = time of concentration, h
= 1.0 for Tc > 1.75
C is the runoff coefficient of the catchment (Table A6)
IC is the rainfall intensity, cm/h
A is the catchment area, km2

Table A6. Runoff coefficient for various catchment characteristics

Type of Catchment Recommended Values of C

Low runoff condition C = 0.0000854 × (100.07)log10 Ic for Ic < 5
(exceptionally well-grassed cm/h
vegetation, sandy soil, flat C = 0.0001465 × (46.54)log10 Ic for Ic > 5
topography) cm/h
Moderate runoff condition
C = 0.0006149 × (17.29)log10 Ic
(good vegetation coverage, light
soil, gently sloping topography)
Average runoff condition
(good to fair vegetation, medium-
C = 0.002521 × (5.909)log10Ic
tectured soil, sloping to hilly
topography)
High runoff condition
(fair to sparse vegetation, heavy C = 0.005601 × (3.285)log10Ic
soil, hilly to steep topography)

A.6 Correlation Method

A.6.1 Select a similar river within the considered area. There shall be no
appreciable difference in the size of the drainage area, watershed characteristic.
There shall be hydrologic similarity in terms of rainfall, soil overcomplex, and
valley storage and geologic similarity with regard to groundwater flow.

A.6.2 Perform frequency distribution analysis for the river with streamflow
records using Gumbel Method.

A.6.3 Determine the C factor.

𝑄
𝐶=
√𝐴
29
 where:

C is the C factor
Q is the magnitude of flood in the gaging station (m 3/s)
A is the drainage area (km2)

A.6.4 Compute for the magnitude of flood in the river with no streamflow
record using the computed C factor.

𝑄𝑥 = 𝐶√𝐴𝑥
where:

C is the C factor
Qx is the magnitude of flood in the river with no streamflow
record (m3/s)
Ax is the drainage area of the river with no streamflow record
(km2)

A.7 Empirical Flood Formula

A.7.1 Determine the rare and occasional flows for the river with streamflow
records.

150𝐴
𝑄𝑟𝑎𝑟𝑒 =
√𝐴 + 17
85𝐴
𝑄𝑜𝑐𝑐𝑎𝑠𝑖𝑜𝑛𝑎𝑙 =
√𝐴 + 9
where:

Qrare is the rare flow for the river with streamflow record
(m3/s)
Qoccasional is the occasional flow for the river with streamflow
record (m3/s)
A is the drainage area of the river with streamflow
record (km2)

A.7.2 Determine the average of these values and use as the flow for the river
with no streamflow record.

A.8 Drainage Area Vs. Discharge-Frequency Curve

A.8.1 Perform frequency analysis using Gumbel Method for the rivers with
streamflow records within the same basin.

A.8.2 Select all 50-year and 100-year flood values and plot on the log-log paper
against their corresponding drainage area.

30
A.8.3 Determine the best-fit line through the plotted points. This line
represents the curve for the basin.

A.8.4 With the drainage area of the river with no streamflow records,
determine the flood value using the curve for the basin.

31
 ANNEX B
(informative)

Determination of Ogee Crest Shape

The following procedure for the overflow ogee crest is designed to fit the
underside of the nappe of a jet flowing over a sharp-crested weir which had been
found to be the most ideal for obtaining optimum discharges governed by the
equation below. Figure B1 shows the elements of an ogee crest profile.

𝑦 𝑥 𝑛
= −𝐾 ( )
𝐻 𝐻
where:

y is the vertical distance from the apex of the crest

x is the horizontal distance from the apex of the crest
H is the difference between the energy elevation and dam
crest elevation
K,n is the constants developed based on the upstream
inclination and velocity of approach (Figure B3)

Figure B1. Elements of an ogee crest profile

B.1 Using the maximum afflux, energy and dam crest elevations, determine
ha/H.

32
B.2 Determine the location of the apex of the crest (X c and Yc), R1 and R2 using
the computed value of ha/H and Figure B2.

B.3 Determine coefficients K and n from Figure B3 and using the formula,
complete the plotting table below.

𝑥 𝑛
𝑦 = −𝐾𝐻 ( )
𝐻

x x/H (x/H)n y

B.4 Plot the values to determine the crest shape as shown in Figure B4.

33
Figure B2. Values of Xc and Yc

34
Figure B3. Values of K and n coefficients

Figure B4. Plotting for the crest shape

35
 ANNEX C
(informative)

Structural Stability Analysis

C.1 The following are the basic assumptions in structural stability analysis:

C.1.1 The bearing power of the foundation can sustain the total loads from the
dam and other external.

C.1.2 The base of the dam is properly installed on undisturbed foundation.

C.1.3 There is homogeneity of concrete in all parts of the structure.

C.1.4 The dam may be considered as a single structure as long as the

construction joints are adequately provided with open slots or shear keys and
properly filled with concrete.

C.1.5 Uplift pressure under the dam is reduced by sufficient filter drains and
properly installed weep holes.

C.1.6 In the event of temporary abnormal loads, such as those produced by

earthquake shocks, adjustments in the allowable stresses and factors of safety
are permissible.

C.1.7 A cross-section of unit width under analysis is assumed independent of

adjoining sections and the beam action in the dam as a whole is disregarded.

C.1.8 The resistance offered by steel sheet piles or cut-offs against sliding and
overturning is disregarded.

C.2 The following factors of safety shall be considered:

C.2.1 Under normal stable condition, the factor of safety against overturning
ranges from 1.5 to 2 and can be reduced to 1 considering seismic forces.

C.2.2 If the ratio of the summation of all horizontal forces to the summation of
all vertical forces is equal to or less than the allowable sliding factor, f, the dam is
considered safe. Table C.1 shows the allowable sliding factor for various
foundation materials.

Table C.1. Allowable sliding factor for various foundation materials

Foundation Material Sliding Factor, f

Sound rock, clean and irregular surface 0.8
Rock, some jointing and laminations 0.7
Gravel and coarse sand 0.4
Sand 0.3

36
Shale 0.3
Silt and clay Laboratory test necessary

C.2.3 A concrete cut-off designed as a cantilever beam loaded with the

horizontal force that is in excess of the foundation’s resistance to sliding, will
prevent dam displacement.

C.3 The following are the conditions with which stability analysis should be
made. In all conditions, the resultant shall be located the middle third of the base
of the dam and the allowable bearing capacity of foundation materials shall be
less than the allowable value as shown in Table C.1.

• During maximum flood condition

• During normal operation condition when the water surface us at the same
level as the dam crest and tailwater is at the same level as the
downstream apron
• During construction

Table C1. Suggested Allowable Bearing Values for Footings of Structures

Appurtenant to Small Dams

Material Condition, Average Allowable Bearing

Relative Standard Pressure (tons/ft2)
Density or Penetration
Consistency Values
(Number of
Blows/Feet)
Massive igneous Sound (minor - 100
metamorphic or cracks
sedimentary rock allowed)
like granite, gneiss
and dolomite
Hard laminated rock - - 35
including bedded
limestone, schist
and slate
Sedimentary rock Shattered or - 10
including hard broken
shales, sandstones
and thoroughly
cemented
conglomerates
Gravel (GW, GP, GM, - - 4
GC)
Cohesionless sands Loose 4 to 8 Requires compaction
(SW, SP) Medium 8 to 40 Requires compaction
Dense 8 to 40 1
Saturated cohesive Soft 4 0.25

37
sands, silts and clays Medium 4 to 10 0.50
(SM, SC, ML, CL, MH, Stiff 11 to 20 1.0
CH) Hard 20 1.5
NOTE: Unsound shale is treated as clay.
GW denotes group symbol for well-graded gravels, gravel-sand mixtures with
little or no fines
GP denotes group symbol for poorly graded gravels, gravel sand mixtures
with little or no fine
GM denotes group symbol for silty gravels, poorly graded gravel-sand-
silt mixtures.
GC denotes group symbol for clayey gravels, poorly graded gravels, poorly
graded gravel-sand-clay mixtures.
SW denotes group symbol for well-graded sands, gravelly sands with
little or no fines.
SP denotes group symbol for poorly graded sands, gravelly sands with
little or no fines.
SM denotes group symbol for silty sands, poorly graded –silt mixtures.
SC denotes group symbol for clayey sands, poorly graded sand-clay
mixtures.
ML denotes group symbol for inorganic silts and very fine sands, rock flour
silty or clayey fine silts with slight plasticity.
CL denotes group symbol for inorganic clays of low to medium plasticity,
gravelly clays, sandy clays, silty clays lean clays
MH denotes group symbol for inorganic silts micaceous or diatomaceous fine
sandy or silty soils, elastic silts.
CH denotes group symbol for inorganic clays o high plasticity, fat clays.
REFERENCE: United States Bureau of Reclamation. 1967. Design of small dams.

C.4 The following forces shall be taken into consideration during stability
analysis:

• Water pressure (external and uplift)

• Silt pressure
• Earthquake
• Weight of the structure
• Resulting reaction of the foundation

C.4.1 Water Pressure

C.4.1.1 Determine the external forces in the dam using Figures C.1 and C.2.

C.4.1.2 Determine uplift pressures using either of the accepted methods:

• Bligh’s Line-of-Creep Theory

• Lane’s Weighted- Creep Ratio Principle
• Flow Net Analysis
• Khosla’s Method

38
C.4.1.3 The procedures described below is based on Lane’s Weighted- Creep
Ratio Principle. Other established methods preferred by the designer may be
used.

C.4.1.3.1 Verify that there is no short path condition such that the distance
between the bottom of two successive cutoffs is greater than or equal to half of
the weighted creep distance between them.

C.4.1.3.2 Determine if the foundation material is safe.

C.4.1.3.3 Determine the uplift head under the assumption that the drop in
pressure head from headwater to tailwater along the contact line of the dam and
the foundation is proportioned to the weighted-creep distance.

C.4.2 Silt Pressure – It may be assumed that a horizontal pressure due to silt
load is equivalent to 85 lbs/ft3 fluid pressure and vertical weight of 120 lb/ft3.

C.4.3 Earthquake Forces

C.4.3.1 The effect of earthquake forces on the gravity dam itseld is determined
by applying 0.15g for horizontal acceleration to the energy formula at the center
of gravity of the dam.

C.4.3.2 The effect of horizontal earthquake on water pressure is determined by

the hydrodynamic pressure using Figures C3 and C4.

C.4.4 Weight of the structure – It includes the weight of the concrete and
appurtenance structure where the unit weight of concrete is estimated at 150
lbs/ft3 and the sectional weights act vertically through the center of gravity of
each subsection.

C.4.5 Reaction of the Foundation – Determine the foundation reaction at the toe
and heel of the dam similar to analysing the stability of retaining walls.

39
 ANNEX D
(informative)

Sample Computation

D.1 Design Data

Drainage Area A 780 km2

Drainage Area at Gaging Station Ag 900 km2
Return Period 100 years
Maximum Allowable Flood q 15 m3/s/m
Concentration
Afflux Elevation ELaff
Upstream Elevation ELU/S
Downstream Elevation ELD/S
Tailwater Elevatiion ELTW
Tailwater depth dTW 7.50 m
Energy Elevation ELenergy
Dam Crest Elevation ELdam crest
Dam Crest Height P 3.00 m
Free Flow Coefficient Co From Figure

D.2 Determination of the Design Flood Discharge

D.2.1 Using Empirical Formula

(A.A. Villanueva and A.B. Deleña’s Flood Formulas for Central Luzon)

150A 150 × 780

Qrare = = = 4170 m3⁄s
√A + 13 √780 + 13
85A 85 × 780
Qocc = = = 2370 m3⁄s
√A + 9 √780 + 9

D.2.2 Using Drainage-Area-Discharge-Frequency Curve

Using Figure A-2

Q = 3700 m3/s

D.2.3 Using Correlation Method

D.2.3.1 Acquire recorded annual peak flow at gaging station

Year Flow Rate (m3/s) Year Flow Rate (m3/s)
1951 2000 1959 950
1952 2500 1960 800
1953 1900 1961 1200

40
 1954 2300 1962 3000
1955 1500 1963 1000
1956 1100 1964 850
1957 1300 1965 900
1958 1800

D.2.3.2 Perform frequency distribution analysis by Gumbel Method

Magnitude ̅
𝐐−𝐐 ̅ )𝟐
(𝐐 − 𝐐
(in descending order)
3000 1460 2132600
2500 960 921600
2300 760 577600
2000 460 211600
1900 360 129600
1800 260 67600
1500 -40 1600
1300 -240 57600
1200 -340 115600
1100 -440 193600
1000 -540 291600
950 -590 348100
900 -640 409600
850 -690 476100
800 -740 547600
∑ = 23100 ∑ = 6482000

23100
̅Q= = 1540 m3 ⁄s
15
D.2.3.2.1 Determine the standard deviation.
∑(Q − ̅Q) 6482000
Ds = √ = √ = 680
N−1 14
D.2.3.2.2 Determine the reduced variate.
Tr 100
y = −ln (ln ) = −2.3 log (2.3 log ) = 4.61
Tr − 1 99

D.2.3.2.3 Using the other equation for reduced variate and values of a’ and C
from Table, determine the discharge at T =100 for the gaging station.

a′ 1.021
̅
y= (Qr − Q) + C = (Q
100 − 1540) + 0.513
Ds 680

Q100 = 4300 m3⁄s

D.2.3.2.4 Using Creager’s Formula,
Q100 9300
C= = = 143
√A √900

41
D.2.3.2.5 Apply the above C-factor to the proposed damsite to determine
discharge.

Q100 = C√A = 143√780 = 4000 m3⁄s

Method Discharge (m3/s)

Empirical Formula 3270
Drainage Discharge-Frequency Curve 3700
Correlation Method 4000
Average 3656.67
Design Flood Discharge = 3700 m3/s

D.3 Determine and Plot the Tailwater Rating Curve

D.4 Determination of the Length of Diversion Dam

D.4.1 Determine the allowable maximum flood concentration, qallow from Table
2.

D.4.2 Calculate minimum required length, Lmin.

D.4.2.1 Using the formula,

Q 3700 m3⁄s
L= = = 247 m
qallow 15 m 3⁄s /m

D.4.2.2 Using Lacey’s formula,

Pw = 2.67Q1⁄2 = 2.67 × (13000 ft 3 ⁄s)1⁄2 = 965 ft or 294 m

D.4.2.3 Take the average as the minimum required length.

L + Pw 247 m + 294 m
Lmin = = = 270.50 m or 270.00 m
2 2

D.5 Determination of Afflux Elevation

D.5.1 Set the required dam crest level and tailwater depth reckoned from the
upstream apron as illustrated below.

42
D.5.2 Compute for q using trial and error method until q = qr.

Q 3700 m3⁄s
qr = = = 13.70 m3 ⁄s /m
Lmin 270 m

D.5.2.1 At afflux elevation of 48.10 m,

da = EL aff − ELD⁄S = 48.10 m − 40.00 m = 8.10 m

qr 13.70 m2 ⁄s
Va = = = 1.70 m⁄s 8.10
d a
Va (1.70)2
ha = = = 0.148 m
2g 19.6

EL energy = EL aff + ha = 48.10 m + 0.148 m = 48.248 m H = EL energy − EL dam crest =

48.428 m − 43 m = 5.248 m

P 3.00 m
= = 0.57; Co = 3.82 (from Figure 8)
H 5.248 m

hd = EL energy − ELTW = 48.248 m − 47.50 m = 6.748 m

hd 0.748 m hd + dsupplied H 0.748 m + 7.50 m

= = 0.142; = = 1.57
H 5.248 m 5.248 m
% Decrease = 16% (from Figure 9)

100 − % Decrease 100 − 16

Cs = × Co = × 3.82 = 3.21
100 100

Cs 3.21 m3⁄s m3⁄s

qs = × H3⁄2 = × 5.2483⁄2 = 21.20 > 13.70 required
1.811 1.811 m m

43
D.5.2.2 At afflux elevation of 47.80 m,

m3⁄s m3⁄s
qs = 17.40 > 13.70 required
m m

d.5.2.3 At afflux elevation of 47.65 m,

m3⁄s m3⁄s
qs = 13.70 = 13.70 required
m m

Thus, afflux elevation = 47.65 m and energy elevation = 47.81 m

D.6 Hydraulic Jump Analysis and Determination of Length of

Downstream Apron

D.6.1 High Stage Flow

D.6.1.1 At d1=1.50 m

q 13.70 m2 ⁄s
V1 = = = 9.15 m/s
d1 1.50 m

v12 (9.15 m/s)2

hv1 = = = 4.27 m
2g 19.6 m/s 2

HE = hv1 + d1 = 4.27 m + 1.50 m = 5.77 m < 7.814 m

D.6.1.2 At d1=1.20 m,

HE = 7.82 m ≅ 7.814 m

D.6.1.3 Jump Height

d1 d 12 2v1 2 d1 1.20 1.202 2(11.40)2 (1.20)

d2 = − + √ + =− +√ +
2 4 g 2 4 9.81

d2 = −0.60 + 5.70 = 5.10 m < 7.50 m

d2 theoretical 5.10 m
= = 0.68; okay
d2 supplied 7.50 m
v1 11.40 m/s
F= = = 3.32; Type 1 Basin
√gd1 √9.81 × 1.20 m

D.6.1.4 Length of Downstream Apron

44
 La = 5(d2 − d1 ) = 5(5.10 m − 1.20 m) = 19.50 m ≈ 20.00 m

D.6.1.5 Summary of Values for High Stage Flow

Parameter Value
Q 3700 m3/s
ELTW 47.50 m
qrequired 13.70 m3/s/m
dsupplied 7.50 m
ELenergy 47.81 m
Co 3.82
Cs 3.21
q 13.70 m3/s/m
d1 1.2 m
d2 5.1 m
F 3.32
La 19.50 m

D.6.2 Low Stage Flow

D.6.2.1 For Q = 2500 m3/s, repeat procedure C.4 to C.6.

Parameter Value
Q 2500 m3/s
ELTW 46.10 m
qrequired 9.25 m3/s/m
dsupplied 6.10 m
ELenergy 46.40 m
Co 3.835
Cs 2.68
q 9.21 m3/s/m
d1 0.90 m
d2 4m
F 3.48
La 17.80 m

D.6.2.2 For Q = 1000 m3/s, repeat procedure C.4 to C.6.

Parameter Value
Q 1000 m3/s
ELTW 43.20 m
qrequired 3.70 m3/s/m
dsupplied 3.20 m
ELenergy 44.55 m
Co 3.835
q 9.21 m3/s/m

45
 d1 0.42 m
d2 2.39 m
F 4.35
La 12.10 m

D.6.2.3 Since the determined La for low stage flow are less than 20 m, use La = 20
m.

D.7 Determination of the Extent of Riprap

L = c × d2 = 5.70 × 5.10 m = 29.10 m

LRa = 1.5(L − La ) = 1.5(29.10 − 20.00) = 13.65 m

q 13.70 m3 /s/m m
v2 = = = 1.83 = 6.00 ft/s
d supplied 7.50 m s

LRb = ( 0.65Ho 3
)2 × v2 = (2 0.65 × 7.65 3
)2 × 6.00 2 = 19.50 ft
d supplied 7.50

Rb L 19.50 ft
LRa +( 3.)28 13.65 m +
LR = = 3.28 = 9.80 m ≈ 10.00 m 2
2

D.8 Determination of the Size of Riprap

D.8.1 Using the average tailwater velocity, V2 and Figure 12, consider 6.3-in
diameter or 12-lb riprap.

For Q = 3700 m3/s,

q 13.70 m3 /s/m m
V2 = = = 1.83
d supplied 7.50 m s

For Q = 2500 m3/s, V2 = 1.52 m/s and for Q = 1000 m3/s, V2 = 1.16 m/s.

D.8.2 From the diameter, the weight can be calculated as follows:

4 4 6.3 3
WR = πr 3 × 165 = π ( ) × 165 = 12.23 lb
3 3 2 × 12

For a greater factor of safety, use 10-in diameter or 50-lb riprap with gravel
blanket underneath.

D.8.3 The riprap thickness is 0.375 m ≈ 0.40 m while the gravel blanket
thickness is 0.20 m.

D.9 Determination of the Depth of the Downstream Cut-off Wall

46
 q (m3/s/m) R (Depth of scour from Required Depth of
Figure 14, m) Downstream Cut-off
Wall (m)
13.70 7.62 0.12
9.25 5.80 -
3.70 3.35 1.15

From the tabulated values, 1.15 m governs. Considering a factor of safety, 8’00”
steel sheet piles will be used.

D.10 Determination of Crest Shape

ha 0.16
= = 0.033
H 4.81

X c /Ho = 0.267 and Yc /Ho = 0.113

X c = 1⁄4 Ho = 0.267 = 1⁄4 (4.81) = 1.20m Yc = 1⁄8 Ho =

1⁄ (4.81) = 0.60m
8
x 1. 855
y = 0.507Ho ( )
ho

Filling up the plotting table,

x x/Ho (x/Ho)n y x x/Ho (x/Ho)n y

0.20 0.041 0.003 -0.007 2.40 0.500 0.280 -0.68
0.40 0.082 0.010 -0.024 2.60 0.540 0.320 -0.78
0.60 0.125 0.020 -0.049 2.80 0.580 0.370 -0.90
0.80 0.166 0.036 -0.087 3.00 0.620 0.410 -1.00
1.00 0.210 0.055 -0.134 3.40 0.700 0.520 -1.27
1.20 0.250 0.077 -0.187 3.80 0.790 0.650 -1.59
1.40 0.290 0.100 -0.244 4.00 0.830 0.710 -1.73
1.60 0.330 0.130 -0.320 4.40 0.910 0.840 -2.04
1.80 0.370 0.160 -0.390 4.80 0.997 1.000 -2.43
2.00 0.420 0.220 -0.490 5.00 1.030 1.056 -2.57
2.20 0.460 0.240 -0.590 5.40 1.120 1.230 -3.01

47
D.11 Stability Analysis

D.11.1 Checking for short path condition,

Length of creep from A to C = 2.00 + 1.80 + 0.80 + 4.48 + (5.00/3) = 10.74 m

Distance ̅BC= √5.002 + 3.482 = √37.20 = 6.10m > 10.74

2

Thus, there is no short path condition.

D.11.2 Analyze under maximum flood condition.

48
D.11.2.1 Length of creep to point F = 2.00 + 1.80 + 0.80 + 4.48 + 1.60 + (13.00 +
20.00/3) = 26.16 m

D.11.2.2 Pressure heads above the ogee crest:

Point Distance from y h

crest/Ho
10 1.25 0.82 Ho 6.80
11 1.17 0.80 Ho 6.70
12 0.56 0.80 Ho 4.50
13 0.215 0.70 Ho 3.40
0 - 0.60 Ho 2.79
1 - - 2.79

D.11.2.3 Calculation for hydrostatic pressure on the dam

Total Length of Creep 26.16

C= = = 174.00
Diff. in Water Levels 0.15

Point Length of Available Head loss, Net head, h Pressure,

Creep, Lc (m) Head Lc/c (m) P = 205 n
(m) (psf)
0 - 2.79 - 2.79 572
1 - 2.79 - 2.79 572
2 - 7.65 - 7.65 1,570
3 5.47 7.85 0.030 7.82 1,600

49
 4 6.27 8.65 0.040 8.61 1,770
4 15.23 8.65 0.09 8.56 1,730
5 15.96 8.65 1.100 8.56 1,750
6 17.36 7.25 0.11 7.15 1,470
7 18.22 7.25 0.113 7.145 1,460
8 19.75 8.65 0.116 8.537 1,750
9 20.28 8.65 - 8.53 1,750
10 - 7.80 - 6.80 1,390
11 - 6.70 - 6.70 1,380
12 - 4.50 - 4.50 920
13 - 3.40 - 3.40 700

This value is a conservative assumption. It would still be safe to assume also that
the pressure head at point (1) be equal to PO + 0.60 m for this particular
example.

50
 External forces (lbs) Lever arm (m) Moment about toe (m-
lbs)
Righting Overturning
= (572)(1.20)(3.28) = 2,251 ↓ 5.80 + (120)/2 = 6.40 14,406

= [(572+1570)/2](3.00)(3.28) 1.00+ 23,818

=10,539 → (3.00/3)[(1570+1144)/2140]=2.26
(1570)(1.00)(3.28)=5150 ↓ 7.00+(1.00/2)=7.50 38,625

[(1600+1770)/2](0.80)(3.28)= (0.80/3)[(1770+3200)/3370]= 1,525

4,421 → 0.345
(1750)(2.20)(3.28)=12,628 ↑ 5.80+(2.20/2)=6.90 87,133
[(1750+1470)/2](1.40)(3.28)= (1.40/3)[(1750+2940)/3220]=0.68 5,027
7,393 ←
[(1470+1460)/2](2.60)(3.28)= 3.20+(2.60/3)[(1460+2940)/2930]=4.50
56,218
12,493 ↑
[(1460+1750)/2](1.40)(3.28)= 1.60+(1.60/3)[(1750+2920)/3210]=2.38
20,047
8,423
[(1460+1750)/2](1.40)(3.28)= (1.40/3)[(1750+2920)/3210]=0.68
5,012
7,370 →
(1750)(1.60)(3.28)=9,184 ↑ (1.60/2)=0.80 7,347
[(1390+1380)/2](90.40)(3.28)=1,817 (0.40/3)[(1390+2760)/2770]=0.20 363
↓
[(1380+920)/2](2.80)(3.28)= 0.40+(2.80/3)[(1380+1840)/2300]=1.71 18,061
10,562 ↓
[{1380+920)/2](2.20)(3.28)= 1.00+(2.20/3)[(1380+1840)/2300]=2.03 5,142
2,533 ←
[(920+700)/2](1.60)(3.28)= 3.20+(1.60/3)[(920+1400)/1620]=3.70 15,728
4,251 ↓
[(920+700)/2](0.65)(3.28)= 3.20+(0.65/3)[(920+1400)/1620]=3.40 5,872
1,727 ←
[(700+572)/2](1.00)(3.28)= 4.80+(1.00/3)[(700+1144)/1272]=5.28 11,014
2,086 ↓
[(700+572)/2](0.15)(3.28)= 3.85+(0.15/3)[(700+1144)/1272]=3.92 1,227
313 ←

∑MR = 115,465; ∑MO = 201,100; ∑FV = 16,611; ∑FH = 10,364

Weight per ft. Strip (lbs) Lever arm (m) Moment about toe
(m-lbs)
Righting Overturning
(1.00)(1.00)(1,615)=1,615 ↓ 7.00+(1.00/2)=7.50 12,113

[(3.40+4.00)/2](1.20)(1,615)= 5.80+(1.20/3)[(4.00+6.80)/7.40]= 45,750

7,171 ↓ 6.38
[(2.60+2.45)/2](1.00)(1,615)= 4.80+(1.00/3)[(2.15+5.20)/5.05]= 21,613
4,078 ↓ 5.30
[(2.45+1.80)/2](1.60)(1,615)= 3.20+(1.60/3)[(1.82+4.90)/4.25}= 22,184
5,491 ↓ 4.04
(1/2)(1.80)(2.40)(1,615)= 0.80+(2/3)(2.40)=2.40 8,371
3,468 ↓
(2.40)(0.40)(1,615)=1,550 ↓ 0.80+1.10=1.90 2,945
(1/2)(1.10)(0.80)(1,615)=710 ↓ 1.60+(1.60/3)=2.13 1,512
(1.60)(1.00)(1,615)=2,584 ↓ (1.60/2)=0.80 2,067
∑W = 26,687; ∑Mw = 116,555

51
SUMMARY: ∑ M = 115465 + 116555 − 201100 = 30920 m − lb ↺

∑ V = 26687 + 16111 − 10,567 lb ↓

∑ H = ΣFH = 10364 lb →
ΣM 30920 800
x̅= = = 2.92m < (within the middle third) OK
ΣV 10567 3
B
e = − x̅ = 4.00 − 2.92 = 1.08m
2

D.11.2.4 Foundation reactions

ΣV 6e 6x1. 08
E = (1 + 10567 (1 + ) = 729 psf < 2000
toe )=
B B 8. 00x3. 28 8. 00
ΣV 6e
f = (1 − ( )( )
heel ) = 403 0.19 = 77 psf OK safe
B B
ΣMR 232020
Factor of Safety against Overturning = = = 1.18 < 1.5 unsafe
ΣMo 201100
ΣM 10364
Sliding Factor = = = 0.97 >
ΣV 10567
0.4 maximum allowable for Gravel and Sand

D.11.2.5 Recommendation

The first trial section should be modified, say lowering further the middle
portion of the base to attain additional weight and then stability analysis shall be
made for new section and repeated if necessary until the dam is found to have
adequate factors of safety.

D.11.3 Analyze under normal operation condition.

For the purpose of illustration, adopt the first trial section in the stability
analysis for this condition in order to have additional factor of safety, it will be
assumed that the weep holes are all clogged up and the upstream water surface
is in level with the dam crest (El. 43.00) while the tailwater elevation is flushed
with the downstream apron floor is, El. 40.00.

52
D.11.3.1 Length of creep to point F = 2.00 + 1.80 + 0.80 + 4.48 + 1.60 + (13.00 +
20.00/3) = 26.16 m

total length of creep 26.16

C= = = 8.72 > 5 for coarse sand OK
diff. in water levels 3.00

D.11.3.2 Calculation for hydrostatic pressure on the dam

Point Length Available Head loss, Net head, Pressure,

Of creep Head (m) Lc/c (m) h (m) P = 205 n
Lc (m) (psf)
0 --- 0.00 --- 0.00 0.00
1 --- 0.60 --- 0.60 123
2 --- 3.00 --- 3.00 615
2 --- 3.00 --- 3.00 615
3 5.47 3.20 0.62 2.58 528
4 6.27 4.00 0.72 3.28 670
4 15.23 4.00 1.75 2.25 460
5 15.96 4.00 1.82 2.18 448
6 17.36 2.60 1.98 0.62 127
7 18.22 2.60 2.09 0.51 104
8 19.75 4.00 2.26 1.74 356
9 20.28 4.00 2.32 1.68 345

External Forces (lbs) Lever arm (m) Moment about toe

(m-lbs)
Righting Overturni
ng
(1/2)(123)(1.20)(3.28)= 5.60+(2/3)(1.20)=6.60 1,597
242 ↓
(1/2)(615)(3.00)(3.28)= 1.00+91/3)(3.00)=2.00 6,050
3,025
(615)(1.00)(3.28)=2,028 ↓ 7.00+(1.00/2)=7.50 15,200

[(528+670)/2](0.80)(3.28)= (0.80/3)[(670+1086)/1198]= 610

1,570 → 0.39
[(460+448)/2](2.20)(3.28)= 5.80+(2.20/3)[(448+920)/908 22,600
3,260 ↑ ]=6.91
[(448+127)/2](1.40)(3.28)= (1.40/3)[(448+254)/575]=0.5 840
1,480 ← 7
[(127+104)/2](2.60)(3.28)= 3.20+(2.60/3)[(104+254)/231 4,450
980 ↑ ]=4.54
[(104+356)/2](1.60)(3.28)= 1.60+(1.60/3)[(356+208)/460 2,730
1,210 ↑ ]=2.26
[(104+356)/2](1.40)(3.28)= (1.40/3)[(356+208)/460]=0.5 605
1,060 → 7
[(356+345)/2](1.60)(3.28)= (1.60/3)[(345+712)/701]=0.8 1,480
1,840 ↑ 0

∑MR = 17,637; ∑Mo = 38,525; ∑FV = 5,028 ↑; ∑FH = 4,175 →

53
∑W = 26,687 lb ↓; ∑MW = 116,555 m-lb

SUMMARY: ∑ M = 17637 + 116555 − 38525 = 95667m − lb ↺

∑ V = 26687 − 5028 = 21659 lb ↓

∑ H = F = 4175 lb →
ΣM 95667
x̅= = = 4.46m (within the middle third)
ΣV 21656

B
e = − x̅ = 4.00 − 4.46m = 0.46m
2

D.11.3.3 Foundation reactions

ΣV 6e
E = (1 − 21659 6x0. 46
)= (1 + )
toe B B 8. 00x3. 28 8. 00
ΣV 6e
f = (1 + ( )( )
heel ) = 825 1.345 = 1110 psf < 2000 safe
B B
ΣM 134192
FS against overturning = = = 4.48 > 1.5 very safe
M 38525
ΣH 4175
Sliding Factor = = = 0.19 < 0.4 safe
ΣV 21656

D.11.4 Analyze under normal operation condition but with seismic forces.

D.11.4.1 Calculation for earthquake forces

D.11.4.1.1 Lateral force Due to Dam Weight Using horizontal Acceleration, or

0.15g

W W
FD = a = (0.15g) = 0.15W = (0.15)(26687) = 4003 lb →
g g

Mtoe = (4003)(1.86) = 7446 m − lb

54
D.11.4.1.2 Lateral Force Due to Hydrodynamic Force

Fw = 0.583H200 /g

Fw = (0.583)(0.15)(62.5)H2 = 5.49H2 = (5.49)(3.00x3.28)2 = 522 lb → Mtoe = (522)(1.00 + 4 x 3.00) = (522)(2.20) =

1148 m − lb ↻

D.11.4.1.3 Combine with D.10.3,

∑ MR = 17637 + 116555 = 134192 m − lb ↺

∑ MO = 38525 + 7446 + 1148 = 47119 m − lb ↻

∴ ∑ M = 87073m − lb ↺

∑ V = 21659 lb

∑ H = 4175 + 4003 + 522 = 8800 lb →

ΣM 87073
∴ x̅= = = 4.02m (within the middle third) OK
ΣV 21659

B
e= − x̅ = 4.00 − 4.02m = −0.02
2

D.11.4.1.4 Foundation reaction

ΣV 6e 6x0.02
E = (1 − 21659 (1 − ) = (825)(1 − 0.015) = 812 psf
toe )=
B B 8. 00x3. 28 8. 00
ΣV 6e
f = (1 + ( )( )
heel ) = 825 1.015 = 835 psf OK safe
B B
ΣMR 134192
FS against overturning = = = 2.86 > 1.5 safe
ΣMO 47119

ΣH 8800
Sliding Factor = = = 0.405 ≈ 0.40 failry OK
ΣV 21656

D.11.5 Analyze under construction condition.

ΣMW 116555
x̅= = = 4.37m (within the middle third) OK
ΣW 26687
B
e= − x̅= 4.00 − 4.37m = −0.37 2
ΣW 6e 26687 6x0.37
fheel = (1 + )= (1 ) = 1310 psf < 2000 safe
B B 8.00x3.28 8.00
ΣV 6e
Ftoe = (1 − ) = (1020) = 730 psf
B B

55
D.11.6 Determine tensile reinforcement at point 7 of the dam section due to
uplift during normal operation condition.

D.11.6.1 Calculate foundation reaction at point 7.

3.20
f7′ = 540 + = (570) = 540 + 228 = 768 psf
5.00
Under D.10.3, hydrostatic pressures at point (7)(8)(9) are as follows:

P7 = 104 psf; P8 = 356 psf; P9 = 345 psf

56
 Forces and/or weights (lbs) Lever arm (m) Moment about toe (m-
lbs)
Righting Overturning
P7-8v (1.60/3)[(104+712)/460]= 1,147
=[(104+356)/2](1.60)(3.28)=1,207 0.95
→
P7-8H (1.40/3)[(104+712)/460]=0.83 876
=[(104+356)/2](1.40)(3.28)=1,056
↑
P8- 1.60+(1.60/3)[(356+690)/701]=2.40 4,414
9=[(356+345)/2](1.60)(3.28)=1,839
↑
P7- (3.20/3)[(768+1080)/1308]=1.51 10,365
9=[(768+540)/2](3.20)(3.28)=6,864
↑
Wa=(1.60)(1.00)(1615)=2,584 ↓ 1.60+(1.60/2)=2.40 6,202
Wb=((1/2)(1.10)(0.80)(1615)=710 1.60-(1.10/2)=1.23 873
↓
Wc=(2.40)(0.40)(1615)=1,550 ↓ 1.30 2,015
Wd=(1/2)(1.50)(1.20)(1615)=1,454 0.90+(1.50/3)=1.45 2,108
↓
We=(1/2)(0.90)(1.20)(1615)=872 ↓ (2/3)(0.90)=0.60 532

∑MR = 11,730; ∑MO = 16,802

M = 16802 − 11730 = 5072m − lb or 16636 ft − lb d = 1.40 − 0.07 = 1.33m

or 52"

D.11.7 Determine the thickness of downstream apron.

26. 16
C= = 8.72
3. 00

57
Point Length of Head Net head, Effective Equation: t-wc =
Creep, Lc Loss, HL Hnet = H- head, h (4/3)wh
(m) = Lc/C HL (ft)
(m)
9 20.28 2.32 0.68 2.23 + t9 t9 x 150 =
(4/3)(62.5)(2.23+t9)
D 21.61 2.48 0.52 1.70 + tD tD x 150 =
(4/3)(62.5)(1.70+tD)
E 24.61 2.85 0.15 0.49 + tE te x 150 =
(4/3)(62.5)(0.49+tE)

4 62. 5
t = x = (2.23 + t ) = 0.555(2.23 + t ) = 1.239 + 0.555t
9 3 150 9 9 9

1.239
0.445t 9 = 1.239; t9 = = 2.78 ft. or 0.85m; USE t 9 = 1.00
0. 445
0.943
t D = 0.555(1.70 + t D) = 0.943 + 0.555 t D ; tD = = 2.12 ft. or 0.65m
0. 445

0. 272
t E = 0.555(0.49 + t E) = 0.272 + 0.555 t E; tE = = 0.61 ft. or 0.19m
0. 445

Use 0.30m (min.)

58
 Technical Working Group (TWG) for the Development of Philippine
National Standard for Design of a Diversion Dam

Chair

Engr. Bonifacio S. Labiano

National Irrigation Administration

Members

Engr. Felimar M. Torizo Dr. Teresita S. Sandoval

Board of Agricultural Engineering Bureau of Soils and Water Management
Professional Regulation Commission Department of Agriculture

Dr. Armando N. Espino Jr. Dr. Elmer D. Castillo

Central Luzon State University Philippine Society of Agricultural Engineers

Dr. Roger A. Luyun Jr. Engr. Francia M. Macalintal

University of the Philippines Los Baños Philippine Council for Agriculture and Fisheries
Department of Agriculture

Project Managers

Engr. Darwin C. Aranguren

Engr. Romulo E. Eusebio

Engr. Mary Louise P. Pascual

Engr. Fidelina T. Flores

Engr. Marie Jehosa B. Reyes

Ms. Micah L. Araño

Ms. Caroline D. Lat

Mr. Gerald S. Trinidad

University of the Philippines Los Baños –

Agricultural Machinery Testing and Evaluation Center