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Cornell University Library
TC 540.B64 1916
Dams and weirs; an analytical and practic
3 1924 004 065 664
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DAMS AND WEIRS AN ANALYTICAL AND PRACTICAL TREATISE ON GRAVITY DAMS AND WEIRS; ARCH AND BUTTRESS DAMS; SUBMERGED WEIRS; AND BARRAGES
By
W.
G.
BLIGH
FORMERLY EXECUTIVE ENGINEER, PUBLIC WORKS DEPARTMENT OF INDIA, AND INSPECTING ENGINEER, DEPARTMENT OF THE INTERIOR, CANADA. MEMBER, INSTITUTE CIVIL ENGINEERS (LONDON) MEMBER. AMERICAN SOCIETY OF CIVIL ENGINEERS MEMBER, CANADIAN SOCIETY OF CIVIL ENGINEERS
ILLUSTRATED
AMERICAN TECHNICAL SOCIETY CHICAGO 1916
%\10 15
7^
/.3?21I5
COPYRIGHT.
1916,
BY
AMERICAN TECHNICAL SOCIETY COPYRIGHTED IN GREAT BRITAIN ALL RIGHTS RESERVED
CONTENTS PAGE
Gravity
dams
2
Pressure of water on wall
2
Method
2
for graphical calculations
Conditions of "middle third" and limiting stress
3
Compressive stress limit
4
Design
of
dams
4 4 8
Theoretical profile Practical profile
Crest width
9
Rear widening
10
Variation of height
13
High and wide crest Graphical method Analytical method
16
Pressure distribution
23
Graphical method for distribution of pressure pressure limit
25 27 29 30 31 31 34 36 37 37 39 41 42 43
13
18
Maximum
Limiting height Internal shear and tension Security against failure
by
sliding or shear
Influence lines
Actual pressures in figures Haessler's
method
Stepped polygon Modified equivalent pressure area in inclined back
Curved back profiles Treatment for broken line profiles Example of Haessler's method Relations of R. N. and
W
Unusually high dams Pentagonal profile to be widened Silt against base of dam Ice pressure. Partial overfall
dam ...
43 ,
47 50 51
dams
52
CONTENTS PAGE
Notable existing dams
53
Cheeseman Lake dam
53 55
Analytical check
Roosevelt
New
dam dam
56
Croton Assuan dam
...
Cross River and Ashokan dams Burrin Juick dam
Arrow Rock dam Special foundations
69
Aprons affect uplift Rear aprons decrease uplift Rock below gravel Gravity dam reinforced against Gravity overfall
58 59 65 65 67
dams
70 71 72 ice pressure
73 75
or weirs
Characteristics of overfalls
75
Approximate
77
width Pressures affected by varying water level Method of calculating depth of overfall crest
79 82
85 85 90 92 92
Objections of '*Ogee" overfalls
Folsam weir
Dhukwa
weir,
Mariquina weir Granite Reef weir Nira weir Castlewood weir American dams on pervious foundations
95 96 97
Arched dams
101
Theoretical and
102
practical profiles
Support of vertical water loads in arched Pathfinder
dams
104
dam
104 107 109 Ill 112
Shoshone dam Sweetwater dam Barossa dam Burrin Juick subsidiary dam Dams with variable radii Multiple arch or hollow arch buttress
112
dams
113
Multiple arch generally more useful than single arch dams. 113 114 Mir Alam dam .
CONTENTS Multiple arch or hollow arch buttress
dams
—continued
117 118
Stresses in buttress
dam Ogden dam
Belubula
120
Design for multiple arch Reverse water pressure Pressure on foundations Flood pressures Big Bear Valley dam
dam
122
124 125 129 131
Hollow slab buttress dams
136
Formulas for reinforced concrete Guayabal dam Bassano dam
137 141
146
Submerged weirs founded on sand Percolation beneath
Governing factor for
151
dam
152
153
stability
Vertical obstruction to percolation
Rear apron Example of design type A Discussion of Narora weir Sloping apron weirs, type B Restoration of Khanki weir Merala weir Porous fore aprons Okhla and Madaya weirs Dehri weir.
PAGE
159 159
164 167
169 171 ,
171
173 177
.
178
Laguna weir
179
Damietta and Rosetta weirs
179
Open dams
or barrages
Barrage defined. .Weir sluices of Corbett dam General features of river regulators Stabihty of Assiut barrage Hindia barrage American vs. Indian treatment .
North Mon canal Upper Coleroon regulator St. Andrew's Rapids dam Automatic dam or regulator
182 182 189 193
194 194 196 200 201
201
205
INTRODUCTION A N
**
unused
waterfall, -no matter
how
beautiful, appeals to
an
engineer mainly as an economic waste, and he fairly aches to
throw a
dam
across the rushing torrents or to utilize the
the water which glides gracefully over the
many
into useless spray
power
and dashes
of
itself
His progress in the past
feet below.
years has, however, in no
falls
way measured up
to his desires, but
with the United States and other governments undertaking gigantic irrigation
projects in order to reclaim vast areas of tillable
lands and with
hydroelectric
companies
acquiring the power
few years have witnessed
rights of our great waterfalls, the last
The use
wonderful progress in this type of engineering work. reinforced concrete as a standard material
many problems
in connection with
it
and the solving
of
of the
has greatly simplified and
cheapened the construction, thus avoiding the greater
difficulties
of
masonry construction usually found
in the older dams.
^
All of this progress in the design of
dams and
weirs, however,
has served to multiply the types of design and has increased the
need for an authoritative and up-to-date treatise on the theoretical
and practical questions involved.
The author
of this
work has
been a designing engineer for more than a generation and has built
He
dams and
is,
therefore,
weirs in India, Egypt, Canada, and this country.
abundantly qualified to speak, not only from the
work but from the modern
historic side of the
practical side as well.
In addition to a careful analysis of each different type of he has given
critical studies of
profile,
the examples of this type, showing
the good and bad points of the designs.
A
wealth of practical
problems together with their solution makes the treatise exceedingly valuable.
^
It is the
satisfy the ject
hope of the publishers that
demand
and that
it will
for a brief
this
modern
treatise will
but authoritative work on the sub-
find a real place in Civil Engineering literature.
.
:
DAMS AND WEIRS PART
I
INTRODUCTION 1.
A dam may
Definitions.
be defined as an impervious wall
of masonry, concrete, earth, or loose rock
water at
its rear,
while
its
which upholds a mass of is free from the pressure
face or lower side
any appreciable extent. The waste water of the reservoir impounded by the dam is disposed of by means of a waste weir, or by a spillway clear of the work, or in rare cases, by sluice openings in the body of the dam. Weirs, or overfall dams, although often confounded with bulkhead dams, differ from the latter in the following points, first, that the water overflows the crest, and second, that the tail water These two differences often modify the is formed below the dam. conditions of stress which are applicable in the design of dams of water to
proper,
and consequently the subject
of weirs
demands separate
treatment. Classification.
2.
Dams and
weirs
may
be
classified
as
follows
Dams
1
Gravity
2.
Gravity Overfalls, or Weirs
3.
Arched
4.
Hollow Arch Buttress Dams Hollow Slab Buttress Dams
5.
Dams
6.
Submerged Weirs
7.
Open Dams,
The
or Barrages
subjects of earth, rock
fill,
and
steel
dams
will
not be
treated in this article, as the matter has been already dealt with in
other volumes.
made
use
of,
Graphical as well as analytical methods will be
the former pronedure being explained in detail as
occasion demands.
DAMS AND WEIRS
GRAVITY DAMS GENERAL DISCUSSION OF DAMS A
gravity
dam
is
one in which the pressure of the water
is
by the weight or ''gravity" of the dam itself. Pressure of Water on Wall. The hydrostatic pressure of the water impounded by a wall or dam may be graphically represented by the area of a triangle with its apex at the surface and its base drawn normal to the back line of the dam, which base is equal resisted 3.
or proportionate to the vertical depth. is
When
the back of the wall
vertical, as in Fig. 1, the area of this pressure triangle will
H being the vertical height. this area will
be
H'H
which equals
surface,
The
When,
as in Fig.
H' being the
H
cosec
2,
the back
weight
is
^c,
is
the
This unit
Fig. 2.
Vertical
symboUzed by
dam
Water Pressure Area with Back of Dam IneUned
Water Pressure Area
Dam
inclined,
.
actual pressure of water per unit length of
with Back of
—
inclined length of the exposed
above area multiplied by the unit weight of water.
Fig. 1.
is
be
which
is
62.5 pounds, or
^
short ton, or
^V long ton, per cubic foot. Unit Pressure, The pressure per square foot, or unit pressure
measured by the corresponding ordinate drawn parallel to its base, and is evidently
on the wall at any point,
is
above triangle, the same in both Figs.
and
of the
1
back as represented by greater than in Fig. 1.
The
total pressure
on the inclined
will,
however, be
Method for Graphical Calculations. For graphical calculawhen forces of dissimilar unit weight or specific gravity are
4.
tions
2.
the triangle in Fig.' 2
DAMS AND WEIRS
3
engaged, as in the case of water and masonry, or earth and masonry, the usual practice to reduce
it is
them
to one
common denominator
by making aUerations in the areas of one or the other, the weight of the masonry being usually taken as a standard. This result is effected by making the bases of the triangles of water pressure jj
F, but to
—
p (rho) being the sign of the specific P gravity of the solid material in the wall. The triangle thus reduced equal, not to
,
then represent a weight or area of masonry
will
which
is
t
Xp
i.
e.,
engaged,
The
1 unit thick,
equiva-
This device enables the item of unit weight,
lent to that of water.
common
to be eliminated as a of the water pressure
and
factor
from the forces
of the weight of the wall.
factor thus omitted has to be multiplied in again at the close
of the graphical operation, only,
however, in cases where actual
pressures in tons or pounds are required to be known.
Value of
The
p.
values ordinarily adopted for p, the specific
gravity of masonry or concrete, are 2\ and 2.4,
i.e.,
equivalent to
weights of 141 and 150 pounds, respectively, per cubic foot, while for
brickwork 2
is
a sufficiently large value.
The value
former case will be .069 ton and in the latter .075, or
of
wp
in the
—3 ton.
In some cases the actual value of p mounts as high as 2.5 and even
2.7,
The
when heavy
granite or basalt
reduction thus
made
is
the material employed.
in the water pressure areas has further
the convenience of reducing the space occupied
The
by the diagram.
areas of the reduced triangles of water pressure in Figs. 1 and
2 are -r— and
Zp 5.
—-—
,
respectively.
Zp
Conditions of ^'Middle Third" and Limiting Stress.
tions of gravity
dams
the "middle third."
are designed
Sec-
on the well-known principle
This expression
signifies
of
that the profile of
the wall must be such that the resultant pressure lines or centers of pressure due
first
to the weight of the
dam
considered alone, and
second with the external water pressure in addition, must both fall
within the middle third of the section on any horizontal base.
These two conditions of (R.E.)
stress are designated.
and Reservoir Full (R.F.).
insures the following requirement:
The
Reservoir
Empty
fulfillment of this condition
The
maximum
compressive ver-
^DAMS
4
tical unit stress (s), or reaction
mean
twice the
AND WEIRS on
compressive unit
the base of
stress, or,
a iam^ shall not exceed
stated syrnbolically,
s^2si
Now the mean vertical compressive unit structure divided
by
its
base length-
stress Si is the
weight of the
i.e.,
W ''=T Hence,
s,
the
maximum vertical
unit pressure, should not exceed
2W
Further comments on the distribution of the reaction on the base of a
dam 6.
"wdll
be made in a later paragraph.
A
Compressive Stress Limit.
second condition imposed
that of the internal compressive stress limit, that "permissible compressive unit stress
which
is
is
The
:
is
maximum
developed in the interior
This value can be masonry of the experimentally found by crushing a cube of the material employed,
dam, must not
of the
be exceeded.
and using a factor of safety of 6 or 8. Cement concrete will crush at about 2000 pounds per square inch, equivalent to 144 tons (of The safe value of s would then be 2000 pounds) per square foot. 144 —= 18
tons per square foot.
For ordinary lime concrete as
8
employed
in the East, the limit pressure
adopted
'long" tons, equivalent to 9 tons of 2000 pounds.
common
tons, or 11.2 "short" tons is also a
generally 8
is
Ten "long"
value.
DESIGN OF DAMS 7.
Theoretical Profile.
The
so-termed "low" masonry dam,
theoretically correct profile of a
i.e.,
one of such height that the limit
under the conditions above outlined, is that having its back toward the water vertical, triangle right-angled of a and its apex at the water surface. It can be proved that the proper base width b of this triangle is expressed bv the formula stress is not attained
This
profile,
shown
gular profile", less based.
in Fig. 3, will be
as on
it
termed the "elementary trian-
the design of
all profiles of
dams
In this expression, // is the vertical height.
is
more or
The base
DAMS AND WEIRS
5
width of -7=- insures the exact incidence of the vertical resultant (W) (R.E.)
and
of the incUned resultant
R
(R.F.) at the inner
and outer
The
edge, respectively, of the central third division of the base.
condition of the middle third
manner
and
thus
fulfilled in
the most economical
a factor of safety of 2 against overturning
is
further, the angle of inclination of the resultant
R
possible,
obtained,
is
with regard to the base
is
usually such as to preclude danger of
by sliding. The fore slope or hypothenuse will be in the ratio 1 "Vp which, when p = 2^, will equal 2:3, a slope very commonly adopted, failure
:
DOUBLE 3CHLE (o)
Elementary Triangular
Fig. 3.
and with p = 2.4 the mentary
Profile for
"Low" Masonry
ratio will be 1:1.549.
while, as
triangle is
we have
Dam
The area
of the ele-
seen, that of the
water
2Vp pressure
is
Q
'
is
the vertical angle between
2p sec S
=
Vp+1 =
1.187 with p
W
and R, and
= 2.4.
In Fig. 3 the resultant pressure hues are drawn to intersect the base so as to afford ocular proof of the stability of the section under the postulated conditions. Graphical Method. explained,
and
The
graphical procedure will
also in the future as fresh
now be
developments
briefly
arise, for
,
DAMS AND WEIRS
6
who
the benefit of those
are imperfectly acquainted with this valu-
able labor-saving method.
There are two forces engaged, P the horizontal, or, it may be Pi, the inclined water pressure acting through the center of gravity of its area normal to the back of the wall, and the weight or area of the
W
Of these two forces the item wp^ or unit weight, has already been ehminated as a common factor, leaving the pressures reprewall.
sented
by
superficial areas.
As, however, the height
H
is
also
com-
mon to both triangles, this can likewise be eliminated. The forces may then be represented simply by the half widths of the triangular areas by which means all figuring and scaling may be avoided. a force polygon has to be constructed.
First, first
drawn horizontally to designate the water
being
made
From
the extremity of P, the load line
equal to the half width of
its
W
In Fig. 3a,
P
pressure, its length
pressure area in Fig. is
is
drawn
3.
vertically, equal
to the half width of the elementary triangular profile, then the closing line
R
according to the law of the triangle of forces will
represent the resultant in magnitude and direction.
Second, the
lines of actual pressures reciprocal to those
on the force polygon
have to be transferred to the resultant water pressure on the back
The
will
the
profile. is
that of a line drawn through
the area of pressure, parallel to
e.g. of
incidence of the
its
base, in this case, at
—o
or one-third the height of the water-pressure triangle, above the base.
normal to the back, in and if prolonged it will intersect the vertical W, which in like manner acts through the center of gravity of Its direction, like that of the base, is
this case horizontal,
force
From
the elementary profile of the wall. the resultant
R
is
this point of intersection
Both and these points be found to be exactly at the inner and outer
drawn, parallel to
its
reciprocal in Fig. 3a.
W and R are continued until they cut the base of intersection will
edges of the middle third division of the base.
line,
It will be seen that
is empty the center of pressure on the base is at Wj when full it is shifted to that of P. Analytical Method. The same proof can be made analytically and P can be repreThe weight of the two triangles follows
when the
reservoir
the incidence of
as
sented
W
:
by their bases which
are -j=
Vp
and
— respectively. ,
P
If
moments
DAMS AND WEIRS be taken about the outer edge of the middle third, the lever arm of the vertical force
W
is
clearly
— or —r^^ and that of P, the horizontal
force, is the distance of the center of gravity of the triangle of
pressure above the base, viz,
—
The equation
.
will
water
then stand
o
(^x4:)-(fx|)or
3p If the actual values of
R and of
3p
W were required, their measured or
calculated lengths would have to be multiplied
to convert
them
by i?and by ivp in order
to tons, pounds, or kilograms, as
may
be required.
In many, in fact most, cases actual pressures are not required to be
known, only the position of the centers of pressures in the profile. Thus a line of pressures can be traced through a profile giving the positions of the centers of pressure without the necessity of converting the measured lengths into actual quantities.
In the
elementary triangle, Fig. 3, the value of the vertical resultant _fi_i^_^.
That
of
R
W
is
required in the older methods of calculation
2 IS
,
P OR
The following values relative to
p will be found useful.
DAMS AND WEIRS When inside
section 8.
and is
the back
W
overhanging, on the other hand,
is
outside the middle third.
The
R
will fall
vertically
backed
consequently the most economical.
Practical Profile.
In actual practice a
dam
profile
must be
provided with a crest of definite width, and not terminate in the
apex of a nite
triangle.
The upper part
of a
dam is
subjected to indefi-
but considerable stresses of an abnormal character, due to extreme
changes in temperature, consequently a solid crest
The
is
a necessity.
imposition of a rectangular crest, as shown in dotted lines on
Practical Pentagonal Profile for
Fig. 4.
Masonry
Dam
"Low"
Fig. 3, transforms the triangular profile into a pentagon.
This has
the effect of increasing the stability of the section (R.F.) so that the base width can be somewhat reduced, at the same time the vertical resultant
W
(R.E.), falls outside the middle third, but to
so small an extent that this infringement of the imposed condition is
often entirely neglected.
In order to provide against
this,
a strip
of material will have to be added to the back of the plain pentagonal profile.
Fig.
4
The dimensions
is
a diagram explanatory of these modifications.
of this
added
strip, as well as its position,
can be
DAMS AND WEIRS conveniently expressed in terms of
The
in Fig. 4.
9
the crest width
(Jc)
—
third at the depth
AD, which
not be produced
here,
to
is
be
AB
i.e.,
line of pressure (R.E.) will begin to leave the
middle
found by calculation which need
Below the point D, the
2/{;Vp.
divergence of the line of pressure will continue for a further depth
DE,
the point E, being close upon S.lfcVp below the crest,
lAh^p below D.
Below point E, the
line
longer diverge outward, but will tend to regain
consequently no further widening
will
of pressure
or
will
no
its original position,
be necessary, and the added
^3.,^ EII6ID.S
Fig. 5.
strip will
and
F
Profile of Chartrain
Showing Crest with Overhang
be rectangular in form down to the base.
The
points
being joined, this portion of the back will be battered.
width of this added strip or
Dam
EF
will be,
D
The
with close approximation,
.06fe.
16 9.
Crest Width.
The
crest
width of a
dam
should be propor-
dam, and in the case of a "high" dam to the limiting height i.e., to that depth measured below the crest at which the maximum stress in the masonry is first tioned to
its
actual height in case of a "low"
—
:
DAMS AND WEIRS
10
reached. Thus in "high" dams the upper part can always be of the same dimensions except where the requirements of cross com-
munication necessitate a wider
The causing
crest.
an abnormally wide crest can be modified by to overhang the fore slope, this widening being carried
efTect of
it
A
-by piers and arches.
good example of this construction occurs dam, Fig. 5. The arches form a stiff but light finish to the dam and have a pleasing architectural effect. The same procedure, but in a less pronounced degree, is carried out in in the Chartrain
the Croton dam, Fig. 27.
The formula
M£\' /frea
width
can
be
for crest
expressed
either in terms of the limit-
ing
height Hij or of the
base
6,
height
where the limiting is
not attained, and
a good proportion
by the
is
given
following empirical
rule
= <'Hi
(2)
h = lhh\
(2a)
k or Fig. 6.
Pentagonal Profile
—Back Vertical
This latter formula makes the crest width a function of the specific gravity as well as of the height, 10.
dam
is
which
is
theoretically sound.
Where the
Rear Widening.
arrangement shown in Fig.
When the body of the dam
line.
(k)
rear widening of a *'low"
neglected or where a uniform batter
and the base
back
is
4,
vertical the
are similar.
h (b),
or
-r-
is
substituted for the
the profile will be pentagonal in out-
two
triangles
If the ratio existing
be designated by
depth of the vertical side in Fig.
6,
r,
composing the
between the crest
then
A;
= Hr and kxh=
= &r, and
A,
the
Hbr^,
In order to find what value the base width b should have, so that the center of pressure (R.F.) will
middle third, the moments of
all
fall
exactly at the edge of the
the forces engaged will have to be
taken about this point and equated to zero.
The
vertical forces
DAMS AND WEIRS W,
consist of
11
the lower, and TT^ the upper triangle; the horizontal
of P, the water pressure. 1 1
Method
.
by the
The
of Calculation.
areas of the prisms involved, the triangle of water pressure
being as usual reduced by dividing
common
ination of
common
is
pressures can be represented
to
all
base by
its
W\
because the actual original value
is
—-—
further elim-
—
by discarding
factors can be achieved
three forces, the area
A
p.
which
being represented by 6r^
The
.
forces then are PF,
TI
represented
by
h;
W\ by
hf^;
and
P by
—
;
the actual value of the
P latter being -——
.
The
lever
arm
distances of the c.g.'s of these three
Ip
from A, the incidence
forces
—o
(6
— br)
,
and of P,
—
-.
of
i?,
W —o
are as follows: of
The equation will then
,
,
of W\,
stand, eliminating -,
o
o
bXb+br^(2b-2XbT)-— = P or
62(l+2r2-2r3)-— = P
whence
b=^x^=l= Vp Vl+2r2-2r3 The value of
b thus obtained will
the base width even
when the back
(3)
prove a useful guide in deciding
of the wall is not vertical, as only
a small increase will be needed to allow for the altered k
—or r = .15 the of r
which
.
reducing coeflBcient works out to
is .981.
= .15, the
.
Thus with a
profile
profile.
When
1 .
the reciprocal
80 feet high with p = 2.5 and
base width of the pentagonal profile will be
6
80
=
.
X .981
V2.5
= 49.64 feet; the decrease in base width below that of the elementary without crest will be 50.60-49.64 = 0.96 feet. The crest width will be 49.64 X. 15 = 7.45 feet. In actual practice, the dimenprofile
sions
would be in round numbers, 50
feet base
and 7^
feet crest
DAMS AND WEIRS
12
width as made on Fig.
by
The
6.
face of the profile in Fig. 6
is
made
joining the toe of the base with the apex of the triangle of water
pressure.
Graphical incidences of
shown on (l)j
ants
Fig. 6.
The
45 square feet and fall
The
Process,
graphical
processes
of
finding
the
W and of R on the base are self-explanatory and are profile is divided into (2),
2000 square
feet.
two triangular areas, The two final result-
almost exactly at the middle third boundaries, TF, as
might be conjectured, a
trifle
outside.
Areas are taken instead of
H
not being a common factor. I widths, owing to Analytical Process. The analytical process of taking
about the heel
is
shown below:
moments
DAMS AND WEIRS The
Variation of Height.
12.
form throughout; bedj so that the
it
while the remainder
height of a
must vary with the
maximum
dam
is
seldom uni-
irregularities of the river
section extends for a short length only,
This situation
of varying height.
is
13
will affect
the relationship between the crest width and the height, and also the base width.
To be
consistent, the former should
in proportion to the height.
This, however,
is
vary in width
hardly practicable,
consequently the width of the crest should be based more on the average than on the
maximum
height,
and could be made wider
wherever a dip occurs in the foundation
High and Wide
13.
wide
crest, i.e.,
Crest.
one carried
level.
In case of a very high as well as
much
higher than the apex of the trian-
water pressure,
gle of
it
is
not
desirable to reduce the base width
much below tary
that of the elemen-
triangle.
The
excess
of
material in the upper quarter of
a *'low"
dam
manipulating
This
latter,
can be reduced by the
which
fore is
ward from the toe
slope.
drawn up-
of the base,
in Fig. 7, can be aligned in three First,
directions.
by a
line ter-
minating at the apex of the trian-
water pressure; second,
gle of
it
CB Fig. 7.
--=r
to the vertical,
and
is,
CBi^SO.O'
it
C
C.
Showing Different Disposition of
can be made parallel to that of the elementary profile, that
Off
Profile
Fore Slope
can be given an inclination of
third, the slope
or batter can be
made
Vp than the last. This latter disposition is only suitable with an abnormally high and wide crest and is practically carried out in the Chartrain dam, Fig. 5, where the base is not reduced at all flatter
H below ~i=.
Vp
Reduction to any large extent, of the neck of the effected
a
dam
is,
is
profile
thus
however, not to be commended, as the upper quarter of
exposed to severe though indeterminate stresses due to
DAMS AND WEIRS
14
changes of temperature, Avind pressure,
masses of
and
etc.,
probably to
also
put in motion by the wind.
The Cross River dam, Ashokan dam, are examples of an abnormally thick upper quarter being provided on account of ice
to be illustrated later, as well as the
\Yhatever disposition of the fore slope
ice.
is
adopted, the profile
should be tested graphically or analjlically, the hne of pressure,
if
necessary, being d^a^^TL through the profile, as will later be explained.
From
the above remarks
the section of a
few
dam
it ^dll
based on the elementary
lines
be gathered that the design of
do^\Ti to the limiting
depth can be
profile which,
dra^^^l
by a
necessary, can
if
be modified by applying the test of ascertaining the exact position of the centers of pressiu-e
on the base.
If
the incidence of these
resultants falls at or close AAithin the edge of the middle third division of the base, the section can be pronounced satisfactory;
other^ase,
it
can easily be altered to produce the desired
if
result.
The crest has to be raised above actual full reservoir by an extent equal to the calculated depth of water passing
Freeboard, level
over the waste weir or through the spillway, as the case
may
be.
which adds considerably to the cost of a work, particularly when the dam is of great length and connected with long embankments, can be avoided by the adoption of auto-
This extra freeboard,
matic waste gates by which means level are
merged
full reservoir level
In addition to the above, allowance the height of
and high flood
into one.
which
is
is
made
for
wave
action,
obtained by the following formula: Ji^
= lMF+{2.5-yY)
(4)
In this expression F is the ''fetch", or longest line of exposure of surface to wind action in miles. Thus if F = 4 miles, the water the extra height required over and above maximum flood level will be
5X2) + (2.5-1.4)=4.1 5i feet. The apex of the (1
feet.
If
f = 10
miles, h^
at this higher level; the crest, however,
is
work out to must be placed
\^-ill
triangle of water pressure
frequently raised
still
higher, so as to prevent the possibility of water washing over it. Example. The working out of an actual example under 14.
assumed conditions cal
methods.
will
now be
given by both graphical and analyti-
Fig. 8 represents a profile 50 feet in height \Adth crest
level corresponding with the
apex of the triangle of water pressure,
DAMS AND WEIRS The assumed value the crest width
X 50 = 33.3 carried
down
here on,
it is
made
is
feet,
of p is 2^.
.15b
is
thus 5
edge of the crest
is
extent that the heel
line joining the toe
is
the
full
The back
77
2
Vp
^
slope
is
-7^ =
a distance of 8 feet,
Profile for
therefore .84 foot. is set
feet.
nearly pentagonal,
and from 1 in 50. The outset at the heel beyond a vertical line drawn through the rear e^
Diagram Showing Suitable
Fig. 8.
If
is
vertically to the point
given a batter of
is
and the base width
the crest width
the axis of the dam, which
by a
The outhne
15
out.
The
The
50-Foot
toe
Dam
is set
face line of the
in the
body
is
same
formed
with the apex of the water-pressure triangle.
the face line were drawn parallel to the hypothenuse of the ele-
mentary
triangle,
i.e.,
to a slope of
1
:
Vp,
it
would cut
off
too
much
material, the area of the wall being then but very httle in excess of
that of the elementary triangle, which, of course, quantity.
As
will
be seen
later,
is
a
minimum
the analysis of the section will show
that the adopted base width could have been reduced below what
DAMS AND WEIRS
16
has been provided, to an extent somewhat in excess of that given in formula (3).
The
Graphical Method.
15.
the resultant lines
graphical procedure of drawing
W (R.E.) and R (R.F.) to their intersection of the
base presents a few differences, from that described in section
7,
Here the profile is necessarily divided 6, into two parts, the rectangular crest and the trapezoidal body. As the three areas (1), (2), and Pi, are not of equal height, the item H with regard to Fig.
page
3.
common
cannot be eliminated as a
factor, consequently the forces
have to be represented as in Fig. 6 by their actual superficial areas, not by the half width of these areas as was previously the case. In Fig. 8a the vertical load line consists of the areas 1 and 2 totaling 844 square feet, which form W. The water pressure Pi is the area will
IT
of the
inchned triangle whose base
is
—
This
.
is
best set out graph-
P
polygon by the horizontal
ically in the force
the horizontal water pressure, which
The
feet.
is
line P,
made
= 555 square
~ 9
-77-= 2p
equal to
water-pressure area strictly consists of two parts corre-
sponding in depth to
(1)
and
but the difference
(2) as
the upper part
is vertical,
not
and considered area of water pressure is as it would be if the the back so of the wall were in one inclined plane. In Fig. 8 the line Pi normal to and it is cut the back of the wall is drawn from the point of origin inclined,
off
by a
vertical
is
so slight as to be inappreciable,
through the extremity of the horizontal
This intercepted length
Oi
is
line P.
clearly the representative value of
the resultant water pressure, and the line joining this point with the
base of the load line horizontal line it will
W
is
cut off an intercept
termination of Pi.
This
W and of
R, the resultant of
^P be drawn from the (
N) from a
line
AB = P,
Pi.
If
lower end of the load line
a
W
drawn through the the vertical comand Pi as well as and are naturally
vertical
and
N.
ponent of P, the latter being the resultant of and P- When the back is vertical, of
N
N
is
W
W
identical in value, their difference being the weight of water over-
lying the inclined rear slope.
The
further procedure consists in drawing the reciprocals of
the three forces Pi, IF,
and
R
on the
profile.
The
first
in finding the centers of gravity of the vertical forces 1
step consists
and 2
in
which
DAMS AND WEIRS the hexagonal profile
is
That
divided.
middle of the rectangle whose base
The
a trapezoid.
is de.
17
of (1) lies clearly in the
The lower
center of gravity of a trapezoid
is
the following extremely simple graphical process.
division (2)
best found
is
by
From d draw
dh horizontally equal to the base of the trapezoid fg and from g, gj is set off equal to de; join /y, then its intersection with the middle line of
the trapezoid gives the exact position of
Thus a few
its
center of gravity.
what would involve considerable calculation by analytical methods, as will be shown later. The next step is to find the combined c. g. of the two parallel and vertical forces 1 and 2. To effect this for any number of parallel lines effect graphically
or non-parallel forces, tw;o diagrams are required, force
and ray polygon and, second,
chord, or funicular polygon.
The load
in the former of these figures.
rays
must be taken.
First,
Its position
line in Fig.
a so-termed
first,
its reciprocal,
the force and
8a can be utilized
a point of origin or nucleus of
can be anywhere relative to the
load line, a central position on either side being the best. point Oi, which
is
The
the real origin of the force polygon at the extremity
Pi can be adopted as nucleus and often is so utilized, in which case the force line Pi and R can be used as rays, only one additional
of
ray being required.
For the sake of
illustration,
both positions for
nucleus have been adopted, thus forming two force and ray poly-
on the same load line, and two funicular polygons, the resultants of which are identical. The force and ray polygon is formed by connecting all the points on the load line with the nucleus as shown by the dotted line a, 6, and c, and a', 6', and c'. Among the former, a and c are the force lines Pi and P, the third, &, joins the termination of force (1) on the load line with the nucleus. These lines a, b, c, are the rays of the polygon. Having formed the force and ray diagram, in order to construct the reciprocal funicular polygon 86 the force lines (1) and (2) on the profile Fig. 8 are congons, both based
tinued
down below
the figure.
anywhere right through
Then a
(1) parallel to
line
marked
the ray
a,
(a) is
from
drawn
its intersec-
drawn parallel to the ray (b) 8b meeting (2); through this latter intersection the third chord (c) is drawn backward parallel to its reciprocal the ray c. This latter is the closing line and its intersection with the initial tion with the force (1), the chord (b) is
in Fig.
fine (a), gives
the position of the eg. of the two forces.
;
DAMS AND WEIRS
18
A JV.
through this center of pressure, which represents is continued on to the. profile until it intersects
vertical line
JV1+W2,
i.e.,
the inclined force Pi drawn through the center of gravity of the
water pressure area. draAA-n
This intersection
the starting point of R,
is
on the
parallel to its reciprocal
force polygon 8a.
This
resultant intersects the base at a point within the middle third.
R
is
the resultant "Reservoir Full", while
vertical forces in the
Empty".
masonry
The intersection of the
edge of the middle third
The
fulfilled.
incidence of
on the base
is
the resultant of the
almost exactly at the inner
latter is
—thus the
condition of the middle third
question of induced pressure and
the base will be considered
The
Wj
wall, is the resultant "Reservoir
iV",
its
it
R
is
later.
W,
naturally not identical with that of
line
fix
the position of
N
on the
the resultant
The
is vertical.
the resultant of both Pi and IF, and of
be required to
is
on
the vertical component "Reservoir Full",
"Reservoir Empty", unless the back of the wall line
distribution
P and N.
profile,
If
a horizontal
should be drawn through the intersection of Pi with the back
of the wall.
This
will represent
water pressure Pi, and
it will
the horizontal component of the
intersect P,
produced upward.
a line drawn vertically through this latter point
Then
will represent
The position of N N is made to fall at the inner
the vertical component (Reservoir Full). necessarily outside of
W, consequently if
edge of the middle third of the base, third.
iV, is
W must
fall
within the middle
This fact will later be made use of when the design of the
lower part of a "high" 16.
dam comes
Analytical Method.
under consideration.
The
analytical
method
of ascertain-
ing the positions of the incidences of
W and of R on the base, which
has just been graphically performed,
will
The
first
step
is
and trapezoid
some
vertical plane,
moments with the
of those
moment
two
forces
of which the profile is composed, and then to equate the sum of the about any fixed point on the base,
of their sum.
The most convenient
point in most cases
this projects a distance {y) is
explained.
to find the positions of the centers of gravity
of the rectangle relative to
now be
beyond the
is
the heel of the base
axis of the
dam, which
axis
a vertical line passing through the inner edge of the rectangular
crest,
-
DAMS AND WEIRS As the
19
areas of the divisions, whether of the
masonry wall or
of the water-pressure triangle, are generally trapezoids, the follow-
enumeration of A^arious formulas, whereby the position of the e.g. of a trapezoid may be found either with regard ing
to a horizontal or to a vertical plane, will 9, if
be found of practical
In Fig.
utility.
the depth of the figure between the
parallel sides
be termed H, and that of
the truncated portion of the triangle of
which the trapezoid
termed
d,
of the e.g.
is
a portion be
and h be the "sertical height above the base, then
Fig. 9. Diagram of Gravity of
Showing Centers Water Pressure
Trapezoids
r_H H + 3d (5)
Thus, in Fig.
9,
^=13
and d = Q ,
^
If
then
feet,
13/13 + 18\ 3
V13 + 12/
=
o.o 7 feet
the base of the triangle and trapezoid ^vdth
decreased in length, the value of h
dependent only on
H and d,
T\ill
it
be increased or
not be thereby affected,
which values are not
altered.
as
it is
If,
however, the base of the triangle be inclined, as shown by the
'l—
rrt,
^Fig. 10.
Diagram
dotted lines in Fig.
Illustrating
9,
be higher than
before,
base through
the
(/,
Height of
Trapezoid above Base
the center of gravity of the trapezoid vrAl
but a
c. g.
will
line dra-^-n parallel to
the inclined
always intersect the upright side of
the trapezoid at the same point, vertically
c. g.
^•iz,
above the horizontal base.
one which
is
h feet distant
:
DAMS AND WEIRS
20
The value
of h can also be obtained in terms of a
parallel sides of the trapezoid,
For example,
If tical 8,
in Fig. 10,
and
R = \2, a =10,
the horizontal distance of the
plane
is
is
and
c. g.
6
H
the two
then
from a ver-
of a trapezoid
can be considered as divided into
its angles;
is
As shown in Fig. two triangles, the
equivalent to that of three equal weights
each weight can thus be represented by one-
third of the area of the triangle in question, or tively,
= 16,
explanatory of the working.
weight of each of which placed at
6,
required, as, for example, that of the trapezoid in Fig.
the following
11, this area
and
is
by --- and ---,
respec-
Let y be the
being the vertical depth of the trapezoid.
projection of the lower corner aff.bh
A
beyond that of the upper one -B. Then by equating the
sum
moments
of the
of
the corner weights about the
point their
A with the moment of sum, the distance (x) of
the e.g. of the whole trapeFig. 11. of
Method of Finding Distance of Center Gravity of a Trapezoid from Heel
zoid from
A
will
be obtained
as follows
{^^Ex = ^ [6(a+6)+a(a+2/)+2/(a+6)]
where
2/
= 0,
the formula becomes
For example,
in Fig. 8, a or de = 5 feet, 6
= 33.3, and
2/
=-84, whence
DAMS AND WEIRS The
21
similar properties of a triangle with a horizontal base, as
in Fig. 12,
may
and are obtained
well be given here
in the
same way
by taking moments about A, thus
-^^ In Fig. 12, 6 = 14
y=S
feet,
Reverting to Fig.
8,
feet,
(8)'
and h=10
feet,
then
the position of the incidence of
W on the
obtained by taking moments about the heel g of the base as base follows Here is the area of the whole profile, equal, as we have is
W
:
844
seen, to ft.
and
arm
lever
of
W
is
2.5+.84 = 3.34
(1) is
by formula
(7)
11.63 feet.
Hence, as the
is
area of the upper component (1)
is
40
sq.
of (2) 804.
The that of
The
sq. ft.
equal to the
by hypothesis feet,
x,
that of (2)
has already been shown to be
sum
moment of the whole moments of the
of the
parts, the equation will
become
844a: = 40x3.34+804xll.63
= 9484.1
^.^^e. 12. Method of t 1 n d 1 n g Genter of Gravity of a Triangular Profile
a:
=11. 23
feet
This
fixes
The
position of the inner third point
The
the position of the incidence of
incidence of
W
is
is
W
—o
relative to the heel.
or
,
—-^ from
the heel.
o
therefore 11.23- 11. 10
= .13
foot within the
middle third, which complies with the stipulated proviso.
The the base.
next step
As
is
to find the position of
R
forces are considered'; the water-pressure area
one,
P
relative to the heel of
in graphical methods, only horizontal is split
and vertical two parts,
into
the horizontal component, the value of which
is
——
,
or 555
2p
feet,
and W3 the reduced area
of the back.
The
of
latter area is
water overlying the rear projection
a trapezoid of which the upper side
:
DAMS AND WEIRS
22 (a) is
8 feet long and
&,
the lower side, 50 feet, the depth being .84
hence the distance of
foot,
byformula
(6),
^^^^ = ,32
'-^
.84 = 24.4 feet; this
masonry base.
The
nearly.
its e.g.
inside the heel of the base will be Its actual area
foot.
ft.,
11.23 — .32 TT'^
and
of
i?,
has to be divided by p or 2i to reduce it to a area will then be 10.8 square feet,
distance of the incidence of
W from the ft.
the distance of the e.g. of the latter
= 10.9,
R be
will
^X
The reduced
base has already been determined to be 11.23 being .32
is
nearly.
termed
W
the distance between the incidences of
If
the equation of
.r,
heel of the
and that of Ws from will be
moments about the
incidence
stand thus
o
or
555X^ = 844x+10.8a^+117.83 i.e.
^^ 9132.2 ^ ^„^„^ ^ 854 8
,
'
R
therefore 10.7+11.23
is
= 21.93
ft.
"^
distant from the heel.
The
§ point being 22.2 ft. from the same point, R falls .3 ft. (nearly) within the middle third. This shows that a small reduction in the
area of the profile could be effected. 17.
it
is
Component.
Vertical
component
of
R
obtained
and
If
the position of N, the vertical
Pi, is required, as is
by the equation
sometimes the
case,
NXx=={WXll.2S) + {w3X.32),
X being the distance from the heel of the base.
Or
in figures,
854.8a:=(844Xll.23) + (10.8X.32) a:
The
incidence of
the middle third.
N
is,
= 11.1
feet
therefore, in this case, exactly
middle third, which refers to the resultant
ponent high is,
N (R.F.)
dam comes
middle
limit of
but, as will be seen later,
W (R.E.) not to the comwhen the lower part of a commonly imposed
to be designed, one condition
that the vertical component third, in
on the
This of course does not affect the condition of
which case
W
N must fall at the inner edge of the will necessarily fall inside thereof.
DAMS AND WEIRS It
may
(/), is
actual value of
PH
force R,
—- =
844 - X9 —
32X4 is
PH
59.3 tons, as
w;
= -—
='^ N^+P^-
R = ^l 60^+39^ = 71.5
That
ton.
i.e.,
by
of the inclined
P NR
and the
in
which
Here iV = 855 square
P = 555
feet,
i?,
feet,
equivalent to 39
tons.
In the design of the section of a
Pressure Distribution.
the distribution of pressure on a plane
pier, or retaining wall,
in the section
The
32
obtained from the triangle of forces
whence
are taken
by the eliminated unit weight,
equivalent to 60 tons, nearly, and
18.
moments
therefore /=-—rr.
1
being the hypothenuse
dam,
if
W in tons of 2000 pounds will be the superficial area,
square feet multiplied
wp, viz,
PH —rr because
Nf=——;
about the incidence of R, then
tons,
N and
here be noted that the space between the location of
R, which will be designated
or 844
23
relations existing
between
maximum
unit
and mean or average unit stress (si) will now be considered. The mean unit stress on any plane is that which acts at its center point and is in amount the resultant stress acting on the plane (the incidence of which may be at any point) divided by the width of the lamina acted on. Thus in Figs. 3 or 8 take the resultant W. This acts on the horizontal base and its stress,
symbolized by
mean unit stress ^i
will
vertical coxnponent of
(s),
be
W
In the same way, with regard to
-7-.
R the mean unit
horizontal base will be
N ~.
stress
The maximum
b
produced by
it
TV",
the
on the
unit stress occurs at
that extremity of the base nearest to the force in question which i?.
Thus the maximum
unit stress due to
that due to a combination of at the toe of the base
P
and
It
is
dence of the center of pressure
is
is
maximum
b.
W
is
is
at the heel while
N acting at the incidence of R
evident that the nearer the incito the center point the less is the
developed at the outer edge of the section, until the center of pressure is actually situated at the center point itself. stress
The maximum pressure at the outer part of the section then equals the average and is thus at a minimum value. The relation between maximum and mean unit stress or reaction is expressed in the fol-
:
DAMS AND WEIRS
24
lowing formula in which
it is
assumed that any tension at the heel
can be cared for by the adhesion of the cementing material or of reinforcement anchored do^vn
-'040
(9)
m equal the expression in brackets,
OFj letting
s
In formula
(9a), q is
= 7nsi
(9a)
the distance between the center point of the
base and the center of pressure or incidence of whatever resultant pressure
and
tion,
under considera-
is
is
.51
the
mean
stress,
or the resultant pressure di-
vided by the base.
In Fig. 8 as explained in section 16, the incidence of i.e.,
center
the
(R.F.), falls .3
of ft.
i?,
pressure
within the
middle third of the base, consequently the value of q will be
^-'
33.3
--.3=5.25 ft., and
6
in formula
= 1-
31.5
(9a)
m=
= 1.95.
+
1
6?
The maxi-
33.3
mum =
/.
reaction
at
the
rSTons
always designated by Fig, 13.
1.95X60
Diagram Showing Pressure Distribution
on a
Dam with Reservoir Empty and
ft.
m = l — .95 = .05,
and
^2
tons per sq.
33.3
Reservoir Full
the heel,
= 3.51
5
toe
= mN
For the reaction (R.F.) at
= .05X60 = .09 tons. The distribu-
tion of pressure due to the vertical
33.3
component of
R is shown hatched
in Fig. 8 as well as in Fig. 13.
From formula
(9)
the facts already stated are patent.
the incidence of the resultant force
is
When
at the center of the base.
DAMS AND WEIRS q
= 0,
consequently
and
s
mean when at one of the
to the
when
m=l
;
= Si,
that
is,
25
the
third points, q
==
maximum
— m= ,
is
equal
and s = 2si
2,
;
W m = 4, and s = 45i, or 4^-.
at the toe,
b If the material in the
strain, the
R
incapable of caring for tensile
is
vertical compression, or
not apply.
Formula
falls, outside
the middle third.
(9) will
ever
maximum
dani
In designing sections
5,
obtained by formula
(24), section 86, should
is
it
be used when-
maneuver the
often necessary to
incidence of the resultant stress to, a point as close as possible to the jcenter of the
base in ord^r to reduce the
possible value,
which
that of
is
thje
and may be
less,
and besides
The
stress.
maximum
njiddle third, insures that the the,;mean,
mean
maximum stress to the least stress
condition of the
cannot exceed twice
insures the absence of tensile
stress at the base.
Graphical
19.
Method
Distribution
for
The
Pressure.
of
graphical method, of ascertaining the distribution of pressure on the ba,se pf
a masonry wall^ which has already been dealt with analytiexhibite4,in Fig. 13, which
cally,, is
The procedure
pFig. 8.
on the base
and
line,
i?,
a reproduction of the base of
Two
semicircles are struck
having their centers at the third division points
their radii equal to
incidence of
is
as follows:
is
—
the line eg
the two semicircles.
From
.
is
drawn
the point marked
to
g,
Again from g a
that of the
e,
the point of intersection of line
gn
is
set off at right
angles to eg cutting the base or its continuation at a point n.
point sure
is
nil in either
is
.clear of ff
termed the antipole of
be continued upward,
latter point will,
The
is let fall,
line
Kg
is
by the
—compressive cutting the it will
or tensile.
mean unit pressure, =1.8 from n cutting the new base line and the liiie continued from the toe of the
lar let fall
is
Below and
now made, and from
new base
in g^ while,
intersect the base at
if
K.
the
This
construction, be the center point of the base.
continued through
the
Joined
This
or the neutral point at which pres-
the projBle a projection of the base
a perpendicular
line
spnse
e,
g'
to h\ g'h' being
tons.
A
made equal
perpendicular
at ni; the points Ui
and
is let
¥ are
tQ
f^H
then
until it intersects another perpendicu-
base.
A third perpendicular is drawn
DAMS AND WEIRS
26
from the heel of the base, cutting off a corner of the triangle. The hatched trapezoid enclosed between the last two lines represents the distribution of pressure scale close ^
upon
W 59,3 = =
oo^^^-7^
"r~
3.52
3.51
on the base. The maximum stress will and the minimum .09 tons. If be considered,
W
tons, the
maximum
stress at the heel
and the minimum .04, at the toe. Examples to Illustrate Pressure
20. is
will
be
In Fig. 14 on the base, due to the
Distribution.
illustrated the distribution of pressure
incidence of R,
first,
at the toe of
the base, second, at the two-third point,
third,
and
the center,
at
an intermediate
fourth, at
position.
,.^-^JC.
In the
case
first
.(i?i),
it
will
seen that the neutral point n\ at the
first
third point.
thirds of the base
and one-third in
mum
is
be
falls
Thus two-
in compression
tension, the maxi-
in either case being propor-
tional to the relative distance of the
-TENSION
neutral point from the toe and heel of the base, the compression at the
toe being four times, while the ten-
twice the
mean
In the second case
{R^)
sion at the heel stress.
is
the two-third point,
intersects at
of n is The whole base
and the consequent position exactly at the heel.
.-.J. Fig.
14.
Pressure Distribution on Dam under Various Conditions
Base of
is
thus in compression, and the max-
imum
is
third
case
double the mean.
drawn at
{Rz)y
The
distance to
line
right angles to fg.
latter is vertical
quently be horizontal.
the
n
is
and gn
In the
gn
will conse-
thus infinite and the
area of pressure becomes a rectangle with a uniform unit stress
In the fourth case {Ra), the neutral point
lies
is
The
s.
well outside the profile,
consequently the whole is in compression, the condition approximating to that of
i?3.
.
DAMS AND WEIRS Maximum
21
Pressure
27
maximum
The
Limit.
pressure
increases with the depth of the profile until a level is reached where
the limit stress or highest admissible stress this level the design of the section of
is
Pown
arrived at.
to
a dam, as already shown, con-
simply in a slight modification of the pentagonal profile with
sists
a vertical back, the base width varying between that of theele77
mentary
profile or --=, or its
reduced value given in formula
(3).
Vp Beyond this limiting depth, which is the base of the so-termed "low" dam, the pentagonal profile will have to be departed from and the base widened out on both sides. Formulas for
22.
in the interior of a
Maximum
N
were
P
a function of
In Fig.
Si.
is
R = ^~W+P^ = diho
and P; therefore vertical,
is
shown composed of the horizontal water pressure, and
sentative triangle of forces
of
The maximum unit stress
dam is not identical with {s), the maximum vertical
unit reaction at the base, but
force (R.F.),
Stress.
N and W would
coincide
N sec
and then
The
a dam.
hitherto
the assumption, see Fig. c
= —^ = m
Another theory which
maximum
=
7
d.
a repre-
resultant
If the
internal
based on
is
unit stress
(10a)
Europe and
East assumes that the
maximum
normal to the direction
of the resultant forces as illustrated
stress lines
stress
on the base of Fig.
due to
R would not
be
-r-
h
be
-^—
-
Oi
mum unit
But
8.
stress is
in the
developed on a plane
According to
but ^— and the ,
back
R—yW^+P^,
—7-sec^
finds acceptance in
still
R the
most prevalent theory
that the
8,
8,
the vertical
maximum
Various views have been current regarding the stress in
N
this,
the
by the mean
maximum stress
will
Oi
— = —\b— and R= N sec
6,
consequently the maxi-
Oi
stress
would be c
= ^sec2^
(10b)
Recent experiments on models have resulted in the formula for maximum internal unit stress being recast on an entirely different principle from the preceding. The forces in action are the maxi-
:
'
;
DAMS AND WEIRS
28
mum
vertical unit force or reaction s
P
shearing unit stress ^s=-r-
-—
water pressure, 6r equally resisted
-
by each
ing stress will thus be
The
combined with a horizontal
shearing force
symbohzed by P, which
the horizontal
is
assumed to be
is
unit in the base of the dam;' the unit shear-
P
These forces being at-right angles to
-7-.
b
each other, the status
is
that of a bar or column subject to compres-
and
sion in the direction of its length
length.
The combination
and
increased compressive stress,
The formula
also to a shear
normal to
recently adopted for
also a tension in ;the naaterial.
maximum
unit compression!^ is
as follows
j
C
In this 5 = ??i5i,
mN z— =—
= is+^is' + Ss'
,
(10)
P
As
.
its
of shear with compression produces ran
before ^3=^-, substituting
we
have ,_,
mN
c=-
>
26
When
m = 2,
as
is
462
the case
^2
when
26
the incidence of
R
is
exactly at
outer boundary of the middle third c
23.
=
—
T
=-7-
(1+
sec 6)
(IO2)
N=W
and sec B=
_
profile
(section 7,
which has a
page
5)
and
vertical
m = 2; then
Vp formula (IO2) becomes
c=y Now
"
Application of All Three Formulas to Elementary Profile.
In the case of elementary triangular back,
iM.Q
'
(1+
sec
W H\
^)=y 0+^l^0 Vp
Hwp
-^H^)
(11)
DAMS AND WEIRS
29
in elementary triangle = 150 feet,
p=2.4, ^P = 7X ton.
Example,
Let
H
When, according
to
c
(11),
=
^^
(1
3
+ 1.187) = 12.3
tons per
square foot.
Taking up formula (10a)
mWT— 2WTsec 6 sec 6
=—
^=—
W = —Hwp -— as above ,
-j-
iU
c=
= Ew^'p^^l HwpJ^ ^
(11a)
P
Example with conditions as before
—
150XlXl.55V3^ — = 'TOAx/iQ^ = 1QQ4
c=-
With formula
7.26X1.84
(10b),
——
c=
c= Hwp
,
13.3 tons
or in terms of
H,
f^^=Hw{p+l)
(lib)
Therefore, with values as above,
150X1X3.4
_„^
32
Ffom
the above
it is
a very high value to
dams appear
t6 have
c.
evident that formulas (10b) and
(1 lb)
give
Tested by this formula, high American
maximum
compressive unit stresses equal to
20 tons per square foot, whereas the actual value according to
However, the stresses in the is more like 14 tons. Assuan dam, the Periyar,'and other Indian dams, as also French dam^ have been worked out from formula (lOb) which is still in use.
formula (iO)
24.
Limiting Height by Three Formulas.
The
limiting height
(Hi) of the elementary triangular profile forms a close guide to that obtaining in any trapezoidal section, consequently a formula will
be given for each of the three cases in connection with formulas (10), Referring to case (10), we have from formula (11) (lOa), and (lOb).
f«(.W^)
DAMS AND WEIRS
30
Whence Hu
2c
the limiting height = -
Exarri'ple,
With
^ mentary 4.
c
= 16
tons and p==2A, Hi, the hmit height of the ele-
2X16X33 = ^^^^^^^
fii -Ilk profile will be
Referring to case (10a),
1024
,^^
^^ = 195
,
we have from formula
c=Hw
^
feet.
(lla)
VpVp + 1
^VpVp+1 Exam'ple,
With data
16x32 —tttt = 180 feet, nearly.^ H = -—z
as above
1.55X1.84
Referring to case (10b),
we have from formula
(1
lb)
c=Hwip+l) Hi=wip+1) Exam'ple.
With same data Hi =
Thus the new formula
—
=
(10) gives
^"7 = 150
much
feet.
the same results as that
formeriy in general use in the United States (10a), while in the
more conservative formula (10b) the difference is marked. Internal Shear and Tension. We have seen that the 25. combination of compressive and shearing stresses in a dam (R.F.) produces an increased unit compression. increase in the shearing stress
and
It further develops
also a tensile stress.
The
formulas are given below.
Compression as before
= y^ + Al^+^.^or
an
three
(10)
DAMS AND WEIRS
31
Shear '^
= y-^+'^"
or
^^'~
'—
(13)
The tensile and shearing stresses are not of sufficient moment to require any special provision in the case of a gravity dam. The tension is This fact suggests
greatest at the heel, diminishing toward the toe.
The
that a projection of the heel backward would be of advantage. direction (a) of c to the vertical
Tan 2a=
p
when m = 2, tan 2a = -^.
^=
is
but
as follows:
is
—sob —^-r-=— =-r
jf
mJy
In Fig.
8,
P = 555and
whence 2a = 33° 00' and
.649
R
not that of
(a)
= 16°
/.tan2a =
iV = 855.
The
30'.
inclination
855 of
R
to the vertical, or
The
direction of
of
c.
Sh
the shear, 26.
lies
t
6, is is
33° 50',
i.e.,
twice as large as that
at right angles to that of
by
on the angle
between
6
sliding
depends on the inclination of
W and R.
Thus tan
.7.
Security
W to P,
6 should be less
angle of friction of masonry on masonry, or less than
same
t,
Security against Failure by Sliding or Shear.
against failure
i.e.,
than the
This
is
the
W
and P must be such that 6 be greater than 35°, or that the complement of be not less
as stating that the relation of
shall not
while that of
c,
at 45° from the directions of either c or
The adoption of the middle third proviso generally insures With regard to sliding on the base, this can be further provided against by indentations in the base line or constructing it inclined upward from heel to toe. than 55°. this.
27.
Influence Lines.
It is
sometimes desirable for the purpose
of demonstrating the correctness of a profile for tentative design,
two conditions and empty, through the profile of a dam. This is far better effected by the use of graphic statics. There are two different systems of graphic construction that give identical results, which will now be explained and illustrated. The first method, which is most commonly adopted, is exhibited to trace the line of pressures corresponding to the of reservoir full
in Fig. 15,
gravity 2|.
elementary
which
is
the profile of a 100-foot high
dam
It thus lies within the limiting depth, profile
would be 190
feet.
with
specific
which for the
DAMS AND WEIRS
32
The
profile is pentagonal,
base width of the elementary
with a vertical back, and has the
—^
profile, viz,
which
full
in this case is
Vp
,—
2
-X 100 = 66.7
The
feet.
crest k
pressure triangle has a base of
is
—
V// = 10
The
feet wide.
The water-
profile, aS well as
the water-
P
pressure triangle,
is
divided into five equal laminas, numbered 1 h-/^H //=/^
Graphical ConBtruction for Tracing Lines of Pressure on
Fig. 15.
to
5'
in
which
one case and is
—-
is,
1'
to 5' in the other.
therefore, a
common
Dam
of Pentagonal Profile
The depth
factor
of each lamina:,
and can be eliminated as
5 well as the item of unit weight, viz, wp.
The
half widths of- all
these laminas will then correctly represent their areas
and
als6 their
weights, reduced to one denomination, that of the masonry. Fig. 15a a force polygon
is
formed*
In
In the verticalload line the
several half widths of the laminas 1 to
^5;
are
first set off,
and
at.
DAMS AND WEIRS
33
right angles to it the force line of water pressure is similarly set
out with the half widths of the areas
Ri to step
i?6
is
are drawn.
5'.
to
Then the
resultant
on marked
This completes the force polygon.
The next
with
1
1', 1, 2,
with
and
1' 2'
to find the combinations of the vertical forces on the profile,
that of
viz,
1'
so
combination of
lines of the
1
and
and
2, 1, 2,
This, as usual,
3, etc.
is
effected
by
constructing a force and ray polygon, utilizing the load line in Fig. 15a for the purpose.
individual areas in section 15,
below the
1
and
the centers of gravity of the several
by the graphical process described
to 5 are found verticals
On
profile.
polygon Fig. 15b
Then
drawn through these points
constructed,
is
are projected
these parallel force lines 1 to 5, the funicular
reciprocal rays in Fig. 15a.
The
its
chords being parallel to their
intersection of the closing lines
of the funicular gives the position of the centroid of the five forces
engaged.
By
producing each chord or intercept backward until
intersects the initial line, a series of fresh points are obtained
it
which
1 and 2; 1, 2, and and so on. Verticals through these are next drawn up on the profile
denote the centers of gravity of the combinations of 3,
so as to intersect the several bases of the corresponding combinations,
thus
1, 2,
and
and
3 will intersect the base of lamina 3;
4 will intersect that of lamina 4; and so on.
1, 2, 3,
and
These intersections
many points on the line of pressure (R.E.). The next step draw the horizontal forces, i.e., their combinations on to the profile. The process of finding the centers of gravity of these areas is rendered easy by the fact that the combinations are all triangles, are so is
to
not trapezoids, consequently the center of gravity of each its
height from the base.
bination l'+2'+3' 3'
is
to the apex, in the
Thus the
is
at J
center of gravity of the com-^
at J the height measured from the base of^
same way
for
any other combination, that
of
being at i the total height of the profile. The back being vertical, the direction of all the combined forces will be hori--
Vi
2', 3', 4', 5',
zontal,
to
and the
intersect
Thus
the
lines are
drawn through, as shown
in the figure,
corresponding combinations of vertical forces.
1' intersects 1,
1' 2' intersects 1 2,
several intersections the resultant lines
i?i,
and so on. R^, to
i?5
are
From these now drawn
do\yn to the base of the combination to which they belong, these last intersections giving the incidence of
many
Ri,
points on the line of pressures (R.F.).
i?2,
The
etc.,
and are
process
is
so
simple
DAMS AND WEIRS
34
and takes as long to describe as to perform, and tage, rest,
it has this advaneach combination of forces is independent of the and consequently errors are not perpetuated. This system
that
can also
be
used where
the back of the
has one or
profile
several inclinations to the vertical, explanation of which will be
given later. Actual Pressures in Figures.
28.
described,
it is
calculation is required. it or to
TF
the actual
If
N scales
tiplied
by
^Xl — -,
^^u that IS,
all
174, to reduce this to tons
is
the procedure.
it
has to be mul-
and wp =
174X20X9 „.. ^ \T =244 tons. N= ^^3^
the value of g
and that
is -7-
R
of
of
o
— —— — = „ ^lm+P^ 244+ V2442+ c:=N + —^-= 667 is
unit stress due to
the eliminated factors, which are -— = 20 5
Assuming the incidence
value
maximum
required to be known, the following
is
In Fig. 15a,
In the whole process above
noticeable that not a single figure or arithmetical
-
therefore
exactly at the third division,
m
157 tons.
;
,
1572
P
2;
is
also scales
112,
Applying formula
534
_
= 667 =
its
(IO2),
,
.,
^ *""^ ^'' ^^- ^'^
As 8 tons is obviously well below the limiting stress, for which a value of 16 tons would be more appropriate, this estimation is practically unnecessary but is given here as an example. roughly.
29.
Analytical
Method.
The
analytical
method
of calculation
now be worked out for the base of the profile only. First position of W, the resultant vertical forces (R.E.) relative to will
heel of the base will be calculated
the the
The back
and next that of R.
of
the profile being in one line and vertical the whole area can be con-
two right-angled
veniently divided into
triangles,
ening of the curvature at the neck be ignored.
has an inclination of (1) is A;Vp in
:
Vp
its
the thickfore slope
the vertical side of the upper triangle area will then be
—r— =
—^ = 75
from the heel of the base, which 20 feet = 6f feet. case corresponds with the axis of the dam, is
sq. feet.
in this
The
length;
1
if
As the
distance of
its c. g.
— o
DAMS AND WEIRS The moment
75X20 — —= -
then be
will
35
With regard
500.
to the lower
o
--= = 5000 X- = 3333.3
triangle, its area is
2Vp
lever
its
arm
the whole
is
one-third of
is
about the axis
will
its
is
of the incidence of
length of
The moment The moment of the parts. The
base, or 22.2 feet.
then be 3333.3X22.2 = 74,000.
sum of the moments of 75+3333 = 3408. Let x be the required
equal to the
area of the whole
The
sq.feet.
3
distance
W from the heel, then a:
X 3408 = 74,500
= 21.9feet x=^^^ 3408 The
inner edge of the middle third
heel; the exact incidence of
W
third, a practically negligible
of
R
—YtFJ
the distance in this
P
is,
is
—o or 22.2
from the
therefore, .3 foot outside the middle
amount.
With regard to the position R and that of is
W
between the incidence of
(/)
—^=2220.
the water-pressure area=
= 21.7
feet distant
The
feet.
then be 21.7+21.9 = 43.6
total distance of
feet; the outer
R
.'./=
from the heel
will
edge of the middle third
is
44.4 feet distant from the heel, consequently the incidence of
is
44.4-43.6 = .8 foot within the middle
-.8 = 10.3
ft,
and
third,
R
^6— .8=— 6
then q = ~
m= A+^'^ = l + .93 = 1.93.
At
this stage it
be convenient to convert the areas into tons by multiplying 2 25 them by pw, or -^. Then and become 239.6, and P becomes will
N
156.3 tons. figures;
Formula
(10) will also
W
be used on account of the high
then
'=f+
6.93
tons,
|6.93'
^ therefore,c=-^+\^+(2.34)'=3.46+Vl7.48 ^5 = 7.64 ,
tons.
— DAMS AND WEIRS
36
17
i*or
1 or -.u the compression at^^u the 1. neel,m •
'
52,
6^ ~^
= 1l
n^ = S)1.
S2
——tttt; = 239.6X0.07 66.7
b
= ^51 ton. The area of base pressure is accordingly drawn on Fig. 15.
W
If
= "^^' 2.03;
239.6X2.03
=
therefore, s " '
66.7
base pressure
is
m=l+
be considered, ^ = 4-+-3 = 11.42, and o
(R.E.)
= 7.30
tons.
The
66.7
b
therefore greater with (R.E.) than with (R.F.);
-^-^annnnjU]], Diagram Showing
Fig. 16.
there
This pressure area Haessler's Method.
line of pressures
16, the
Method
also a slight tension
is
quantity. 30.
Haessler's
same
system, which
which
is
on a
Dam
at the toe of .11 ton, a negligible is ,
shown on
A
Fig. 16.
second method of drawing the
termed "Haessler's" used as in the
profile being is
for Locating Lines of Preaaure
is
last
exhibited in Fig.
example.
In this
very suitable for a curved back, or one composed
of several inclined surfaces, the forces are not treated as independent entities as before,
but the process of combination
from the beginning.
They
polygon, Fig. 16a and are 1',
is
1,
2\ the
1'
can readily be followed with
1
last resultant being the
then combined with 2 producing
is
continuous
on the force
producing Ri] Ri with dotted reverse hue.
i?2,
and
so on.
2', i.e.,
This
last
DAMS AND WEIRS The
reciprocals
on the
c.g.'s of all the laminas
profile are
drawn
37 as follows:
1, 2, 3^ etc.,
First the
are obtained by
1', 2', 3', etc.,
Next the water-pressure lines, which in this case are horizontal, are drawn through the profile. Force line (1') intersects the vertical (1), whence Ri is drawn parallel to its reciprocal
graphical process.
in Fig. 16a through the base of lamina (1), until
zontal force line
(2')
J
a line
line in Fig.
From
is
reaches the hori-
Its intersection with the base of (1) is a point
(2').
in the line of pressure (R.F.).
with
it
Again from the intersection of Ri
drawn backward
16a until
this point ifa is
the horizontal force
it
parallel to its dotted reciprocal
meets with the second vertical force
then drawn downward to
its intersection
repeated until the intersection of
i?5
the operation for (R.F.).
It is evi-
dent that Rs as well as
the other
all
with
intersection with the base of lamina
(3'), its
giving another point on the line of pressures.
(2)
(2).
with the
final
This process
is
base coiripletes
resultants are parallel to the cor-
responding ones in Fig. 15, the same result being arrived at
by
different
graphical processes.
Stepped
31.
16b
Polygon.
Fig.
a representation of the so-
is
Fig.
called IS
^'stepped"
17. Transformation of Inclined Pressure Area to Equivalent with Horizontal Ease
which
employed; the form
also often
differs,
polygon,
but the principle
is identical
with that already described.
Inspection of the figure will show that
all
the resultant lines are
drawn radiating to one common center or nucleus (0). The process of finding the incidence of , on the bases
W
several lamina
is
of the
identical with that already described with regard
to Fig. 15, viz, the same combination of
1+2, 1+2+3, and so on, and then projected on to the profile. 32. Modified Equivalent Pressure Area in Inclined Back Dam. When the back of a dam is inchned, the area of the triangle of water are formed in the funicular 16c
pressure
with
ABC, in Fig.
its half
17, will^not equal the
width, which latter
consequently the factor
H
is
product of H, but of Hi
measured
parallel to the base,
cannot be eliminated.
The triangle can, however, be altered in outhne so that while containing the same area, it will also have the vertical height as a factor itself
^
38
DAMS AND WEIRS
^HIPJM f
DAMS AND WEIRS This
in its area.
is
and subsequently repeated is
by the device
effected
illustrated in Fig. 17,
in other diagrams.
By
the triangle of water pressure.
to the back of the wall
39
AB,
a.
point
D
In this figure
drawing a is
line
CD
ABC
parallel
obtained on the continu-
dam. A and D are then joined. The triangle ABD thus formed is equal to ABC, being on the same base AB and between the same parallels. The area ation of the horizontal base line of the
of
TiD
ABD
equal to
is
—-XH,
and that
of the wall to half width
ji
EF X H. triangle
Consequently we see that the half width
BD -^
of the
ABD can properly represent the area of the water pressure,
H
and the half width EF that of the wall. The vertical height may, therefore, be eliminated. What applies to the whole triangle would also apply to any trapezoidal parts of it. The direction of the resultant line of water pressure will still be as before, normal to the surface of the wall, i.e., parallel to the base BC, and its incidence on the back will be at the intersection of a line drawn through the e.g. of the area in question, parallel to the base. ally
This point
will natur-
be the same with regard to the inclined or to the horizontally
based area.
Curved Back
33.
Profiles.
In order to illustrate the graphical
procedure of drawing the line of pressure on a profile having
a curved back, Figs. 18 and 19 are put forward as illustrations merely
—
as models of correct design.
In these profiles the lower two laminas of water pressure, 4' and 5', have inclined ^not
Both are converted to equivalent areas with horizontal the device explained in the last section. Take the lowest lamina acdb; in order to convert it into an equivalent trapezoid with a horizontal base, de is drawn parallel to ac; the point e is bases.
bases
by
joined with Ay the apex of the completed triangle, of which the trapezoid
is
a portion.
When af is drawn
then be the required converted half width of
which multiplied by
H
horizontally, the area acef will
figure, the horizontally
—o
will
measured
equal the area of the original
trapezoid acdb; —- can then be eliminated as a 5
common
factor
and
the" weights of all the laminas represented in the load line in
40
DAMS AND WEIRS
DAMS AND WEIRS by the
Fig. 18a,
41
The lamina
half widths of the several areas.
4' is
treated in a similar manner.
The
graphical processes in Figs. 18 and 18a are identical with
those in Fig. 15.
of in
In the force polygon 18a the water-pressure forces
drawn in directions normal to the adjoining portion the back of the profile on which they abut, and are made equal length to the half widths of the laminas in question. The back
V, 2\
3', etc. J
are
of the wall is vertical
down
to the base of lamina 3, consequently
and 3', will be set out on the water-pressure load line in Fig. 18a from the starting point, horizontally in one line. In laminas 4 and 5, however, the back has two inclinations; these forces are set out from the termination of 3' at their proper directions, i.e., parallel to their inclined bases to points marked a and b. The direction of the resultants of the combinations, 1' and 2', and the forces,
1', 2', 3',
1', 2^,
will clearly
be horizontal.
the directions of the combination resultant line
Thus the
Aa and
that of
If
1^ 2' 3' 4' 5' will
inclination of the resultant of
forces placed
on end, as
Aa and Ab
1^ 2' 3' 4' will
be joined, then be parallel to the be parallel to Ab.
any combination
of inclined
in the water-pressure load line, will always
be parallel with a line connecting the terminal of the last of the forces in the combination with the origin of the load line. Profiles. The method of and directions of the resultants of water pressure areas when the back of the wall has several inclinaThis system involves tions to the vertical is explained as follows the construction of two additional figures, viz, a force and ray polygon built on the water-pressure load line and its reciprocal funicular polygon on one side of the profile. These are shown conof the vertical force structed, the first on Fig. 18a, the nucleus and ray polygon being utilized by drawing rays to the terminations of 1', 2', 3', 4', and 5'. In order to construct the reciprocal funic-
Treatment for Broken Line
34.
ascertaining the relative position
:
ular polygon, Fig. 18c, the
trapezoidal laminas which 1'
to
first
step
is
to find the c.g.'s of all the
make up the
w^ter-pressure area, viz,
This being done, lines are drawn parallel to the bases of
5'.
the laminas (in this case horizontal lines), to intersect the back of the wall. etc.,
are
From
drawn at
the points thus obtained the force lines
on which they abut.
On these
force lines,
1', 2', 3',
back of the wall which are not all parallel,
right angles to the portions of the
DAMS AND WEIRS
42
the chord polygon (18c) line
is
AO is drawn anywhere parallel to
From
its reciprocal
the intersections of this line with the force
marked
OV
is
drawn
First the initial
constructed as follows:
parallel to
OV
in Fig. 18a
AO,
in Fig. 18a.
line 1'
and
the chord
intersecting
Again from this point the chord '02' is drawn inter3' whence the chord 03^ is continued to force 4 and secting force 04' up to the force line 5\ each parallel to its reciprocal in Fig. 18a.
hne
force
2'.
,
The
The intersection of the initial and the
closing line is 05.
lines of the funicular line 1' 2' 3' 4' 5',
reciprocal
Ob
closing
polygon gives the position of the final resultant is then drawn from this point parallel to its
which
in Fig. 18a to its position
on the
The
profile.
resultants are obtained in
other
a sim-
manner by projecting the sevbackward till they intersect the initial line OA^ these
ilar
eral chords
intersections being the starting
points of the other resultants, viz, r-4^ l'-3', l'-2', and 1'. These resultant lines are drawn
parallel
to their reciprocals in
18a, viz, l'-4' is parallel to Aa,
while the remainder are horizontal in direction, the same as Diagram Showing Third Method Determining Water Pressure Areas
Fig. 20.
of
their reciprocals.
This procedure
is
identical
with that pursued in forming the funicular 18b, only in this case the forces are not all parallel.
35.
used
is
Example
of Haessler's
Method.
In Fig. 19 the
similar to Fig. 18, except in the value of p,
not 2.4 as previously.
The graphical system employed
profile
which is
is 2|,
Haessler's,
each lamina as already described with reference to Fig. 16 being independently dealt with, the combination with the others taking place
on the
profile itSelf.
In this case the changes of batter coin-
cide with the divisions of the laminas, consequently the directions of the inclined forces are their areas abut.
normal to the position of the back on which
This involves finding the
pressure trapezoids, which
is
c. g. 's
of each of the water-
not necessary in the
first
system, unless
the funicular polygon of inclined forces has to be formed.
In spite
DAMS AND WEIRS most cases Haessler's method
of this, in
will
43 be found the handiest
to employ, particularly in tentative work.
Example
36.
of Analytical Treatment.
In addition to the
two systems already described, there is yet another corresponding In to the analytical, an illustration of which is given in Fig. 20. this the vertical and horizontal components of R, the resultants (R.F.), viz, N and P are found. In this method the vertical component of the inclined water pressure Pi is added to the vertical
dam
weight of the
itself,
and when areas are used to represent
weights the area of this water overlying the back slope will have to
be reduced to a masonry base by division by the specific gravity of the masonry. Relations of R. N. and
37.
W.
The diagram
in Fig.
20
further illustration showing the relative positions of R, P, Pi,
and W, force
that
R a',
is
The
P with it is
R
from
a,
the intersection of the horizontal
all
the vertical forces, for the reason
starts
the resultant of the combination of these two forces; but
also the resultant of Pi
and W, consequently
The
the intersection of these latter forces.
consequently in the resultant position of
by the
a
N,
the resultant of
line iV,
is
R
is
known, that of
intersection of
P
R
N
and and
it
it will
pass through
points a
and
follows as well that
ai are if
the
W can be obtained graphically
or Pi with R.
These
lines
have already
been discussed.
UNUSUALLY HIGH DAMS 38.
"High" Dams.
An example
the design of a high dam, before stated.
i.e.,
will
now be
given, Fig. 21, of
one whose height exceeds the limit
As usual the elementary
triangular profile forms
We have seen in 24 that the limiting depth with p = 2.4 and c=16 tons = 195 feet, whence for 18 tons' hmit the depth will be 219 feet. In
the guide in the design of the upper portion. section
down to a depth of 180 feet. made 15 feet wide and the back is battered 1 in 30; the base width is made 180X.645 = 116 feet. The heel projects 6 feet outside the axis line. The graphical procedure requires no special Fig. 21 the tentative profile is taken
The
crest is
explanation.
It follows the analytical in dealing
with the water
pressure as a horizontal force, the weight of the water overlying
the back being added to that of the solid dam.
For purposes of
44
DAMS AND WEIRS
DAMS AND WEIRS
45
divided into three parts (1) the water on the sloping back, the area of which is 540 sq. ft. This has to be calculation the load
is
reduced by dividing
it
by
p and so becomes 225 sq.
not areas, will be used, this procedure
is
for the sake of uniformity in treatment to
As
ft.
not necessary, but
is
avoid errors.
tons,
adopted
The
the heel of the base of (3),
e.g.
about
is clearly 2 feet distant from which point moments will be taken. That of the crest (2) is 13.3 feet and that of the main body (3) obtained by using formula (7) comes
of (1)
to 41.2 feet.
No.
The statement
of
moments
is
then as follows:
DAMS AND WEIRS
46
The
middle third. feet,
and
value of q (R.F.)
is
—-.7=
19.33 -.7
=
18.6
m =04)=
Then by formula
(Id),
^+TI^--N being 815 and P, 506
^_ 1.96X815+V(1^96x815)^+4
(506)^
232
_ 1597+V2551367+1024144 232
= 15.03 tons per square foot Extension of Profile, This value being well below the limit of 18 tons and both resultants (R.F.) and (R.E.) standing within the
middle third another 30
now be a 3769
it is
ft.
trifle
sq. ft.
deemed that the same
over 135
=283
tons.
The
feet.
The
can be carried down The base length will the new portion (4) is
profile
in depth without widening.
area of
distance of its e.g. from the heel
W
by
will be The position of (7) is found to be 63.4 feet. obtained as follows, the center of moments being one foot farther
formula
to right than in last paragraph. No.
—
DAMS AND WEIRS being
—7
moment about the heel will be 23 X2..3 = 52.9,
or 2.3 feet the
,
47
*^
This amount added to the
say 53.0 ft.-tons.
represent that of iVi
Ni
is
The
Wi+
that of
Ni from the heel
distance of
moment
of
Wi
will
and will be 50,864+53 = 50,917. The value of the water on back or, 1081+23 = 1104 tons. then
is
=46.2
^
To
feet.
1104 obtain that of
i?i
the value of
—
f=~
to 46.2
= 89.9
,
the incidence of Ri
ft.,
=43.7
added
feet: this
3X1104
-^
is
^^~
therefore
=
89.9
o
90— 89.9 =
.
1 ft.,
-.1=22.5-.1=22.4
and
ft.,
m = l+^ = 1+^^^ = 2.00,
m .
find „
formula (10) will be used, the quantities being
c,
,-„,
,
iormula
P
(lOi).
689
-
— T~=7TT7=^-1 ,
„ Here
5
=
nearly.
135
b
To
q = -^
Then
within the middle third boundary.
mN'= 2.00X1104 _ — =16.4 ^— 135 .
less
^
than ,
tons and
Ss
™ inen
^ tons,
135
2^4
'=^vV' The
(16.4)2
limit of 18 tons, being
,
(5j)2^8_2+9.7 = 17.9tons
now
reached, this profile will have to
be departed from. 39.
Pentagonal Profile to Be Widened.
be adopted
is
The method now
to
purely tentative and graphic construction will be
its solution. A lamina of a depth of 60 ft., be added to the profile. It is evident that its base width must be greater than that which would be formed by the profile being continued down straight to this level. The back batter naturally
found a great aid to will
will
be greater than the
fore.
From examination of
other profiles
appears that the rear batter varies roughly from about in 8 while the fore batter is
extra offset at the back
about
1
to
1.
As a
1 in
first trial
was assumed with a base
5 to
an 8
it
1
ft.
of 200 feet; this
would give the required front projection. Graphical showed that N would fall without the middle third, and
trial
lines
W as well;
A second trial was was increased and the base
the stress also just exceeded 18 tons (R.E.).
now made
in which the back batter
shortened to 180 feet.
In this case c exceeded the 18 ton
limit.
DAMS AND WEIRS
48 Still
further widening
was evidently required
at the heel in ordei
to increase the weight of the overlying water, while
that the base width would not bear reduction.
was
it
The
clear
rear offset
was then increased to 15 feet and the base width to 200 feet. The stresses now worked out about right and the resultants both fell within the middle third. e.g. of the trapezoid of
By
using formula
(6)
the distance of the
water pressure, which weighs 112 tons, was
found to be 7.2 feet from the heel of the base, and by formula that of the lowest lamina
(5)
from the same point
91.8 feet; the weight of this portion vertical forces
tion are ft.
is
754 tons.
is
These two new
can now be combined with Ni whose area and posi-
known and thus that
of
N2 can be
ascertained.
distant from the heel of the upper profile; its lever
therefore,
(7)
found to be
be 46.2+15 = 61.2
the heel will then be
feet.
Ni is 46.2 arm will,
The combined moment about
— DAMS AND WEIRS P2 1139 .^^ tons Ss= —;— = -^^r:r =5.7 200
Now Whence by formula
(10),
16.3
which
is
-^1^+32.5 = 8.15+9.9 = 18.05
tons
the exact Umit stress.
The value
of
52
(the pressure at the heel)
same formula, using the minus fore,
49
52=—7— =
"
—
tions are set out
posed of PFi+(5).
= 3.4
sign, viz,
m=l
tons, nearly.
is
obtained by the
——-^ = 130 8
These
.34, there-
vertical reac-
^uu
below the
The
With regard to W2 moments is as follows:
profile.
table of
it is
com-
DAMS AND WEIRS
50
N
In the force diagram the water part of is kept on the top of W. This enables the lengths of the series to be clearly shown. The effect is the same as if inclined water pressure
N
the load Hne
were drawn, as has already been exhibited in several cases.
lines
Base of Dam.
Silt against
40.
In Fig. 21, suppose that the
water below the 210-foot depth was so mixed with specific gravity of 1.4 instead of unity.
shown graphically without
The
silt
as to have a
effect of this
alteration of the existing work.
can be In the
trapezoid lying between 210 and 270 the rectangle on ab represents
the pressure above 210 and the remaining triangle that of the lower
60 feet of water.
The base
of the latter, be
is,
therefore,
= — = 7rT 2.4 p
25
Now the weight of the
feet.
water
is
increased in the proportion
H'xlA — = '
of 1.4
:
consequently the proper base width will be
1,
60X1 4 — 7r7~="^^
The
f^^t.
The normal pressure on the back of the silt is shown graphically by the triangle whose base = cd= 10 feet; its area is 310 square feet,
pressure area due to
dam due
triangle acd then represents the additional
silt.
to the presence of
attached,
combined with R2 at is Rs] on the profile the reciprocal inclined force is run out to meet R2 and from this intersection R3 can be drawn up toward P2. This latter intersection gives the altered position of iV^2, which is too slight to be noticeable on this scale. The value of c and the inclination of R both increased, which is detrimental. are If the mud became consolidated into a water-tight mass the pressure on the dam would be relieved to some extent, as the earth equivalent to 23 tons.
This inclined force
is
the top right-hand corner of Fig. 21a and the resultant
will
not exert liquid pressure against the back.
Liquid
mud
pres-
sure at the bottom of a reservoir can consequently be generally
neglected in design. 41.
dam
Filling against
be considered.
immersed 's. case.
g. of 1.8 is
Then a
45 feet
is
Toe
of
Dam. Now
let
the other side of the
Supposing a mass of porous material having an deposited on the toe, as
is
often actually the
pressure triangle of which the base equals
drawn;
its
HX-^8 = 1
area will be 1755 and weight 132 tons; the
DAMS AND WEIRS resultant P4 acting through its
c. g. is
51
run out to intersect
At the
i?2.
same time from the lower extremity of i?2, in the force diagram, a reciprocal pressure line Pa is drawn in the same direction equal in length 132 tons and its extremity is joined with that of P2; the resulting line -R4 is then projected on the profile from the previous intersection until it cuts the force Une P2; this gives a new resultant R4, also and a new position for N, viz, N4., which is drawn on the profile dam increases The load on the toe of the will be similarly affected. ;
its
stabiUty as the value of 6
is
improved, but that of R^, which
To adjust matters,
the
c. g.
lessened, the position of is
nearer to the toe than
of (5) requires
W
W
is also
i?3, is
not.
moving to the right which
fee Pressure
J Diagram Showing
Fig. 22.
is
affected
by
Z9£ Tons
Effect of Ice Pressure
shifting the base line thus increasing the
back and same as
decreasing the front batter, retaining the base length the before.
42.
Ice Pressure,
Ice pressure against the back of a
has sometimes to be allowed for in the design of the rule,
however, most reservoirs are not
expansive pressure lower
is
full in
dam
profile; as
a
winter so that the
summit but at some distance negligible. In addition to this when
exerted not at the
down where the
effect is
the sides of a reservoir are sloping, as
can take place and so the
is
dam
generally the case,
movement
from any pressure. In the estimates for the Quaker Bridge dam it is stated that an ice pressure of over 20 tons per square foot was provided for. No of ice
is
relieved
DAMS AND WEIRS
52 definite rules
Many
The
suitable.
is
run on a hundred-
effect of a pressure of ten tons per foot
dam
foot
seem to be available as to what allowance
authorities neglect it altogether.
acting at the water level
For
in Fig. 22.
is illustrated
this
purpose a trapezoidal section has been adopted below the sum-
mit
level.
The
crest is
made
15 feet wide and 10 feet high.
solid section is only just sufficient, as will
of R'
on the
The
base.
area of this profile
is
of the ordinary pentagonal sections as dotted
Two
Fig. 23.
contain but 3325 sq. therefore 825 sq.
ft.
W the
weight of the
The is
43.
An
Partial Overfall
Dams.
p
is first
their resultant
R
combined
P
cuts
It falls just within the
actual example It
is
is
graphical procedure
at a
run down to the base parallel to
reciprocal in the force diagram.
third of the base.
The
ice pressure
dam and
point from which the final Ri
Dams
increase due to the ice pressure
or about 25 per cent.
hardly needs explanation.
with
4150 sq. ft., while one on the drawing would
Profiles for Partial Overfall
The
ft.
This
appear from the incidence
its
middle
given in section 56.
not infrequently happens that
the crest of a dam is lowered for a certain length, this portion acting as
a waste weir, the crest of the balance of the the water level.
dam
being raised above
In such cases a trapezoidal outline
is
generally
and the section can be continued upon the same lines to form the upper part of the dam, or the upper part can be a vertical crest resting on the trapezoidal body. In a preferable for the weir portion
DAMS AND WEIRS trapezoidal
dam,
the ratio of
if
k
be
-r-
53
the correct base width
r,
is
b
obtained by the following formula: h
1
= JL
-
(14^)
This assumes the crest and summit water level to be the same. In Fig. 23j p vertical
back
is
taken as 2.4 and then be
will
-^X Vp
feet,
and the
crest
width k
will
as
r
The base width with
.2.
^ .
,
V1 + .2-.04
a
= 5QX.645X.935 = 31.3
be 31 .3 X .2 = 6.3
In the second
feet.
shown canted forward, which is desirable in weirs, and any loss in stability is generally more than compensated for by the influence of the reverse pressure of the tail water which
figure the profile is
influence increases with the steepness of the fore slope of the weir.
The base width is, however, increased by one foot in the second figure. As will be seen in the next section, the crest width of a weir should not be less than
ViZ+Vd;
in this case // = 45
This would provide a crest width of
6.7+2.2 = 9
feet,
and d = 5. which it nearly
NOTABLE EXISTING DAMS Cheeseman Lake Dam. Some actual examples of dam sections will now be exhibited and analyzed. Fig. 24 is the section of the Cheeseman .Lake dam near Denver, Colorado, which is one 44.
of the highest in the world.
It is considered a gravity
however, and will be analyzed as such. into three unequal parts
1, 2,
a curvature of 400 feet
It is built to
radius across a narrow canyon.
and
3,
The
and the
dam,
section can be divided lines of pressure (R.F.)
and (R.E.) will be drawn through the bases of these three divisions. Of the vertical forces (1) has an area of 756 sq. ft., (2) of 3840, and being 17,952 sq. ft., which is marked (3) of 13,356 the total value of
W
on the load line in Fig. 24a. With regard to the water-pressure most convenient method, where half widths are not used, which can only be done with equal divisions, is to estimate the areas of the horizontal pressures only and set them off horizontally, off
areas the
the values of the inclined pressures being obtained
For
this purpose the triangle of horizontal
adjacent to, but separate from, the
profile.
by
construction.
water pressure
The
is
shown
three values of
P
=
DAMS AND WEIRS
54
—H2 —
which are equal to
will
be 270, 2631, and 7636, respectively, the
2p ft. In this computation the value of p is assumed to be 2.4. These several lengths are now set out horizontally from the origin in Fig. 24a, and verticals drawn upward intercept the chords, 1', 2', 3', which latter are drawn from the origin 0, parallel to their respective directions, i.e., normal to the
total being 10,537 sq.
The
adjacent parts of the wall.
rest of the process is similar to that
already described, with reference to Figs. 16 and 18, and need not
In Fig. 24a
be repeated.
N scales
19,450, equal to 1457 tons,
and
Radius = 400
p=£.4
Fig. 24.
on the 1.51.
Profile of
CheeserQan Lake
profile q scales 15 feet, therefore, in
Therefore, 5 =
4.45; then
12.5
>
With regard
to
12.5 tons,
W^
90 m = 1+-—
P
783
and 5s =-r-= ^„^
176
(10)
q scales about 20
^=~r"~
tons, approx.
m
then works out to
1.7,
— = .„„^
1.7X1346
7^
ft.,
ld.Utons.
exercise the inchned final resultant
This
(9),
4
mW
,
and
As an profile.
176
formula
^^^^^'+(4.45)2 = 6.25+V59 = 13.9
2
nearly,
1.51X1457
h
by formula
c=-
.
mN
Dam
line is parallel to
Oc in Fig. 24a,
P
its
is
drawn on the is worked
location
out by means of the funicular polygon, the construction of which
need not be explained after what has gone before.
DAMS AND WEIRS
55
In order to check this result analytbe, first, calculate the position of the e.g. of
Analytical Check.
45. ically the
procedure will
the trapezoids (2)
and
to the rear corner of their bases
(3) relative
(7) and also the positions of the resultants of the vertical components of the water pressure overlying the back with regard
by formula
same points by formula (6). Second, convert the areas into tons by multiplying by A- The statement of moments about the to the
heel of the base, with the object of finding the position of
W
is
given
below.
Moment
of (1)
Moment
of (2)
288
Moment
of (3)
1001
56.7X32.5
W=
Total
The distance
X47.9
order to obtain N, the
moments
=
13795
:
90713
1346 tons
will
1843
= 75075
X75
W from the heel
of
=
then be
of the
90713 ,^,^ =67.5 1346
water weights
will
ft.
In
have to
be added as below.
Moment
of
W
1346X67.5
Moment
of Wi
10X21.6
Moment
of wz
Total
107 X9
N=
90713
= =
216 963
91892
1463 tons
and x=-
91892 ''
=
62.8 feet
1463
R and its distance {q) from the center position of N must be computed from known the from point, that To
find the incidence of
the formula
—
—=
/=^ = ^^^ = 40
14.8 feet.
This
which was taken as 15
is
feet.
to that obtained graphically. is
as follows, g =
—
67.5
ft.,
therefore,
g=(62.8+40.0)
close to the value obtained graphically
The value of iV is also seen to be close (R.E.) The value of q with regard to
= 20.5 feet,
W
almost exactly what it scales on
DAMS AND WEIRS
56 the diagram.
up
In this profile the upper part
is
Hght, necessarily
made
for in the lower part.
At the upper base
line of (2)
the middle third edge, while position of
N
= 58.6
IS
is
the incidence of
R falls
within
62.8 distant from the heel
it.
W
At the
is
exactly at
final
base the
and the inner third point
distant, consequently the incidence of TV lies 4.2 feet
3 within the boundary. If
N were
the position of
the middle third, the value of
made
obligatory at the inner edge of
W would be increased,
but
jR
would
P/iP/t/S 410' v-s^.i-i^e!^
4.
lOOOdOOO
Fig. 25.
be decreased.
maximum by the
Profile of Roosevelt
There
SOOO
Dam
IO.00O3.Fr.
across Salt River, Arizona
may have been special reasons for limiting the On Fig. 24 the position of iV is obtained
stress (R.E.).
intersection of the horizontal resultant
upward. structure
If
Roosevelt
Dam.
it
R
prolonged
would amount to 21^ tons by the
In Fig. 25
Roosevelt dam, and Fig. 26
is
For some years, the Roosevelt existence.
with
the stress were calculated on the supposition that the
was an arched dam,
"long" formula, given in section 78, Part 46.
P
It spans a very
is
II.
given the profile of the
the site plan of that celebrated work.
dam was
the highest gravity
dam
in
deep canyon of the Salt River in Arizona
DAMS AND WEIRS
57
and impounds the enormous quantity of 1 J million which will be utilized for irrigation. This work
acre-feet of water, is
part of one of
the greatest of the se\'eral large land reclamation projects under-
taken by the U. S. Government for the watering and settling of arid tracts in the
The
dry zone of the western
profile is
states.
remarkable for the severe simplicity of
It closely follows the elementary profile right
line.
its
down
out-
to its
extreme base and forms a powerful advocate for this simple style of design.
example. first
The graphical procedure is similar The section is divided into three
two are comparatively
to that in the last
As the
divisions.
small, the triangle of forces in Fig.
25a
^'^
Fig. 26.
is first
Site
Plan for Roosevelt
plotted at a large scale in pencil
Dam
and the
inclinations of the
resultants thus obtained are transferred to the profile ; this accounts for the long projecting lines
near the origin of the force diagram
A
which also appear in some previous examples. overcoming this
for
when the
forces (1)
difficulty is that
and
(2) are
first
neater
method
adopted in the next
figure,
amalgamated into one before
being plotted on the force diagram.
In Fig. 25a,
N
scales roughly 19,000 sq.
tons, q also measures approximately 20 ,
and
mN
s=—r— = b
formula (10)
ft.,
ft.,
equivalent to 1425
then m=l-\
120
= 1.75,
160
1425X1.75 rr-r
loO
= __^ 15.5 tons.
P
5s
826
.^^tons. = -— = —-=5.1 160
^ By
DAMS AND WEIRS
58
c
= ^;^+\^^^+(5.1) = 7.75 + V86 = 17
With regard .,
tons roughly
=
to
W
q measures 23
,
ft.
m
and
works out to 1.86
mW ,„ —=— — = 1.86X1378 =16 tons per sq.
.
thereiore s
—
„^
^
it.
160
6
dam is built on a radius of 410 feet, measured from the measured from the extrados of the curve at the base it will be 420 feet and the arch stress as calculated from the "long" formula This
axis;
if
used in ''Arched
The
site
Dams"
will
amount
to 23.3 tons.
plan given in Fig. 26 forms an instructive example of the
arrangement of spillways cut in the sohd rock out of the shoulders of the side of the canyon, the material thus obtained being used in
the dam.
down
These spillways are each 200
feet
wide and are excavated
to five feet below the crest of the dwarf waste weir walls which
This allows of a
cross them.
much
greater discharge passing under
a given head than would be the case with a simple channel without a drop wall and with bed at the weir crest or afflux,
is
by
this
the height given to the
New
47.
level.
means diminished and that
dam
The "heading
up,
is
a matter affecting
profile of the
New Croton dam New York City
crest.
Croton Dam.
The
constructed in connection with the water supply of is
This
given in Fig. 27.
dam
has a straight alignment and
is
1168
accommodated by an overfull weir 1000 ft. in length, which is situated on one flank forming a continuation of the dam at right angles to its axis. The surplus water falls into the Rocky River bed and -is conveyed away by a separate channel. An elevation and plan of this work are given in Figs.
feet long.
Waste
fiood water
is
28 and 29.
The system
of graphical analysis
employed
ent from that in the last two examples and 18,
where independent combinations of
are used.
The
is
in this case
is differ-
that illustrated in Fig.
vertical
and
inclined forces
profile is divided into four divisions, the first being
a combination of two small upper ones. The further procedure after the long explanations already given does not require any special notice except to point out that the directions of the combined forces
V, l'+2', l'+2'+3' lines
on
tively.
Fig. 27a,
The
etc., in {d)
are
drawn
parallel to their reciprocal
namely to the chords Oa, Oh, Oc, and Od, respecresultants are R^ (R.F.) and (R.E.). The
final
W
DAMS AND WEIRS
W
value of
is
fflTF
out to 1.82 therefore
Wj
s
and
s '
1484 tons, consequently
case scales 20 feet
first
^ 1.82Xl380 190
b
=
and
m works
13.2 tons =c, as with
'
c are identical.
With regard
to
iV,
1.22X1484 1.22,
N is
1380 tons and that of
applying formula (10), g in the
59
q scales 7 feet, consequently
tons
9.5
s
As
only.
m=1+
42 190
P = 10,010 = 750
tons,
190
P
9.5
4 tons; "' therefore
moderate.
Fig. 27.
+^
= X1
which
is
very
It is probable that other external pressures exist
due
c "
b
Diagram
2
of Profile of
New
-
>
Croton
4
Dam
(4) 2
tons,
Showing Influence Lines as in
Fig. IS
front and rear, as also ice pressure, which would materimodify the result above shown. This dam, like the Cheeseman, is of the bottleneck profile, it is straight and not curved on plan.
to
filling in
ally
48. in Egypt,
the
Assuan Dam. The section. Fig. 30, is of the Assuan dam which notable work was built across the Nile River above
first cataract.
As
it
stands at present
it is
not remarkable for
DAMS AND WEIRS its
height, but
works,
is
what
made up
it
61
lacks in that respect, as in
in length,
which
latter is
6400
most eastern
feet.
No
single
modern times has been more useful or far-reacl:ing in beneficial results upon the industrial welfare of the people than this dam. Its original capacity was 863,000 acre-feet and the back water extended for 140 miles up the river. The work is principally
irrigation
work
of
remarkable as being the only solid
dam which
passes the whole
discharge of a large river like the Nile, estimated at 500,000 secondfeet,
through
its
low and 40 high
body, for which purpose sluices.
it is
provided with 140
These are arranged in groups of
ten,
each
STLEL RODS COfifiCCTlNO HLW WORK WITH OLD
40 H/6H LE\/EL VEtfTJ 6-^6'x //.4d'
Assuan
Fig. 30.
low
sluice is
Dam
across the Nile
Showing Old and
New
Profiles
23 feet deep by 6| feet wide with the dividing piers of the weight of the dam due to an excess of width over what would be sufficient in addition to which the maximum pressure in the
The diminution
16J feet wide.
sluices necessitates
for a solid
dam
;
piers is liniited to the extremely
The
low figure of 5 tons of 2000 pounds.
designers have thus certainly not erred on the side of boldness;
the foundation being solid granite, would presumably stand, with perfect safety, pressure of treble that intensity, while the masonry,
being also granite, set in cement mortar, carrying a safe pressure of 15 tons, as
is
certainly capable of
many examples
prove.
'"'iiii^'hiiuii'i'iii^K;^''^".
DAMS This
AJND WEIRS
63
dam has proved such a financial success that it has recently
been raised by 23 feet to the height originally projected. water thus impounded
is
nearly doubled in quantity,
The
i.e.,
to
over IJ million acre-feet; exceeding even that of the Salt River
As
was decided not to exceed the low unit by 16J feet throughout. A space has been left between the new and the old work which has been subsequently filled in with cement grout under pressure, in addition to which a series of steel rods has been let reservoir in Arizona.
it
pressure previously adopted, the profile has been widened
Fig. 32.
View
of
Assuan
Dam before Being Heightened with Sluices in Operation
by boring, and built into the new work. The enlargeshown in the figure. The sluices are capable of discharging 500,000 second feet; as their combined area is 25,000 square feet
into the old face
ment
is
this will
mean a
velocity of 20 feet per second.
the possibility of adjustment of level, gates, they will never be
A
sluices
in process, Fig. 33,
put to so severe a
of the sluice
test.
and longitudinal section shown in Fig. 31, a in operation. Fig. 32, and a view of the new work will give a good idea of the construction features.
location plan
view of the
Owing, however, to
by manipulation
64
DAMS AND WEIRS
DAMS AND WEIRS
Two further sections are
Cross River and Ashokan Dams.
49.
given in Figs. 34 and 35, the
dam
second of the Ashokan
of the Cross. River
first
New
in
65
dam, and the
Both are of unusually
York.
thick dimensions near the crest, this being specially provided to
enable the
dams
to resist the impact of floating
are left to be analyzed
vided with a vertical
a German innovation, which enables any
Dam
Fig. 35.
Fig. 36,
which
is
Dam.
New
feet.
The
It
South Wales not
784
feet
on the
is
batter 20 vertical to
far
crest,
1
from the new Federal Capital. the
maximum
height being 240
and the back
horizontal, both identical with those adopted crest
width
curve to a radius of 1200
feet.
This
The
a close copy of
Murrumbidgee
built across the
is
dam; the
acre-feet.
is
a further corroboration of the
fore batter is 3 vertical to 2 horizontal,
in the Roosevelt
Dam
thereby guarding against
generally termed "Barren Jack",
excellence of that profile.
Its length is
Ashokan
is now frequently adopted. The Burrin Juick dam in Australia,
the Roosevelt dam, Fig. 25, and
River in
off,
Profile of
This refinement
Burrin Juick
50.
pro-
is
leakage through the wall to be drained hydrostatic uplift.
is
line
Profile of Cross River
Fig. 34.
profiles
of porous blocks connected with two inspec-
This
tion galleries.
by
These
ice.
The Ashokan dam
the student.
material of which the
is
18 feet.
It is built
dam wdll impound dam is composed is
on a
785,000
crushed
sandstone in cement mortar with a plentiful sprinkling of large
'plums" of
granite.
The
ultimate resistance of specimen cubes
DAMS AND WEIRS
66
was 180 "long" tons, per square foot; the high factor of safety of 12 was adopted, the usual being 8 to 10. The maximum allowable stress will, therefore, reduce to 15 "long" tons = 16.8 American short tons.
With regard
to the
maximum
16,100, equivalent to 1210 tons,
/
S63-
stresses, for Reservoir Full,
and
N=
q scales. about 15 feet, conse-
DAMS AMD WEIRS The above proves that (R.F.).
the stress (R.E.)
is
67 greater than that of
Probably allowance was made for masses of porous fiUing and to be dam, which would cause
N
lying at the rear of the
W
and so equalize the pressure. It will be noticed that the incidence of N, the vertical component (R.E.) falls exactly shifted forward
at the edge of the middle third, a condition evidently observed in
the design of the base width.
^<:o5g ff
2000 Tons
Fig. 37.
Profile of
The dam
Arrow Rock Dam, Idaho, Showing Incidence
is
which
Centers of Pressure on Base
provided with two by- washes 400 feet wide; the
reservoir will be tapped sluices of
of
will
by a tunnel 14X13
feet,
the entrance
be worked from a valve tower upstream, a
dam. It is interesting to note that an American engineer has been put in charge of the construction of this immense work by the Commonwealth Governsimilar arrangement to that in the Roosevelt
ment. 51.
just
Arrow Rock Dam.
completed (1915)
project, a
U,
S.
is
The
reclamation work.
for^ curtain is 351 feet.
highest
the Arrow
A
dam
in the world
Rock on the
From
now
Boise, Idaho,
the crest to the base the
graphical analysis of the stress in th^
DAMS AND WEIRS
68 base
given in Fig. 37.
is
For Reservoir Empty, 17 = 2609 tons, and
=2
m=l-\
q measures 38 feet; therefore
222
= 23.5 2X^^ 222 1.73X2609
tons.
For Reservoir
^.^^^ =20.2tons.
222
4
(20.2)2
+ (7.3) = 22.6 -
P
Full,J qi
1610
^^"^^"=^22^ tons.
nearly and 5 =
—^ = b
= 27, and m = l+^^ =
^ ^
222
^
,
^
These values
20.2 .
^""Y"^
'
are,
of
course,
but
approximate.
Thus the compressive stresses (R.F.) and (R.E.) and the incidence of and also of N is close
W
equal,
Fig. 38.
are practically
to the edge of
Location Plan of Arrow Rock Dani
CouTUfiy of "Engineering Record^'
the middle third. crest.
The high
The dam
is
built
on a radius of 661
feet at the
stresses allowed are remarkable, as the design is
the gravity principle, arch action being ignored.
on
The curvature
doubtless adds considerably to safety and undoubtedly tends to
reduce the compressive stresses by an indeterminate but substantial
amount.
It is evident that
formula (10) has been applied to the
DAMS AND WEIRS design.
69
dam
Reference to Figs. 38 and 39 will show that the
divided into several vertical sections by contraction joints. also provided with inspection galleries in the interior
weeper drains 10
which
is
feet apart.
carried to a
and
is
It is
vertical
These intercept any possible seepage,
sump and pumped
out.
TopB.jns
These precautions are ^'JJS^.
^^
^
Eorih Surface'^
Diversion Tunnel
Fig. 39.
Elevation of Arrow
to guard against hydrostatic uplift.
The
resembling that of the Roosevelt dam,
52.
Rock
is
Dam
simplicity of the outline,
remarkable.
SPECIAL FOUNDATIONS Dams Not Always on Rock. Dams are not always founded
on impervious roclcbut sometimes, when of low height, are founded
on boulders, gravel, or sand. These materials when restrained from spreading, and with proper arrangements to take care of subpercolation, are superior to clay,
ous material to deal with. base of a dam,
it
which
When
latter is
always a treacher-
water penetrates underneath the
causes hydrostatic uplift, which materially reduces
Fig. 40 represents a wall on a pervious stratum and upholding water. The water has ingress into the substratum and the upward pressure it will exert at c against the base of the wall will be that due to its depth,
the effective weight of the structure. resting
in this case
30
feet.
Now the
point of egress of the percolation will
and, as in the case of a pipe discharging in the open, pressure
be at
6,
is nil
at that point; consequently the uplift area below the base will
be a triangle whose area equals
HXh
The diagram,
the combinations of the horizontal water pressure
V
P
Fig. 40,
w^ith the
and with the weight of the wall W, P bined with F, Ri resulting, whose direction is upward. static uplift
shows
hydro-
is first
com-
Ri
then
is
DAMS AND WEIRS
70
combined with W,
i?2 being their resultant. The conditions without uphft are also shown by the dotted line drawn parallel to dc in
Fig. 40.
The
piezometric
termed the hydraulic gradient;
ab
is
line, i.e.,
a
line
line
it is
also the
connecting water levels in piezometer
tubes, were such inserted. Fig. 41 shows the same result produced on the assumption that the portion of the wall situated below the piezometric line is reduced in weight by an equal volume of water, i.e., the s.g. of this part may
be assumed reduced by unity,
Fig. 40.
i.e.,
Effect of Uplift on
from 2.4 to
Dam Shown
The combination
with that shown in Fig. 40.
of
1+2
with
The
wall
is
Graphically
thus divided diagonally into two parts, one of of s.g. 1.4.
1.4.
P
s.g.
2.4
and the other
is identical
in result
In the subsequent section, dealing
with "Submerged Weirs on Sand", this matter of reduction in weight
due to
flotation is frequently referred to.
53.
Aprons Affect
principle involved in
onal profile abCj uplift is absent.
is
Uplift.
Fig. 42
dams with porous
is
further illustrative of the
foundations.
The pentag-
of sufficient base width, provided hydrostatic
Supposing the foundation to be porous, the area
DAMS AND WEIRS of uplift will be
ctifec,
which
in
bai,
equals ab.
71
This area
is
equal
to abc, consequently practically the whole of the profile lies below
the hydraulic gradient, loses weight;
H
=
b
may
be considered as submerged, and hence
can thus be assumed as reduced by unity,
p — 1. The
from p to
making
its s.g.
correct base width will then be found
H instead of —r=
Vp-1
The new
profile will
by
then be adb;
\p
the base width having been thus extended, the uplift increased in the
i.e.,
same proportion.
Now
is
likewise
supposing an impervious
apron to be built in front of the toe as must be the case with an overfall
dam; then the area
Diagram Showing
Fig. 41.
of uplift
becomes
Weight Submersion
Identical Result If
Due
to
baie,
and the
Is Considered
piezo-
Reduced
all these cases happen Under these circumstances the
metric line and hydraulic gradient, which in to be one and the same
line, is ae.
comparatively thin apron
is
subjected to very considerable uplift
blow up unless sufiiciently thick to resist the hydrostatic pressure. The low water, or free outlet level, is assumed to be at
and
will
the level
e,
consequently the fore apron
lies
above this
level
and
is
considered as free from flotation due to immersion. 54.
Rear Aprons Decrease
apron substituted. ing water
Uplift.
Another case
will
now be
In Fig. 42 suppose the fore apron removed and a rear
examined.
is
In this case the point of ingress of the percolat-
thrown back from
b to 6' the hydraulic gradient is a'cj
DAMS AND WEIRS
72
the triangle of hydrostatic upHft is
more than neutralized by the
is b'azc.
This uplift from
6'
to b
rectangle of water a'ahV ^ which
overlies the rear apron; the latter is therefore not subject to
any
from erosion by moving water, consequently it can be made of clay, which in this position is water-tight as concrete masonry. A glance at Fig. 42 uplift and, oA\ing to its location, is generally free
will
demonstrate at once the great reduction of uplift against the
base of the wall effected by the expedient of a rear apron, the uplift being reduced from a^c to fhc, more than one-half.
Fig. 42.
apron
is
DIasram Showing
Uplift with
as
heavy slabs with open joints.
ing clay
is
55,
it
if
solid,
should
should be formed of open work,
In the rear of overfall dams stanch-
by natural Many works owe
often deposited
effective rear apron.
although
else
rear
and without Fore and Rear Aprons
a sure remedy for uplift while the fore apron,
be made as short as possible, or
Thus a
process, thus forming an their security to this fact
often passes unrecognized.
Rock Below Gravel.
Fig. 43 represents a
dam
founded
on a stratum of pervious material beneath which is solid rock. A fore curtain wall is shown carried down to the impervious rock. The conditions now are worse than those resulting from the imper-
DAMS AND WEIRS
73
vious fore apron in Fig. 42 as the hydraulic gradient and piezometric line are
now
pressure
is
is,
The reduced
horizontal.
area of vertical hydrostatic
1066 against which the wall can only furnish 1200; there
an
therefore,
effective area of only 134 to resist
overturning as the graphical stress lines clearly
diaphragm curtain wall
position of a
the dam; in this location
is
a water pressure
by sliding or prove. The proper
must
at the rear of 800, consequently the wall
fail
at the heel, not at the toe of
prevent
it will effectively
all uplift.
In
the case where an impervious stratum does not occur at a reasonable
depth the remedy
is
to provide a long rear apron which will reduce
may be desired, or else a combination of a vertical diaphragm with a horizontal apron can be hydrostatic uplift to as small a value as
PIEZOMETRIC LINE
SOLID Fig. 43.
used.
In
many
be provided
H
= 60'
ROCK
Effect of Impervious Fore Curtain Wall on Uplift
cases a portion only of the required rear apron need
artificially.
With proper precautionary measures the
deposit of the remaining length of unfinished apron can safely be left for
the river to perform
by
silt
deposit,
if
time can be afforded
for the purpose.
56.
Gravity
Dam
Reinforced against Ice Pressure.
tion will be concluded with a recent example of a gravity
forced against ice pressure, which St.
Maurice River
dam
is
This sec-
dam
rein-
given in Fig. 44, viz, that of the
situated in the Province of Quebec.
The
taken as 25 tons per foot run, acting at a level corresponding to the crest of the spillway, which latter is shown in Fig. ice pressure is
58.
The
profile of Fig.
44
is
pentagonal, the crest has been given
DAMS AND WEIRS
74
the abnormal width of 20 feet, while the base
which
The
is
f of the height, about the requirement, were ice pressure not considered. is
horizontal ice pressure, in addition to that of the water upheld,
will cause the line of pressure to fall well outside the
middle third,
thus producing tension in the masonry at the rear of the section.
To
obviate this, the back of the wall
is
reinforced with steel rods to
the extent of IJ square inches per lineal foot of the dam. If the safe tensile strength of steel be taken at the usual figure of 16,000
Tee F^ressure
liiiil
Fig. 44.
Profile of Saint
Maurice River
Dam
at
Quebec
pounds, or 8 tons per square inch, the pull exerted by the reinforcement against overturning will be 12 tons per foot run. This force
can be considered as equivalent to a load of
like
the back of the wall, as shown in the figure.
dam
is
amount applied at The section of the
divided Into two parts at El 309 and the incidence of the and at the base is graphically obtained.
resultant pressure at this level
The file.
line of pressure connecting these points is
The
line falls outside
drawn on the pro-
the middle third in the upper half of
the section and within at the base, the inference being that the
DAMS AND WEIRS section
75
would be improved by conversion into a trapezoidal outline
with a narrower crest and with some reinforcement introduced as has been done in the spillway section, shown in Fig. 58. It will
be noticed that the reinforcement stops short at El
allowed for by assuming the imposed load of 12 tons removed at the base of the load line in the force polygon. The line jRs starting from the intersection of Ri with a horizontal through El 275.0 is the final resultant at the base. This example is most
This
275.
is
instructive as illustrating the combination of reinforcement with a
gravity section in caring for ice pressure, thus obviating the undue
enlargement of the
profile.
GRAVITY OVERFALL DAMS OR WEIRS Characteristics of Overfalls.
57. crest of a
dam
it is
termed an overfall
When water overflows the dam or weir, and some modi-
fication in the design of the section generally
Not
becomes necessary.
only that, but the kinetic effect of the falling water has to be
provided for by the construction of an apron or floor which in
by
many
most important part of the general design. This is so pronounced in the case of dwarf diversion weirs over wide sandy river beds, that the weir itself forms but an insignificant part of the whole section. The treatment of submerged weirs with aprons, cases forms
will
far the
At present the
be given elsewhere.
section of the weir wall
alone will be dealt with.
Typical Section,
Fig.
45
is
a typical section of a trapezoidal
weir wall with water passing over the crest. as before, will be designated crest
by
upper
d,
still
and that water
of river
by H, that below by D.
level, will therefore,
be
The
height of the crest
of reservoir level above
The
total height of the
H+d,
The depth
of water passing over the crest should be measured upstream from the overfall just above where the some distance
break takes place; the actual depth over the crest
is less
of the velocity of the overfall being always greater
approach. reach.
On
by reason
than that of
This assumes dead water, as in a reservoir, in the upper a river or canal, however, the water
is
in
motion and has In order to
a velocity of approach, which increases the discharge. allow for this, the head
Qi)
corresponding to this velocity, or
— ^9
DAMS AND WEIRS
76 multiplied
by
1.5 to allow for impact, or
added to the
/j=.0233F^ should be
Thus supposing the mean
reservoir level.
velocity
river in flood to be 10 feet per second 100 X. 0233 or 2.3 would have to be added to the actual depth, the total being 15 feet in Fig. 45. The triangle of water pressure will have its apex at the surface, and its base will, for the reasons given previously, be taken as the depth divided by the specific gravity of the feet
of the material of the wall.
S.S -lSh-s
AFFLUX
AFFLUX^ IJ
The
triangle of water pressure will
h,
4
\
Fig. 45.
Typical Section of Trapezoidal Weir Wall
be truncated at the crest of the
overfall.
The water
pressure acting
by a
against the back of the wall will thus be represented
not a triangle, whose base width
level
—
.
is
and
its
Its area therefore (back vertical) will
If
the back
is
inclined the side of the trapezoid
general formula
is
top width at crest
be (
p
trapezoid,
\
—J
|~-
1
9
becomes
9
J^i.
therefore
^ = (/f
.(iy+2(i) or 7/x)X-
2p
^
The
(15)
DAMS AND WEIRS Hi being the distance of
formula
back of the wall. The vertical application above the base according to
inclined length of the
its
point of
page 19
(5)
77
is
A
= —-I 3
whether the back
is
'
and
„| \H+2a/ ,
will
be the same
vertical or inclined.
Approximate Base Width. With regard to the drop wall owing to the overfall of water and possible impact of floating
58. itself,
timber,
ice,
or other heavy bodies, a wide crest
further strengthening
is
effected
The ordinary approximate
by adopting the
rule for the base
is
a necessity.
A
trapezoidal profile.
width of a trapezoidal
weir wall will be either ,
(H+d) (16)
or
(H+M) (16a)
The
correctness of either will depend
such as the value of
head and
also of
on various considerations,
the depth of the overfall, that of
d,
D, the depth of the
tail
h or velocity
water; the inclination given
and lastly, whether the weir wall is founded on a porous and is consequently subject to loss of weight from uphft.
to the back,
material
Hence the above formulas may be considered as approximate only and the base width thus obtained subject to correction, which is easiest studied by the graphical process of drawing the resultant on to the base, ascertaining
its
position relative to the middle third
boundary. 59. it
may
Approximate Crest Width.
With
reference to crest width,
be considered to vary from t
= Vi/+cZ
(17)
fc
= V^+Vd
(18)
to
the former gives a width sufficient for canal, or reservoir waste weir walls,
but the latter
when the weir In
many
wall
is
is
cases,
more
suitable for river weirs,
and
is
quite so
submerged or drowned.
however, the necessity of providing space for during times when the weir is not
falling shutters or for cross traffic
— DAMS AND WEIRS
78
acting, renders obligatory the provision of
With a moderate width, a
an even wider
order to give the requisite stability to the section.
by joining the edge
This
of the crest to the toe of the base
the base width of
line,
^^
crest.
trapezoidal outline has to be adopted, in
by a
-=— being adopted, as shown
formed
is
straight
in Fig. 45.
Vp
When of the of
the crest width exceeds the dimensions given in formula
the face should drop vertically
(18),
elementary
An
dams.
The
weir.
profile, as is
example of this
is
if
meets the hypothenuse
given in Fig. 52 of the
tentative section thus
graphical process and
till it
the case with the pentagonal profile
Dhukwa
outlined should be tested
by
necessary the base width altered to conform
with the theory of the middle third.
In Fig. 45 with
d=15
is
feet.
given a diagram of a trapezoidal weir 60 feet high
According to formula (17) the crest width should
be V75 = 8.7feet, and according to
(18),
7.74+3.87 = 11.6.
age of 10 feet has been adopted, which also equals
—
^
An The
aver-
profile
P therefore, exactly corresponds with the elementary triangle canted
forward and truncated at the overfall 60.
Graphical Process.
crest.
In graphical diagrams, as has already
been explained, wherever possible half widths of pressure areas are taken off with the compasses to form load lines, thus avoiding the
and calculating the areas of the which is always liable to error. There
arithmetical process of measuring several trapezoids or triangles, are,
however, in this case, three areas, one of which, that of the
an altitude of only half of the others. overcome by dividing its half width by 2. If one
reverse water pressure, has
This difficulty height
is
is
not an exact multiple, as this
is
of //, a fractional value
given to the representing line in the polygon will often be found to obviate the necessity of having to revert to superficial measures.
The
application of the reverse pressure Pi here exhibited
is
similar
to that shown in Fig. 16; it has to be combined with R, which latter This combination is effected in is obtained by the usual process. the force polygon by drawing a line Pi equal to the representative .
width of the back pressure, in a reverse direction to P. The closing line i?i is then the final resultant. On the profile itself the force line Pi is continued through its center of gravity till it area, or half
DAMS AND WEIRS intersects R,
from which point Ri
is
drawn
tion of the face of the weir is very flat, as
may
be so deflected as to intersect
R
79
to the base.
If this por-
sometimes the case, Pi
is
below the base altogether as
shown in Fig. 50. In such event, i?i is drawn upward instead of downward to intersect the base. The effect of Pi is to throw the It imresultant Pi farther inward but not to any great extent. is
proves the angular direction of P, however.
A dam is usually, but not invariably, exempt
Reverse Pressure.
from the
effect of reverse pressure.
This reverse water pressure
generally, as in this case, favorable to the stability of the weir,
there are cases
when
its
action
is
either too slight to
is
but
be of service or
This occurs when the face of the weir wall is which points to the equiangular profile being most suitable. An example illustrative of the above remarks is given later in Fig. 50 of the Folsam dam.
is
even detrimental.
much
inclined,
As the moments side of the weir wall is
of the horizontal pressure of water
small influence and
may
adopted, the slope
is
tail
given to the face;
given a
downward
it
water will have but
When
well be neglected.
is all
reverse water pressure its
either
vary almost with the cubes of their height,
evident that a comparatively low depth of
is
on
a vertical back
by which the normal
inclination that reduces
capacity for helping the wall. 61.
of the
Pressures Affected by Varying Water Level-
Calculations
depths of water passing over the weir or rather the height of
reservoir level
above the weir
corresponding depth
D
crest,
by
designated
d,
and of the
in the tail channel, are often necessary for
the purpose of ascertaining
what height
of water level upstream,
on the weir wall. In low submerged or drowned weirs, the highest flood level has often the least effect, as at that time the difference of levels above and or value of d, will produce the greatest effect
below the weir are reduced to a minimum.
This
is
graphically
weir wall,
which represents a section of the Narora dwarf to which further reference will be made in section 124,
Part
In this
shown
in Fig. 46,
11.
profile
shown, of which Pi, due to
under comparison,
The Narora
falls
two resultant
much
pressures,
P
Pi, are
nearer the toe of the base.
weir, the section of the weir wall of
insignificant, is built
and
lower water level of the two stages
across the Ganges River in
which
is
so
Upper India at the
DAMS AND WEIRS
80
head of the Lower Ganges Canal, Fig. 93. The principal part of which is founded on the river sand, consists not in the low
this work,
weir wall, although that
is f mile long, but in the apron or floor, which has to be of great width, in this case 200 feet. As will be seen in Fig. 46 the flood level of the Ganges is 16 feet
above the
two
floor level, while the afflux, or level of the
The
feet higher.
when
river discharges
When
in flood.
full
less
is
when
AFFLUA
Fig. 46.
much tail
it will
this occurs
be seen that the stress on
than when the head water
is
l8'-0
Section of Narora
lower.
is
flood occurs, the weir is completely
drowned, but from the diagrams the wall
head water,
about 300,000 second-feet
Dwarf Weir Wall
This result
is
across
Ganges River in Upper India
due to the reverse pressure of the
water.
The
rise of
the river water produces, with regard to the stress
induced on the weir, three principal situations or "stages" which are
enumerated below. (1)
When
the head water
where a water cushion is
exists,
is
at weir crest level; except in cases
natural or
artificial,
the
tail
channel
empty, and the conditions are those of a dam. (2)
When
the level of the
tail
water
lies
but above half the height of the weir wall. rocal depth of the head water above crest
below weir crest level In this case the recip-
is
found by calculation
:
.
DAMS AND WEIRS At highest
(3)
and
tail
fall
dam
water
minimum.
the greatest stress
is
In an unsubmerged weir or over-
generally produced during stage (3).
In a submerged weir the greatest stress
Moments
62.
between the head
flood level, the difference
at a
is
81
is
produced during stage
The moments
of Pressure.
(2).
of the horizontal
water pressure on either side of a wall are related to each other in
In cases where
proportion to the cubes of their respective depths. the wall
is
latter, as
(M) of
overflowed
we have
by the
seen,
is
water, the triangle of pressure of the
this trapezoidal area of pressure will
area with
formula
h,
(5)
The moment
truncated at the weir crest.
be the product of
its
or the product of the expressions in formula (1) and as follows
or
Hw ^=-^
iH+U)
(19)
That
of the opposing tail water will be
these
two being the resultant moment.
shown
in Fig. 46, during stage (1)
=1^=166.6—. 6p
In stage
— M = —-Dho op
= 10,
the difference of
For example,
if = 10,
(2) //
,
Z* = 0,
(^
in the case
unbalanced moment
= 3.5, and D = 10.
Then
p
the unbalanced
In stage
(3)
moment
will
H = 10, D = 1Q;
be ;^ [(100 op
d^S, and
X 20.5) - 1000] =175— .
D- if = 6 feet.
There
p will
thus be two opposing trapezoids of pressure, and the difference in their
moments
will
be
w(100X28) ^
w(100x34)
6p
Qp
Thus (3),
^w_ p
stage (2) produces the greatest effect, the least being stage
In this expression
symbolizes, as before, the unit weight
(to)
of water, per cubic foot or,
^
ton.
many weir wall sections have been under the erroneous supposition designed that the overturning moment is greatest when the upper water is at crest level and the tail In spite of this obvious
channel empty,
i.e.,
fact,
at a time
when the
difference of levels
above and
DAMS AND WEIRS below the weir
is
at a
maximum, or at full flood when the difference is at a minimum. 63. Method of Calculating Depth of
During the
Overfall.
second stage of the river the value. of d,
the depth of the over-
have to be
will
fall,
To
calculated.
effect
this the discharge of
the river must
estimated
first
when the
reaches
surface
be
the
crest level of the weir,
which .a
is
done by use
Q=
of the formula,
Ac^rs, given in secO
page 47 of
.3
tion
"a
"Hydraulics,
<
35,
Ameri-
can School of Correspondence",
A
being
the area, equal to d
times length of weir (c)
Kutter's
coeffi-
cient, (r) the hydraulic
mean
and
radius,
5,
the surface grade or slope of the river.
The
discharge for the whole river should
divided
now be
by the
length
of the weir crest, the
quotient
giving
the
unit discharge, or that SS3S9NI JO J-NlOd
per foot run of the weir.
DAMS AND WEIRS The depth is
83
required to pass this discharge with a free overfall
found by use of Francis' formula of 3.33d^ or a modification of it
for
wide crest weirs for which tables are most useful.
See ''Hydrau-
lics", section 24, p. 30.
For example, supposing the
up
to crest level
3.33(i^
= 20. Whence
rf^
= 6 and d=^lQ^ = 3.3
64.
by
feet.
Illustrative
Example.
47
Fig.
illustrates
(2)
head water
With
(3)
tail
is
assumed as 4
reciprocal depth of the
and and head water assumed 7
feet
crest.
From
three resultants have been worked out graphically.
their location
case.
feet;
water at crest level
deep above
The
an assumed
15 feet high, 3 stages are shown:
is
When head water is at crest level; When tail water is 7^ feet deep, and the
(1)
This ignores
the velocity head, or by .0155 T^.
(hi)
Here the weir
Then
which would be to
velocity of approach, a rough allowance for
decrease d
water
river discharge with tail
20 second-feet per foot run of the weir.
is
on the base the greatest
stress is
due to
i?2, i.e.,
stage (2).
The hydraulic gradients of
all
weir body
lies
shown with more than half the which here corresponds
three stages have been
an assumed rear and fore apron on
floor.
below the piezometric
with the hydraulic gradient, while in
In
line,
(2)
(1)
nearly the whole
lies
below
and in (3) entirely so. Owing to this uplift it is well always to assume the s.g. of a weir wall under these conditions as reduced by immersion to a value of In these cases the triangles of water pressure are shown with p—1 this line
.
their bases
made
p—1
,
or •-— instead of ,
—
-.
Actually, however, the
2.4
1.4
resistance of the weir wall to overturning relative to its base at floor level is
not impaired by flotation, but as weight in these cases
desideratum, the weir wall should be designed as
The rear apron is evidently and
its resisting
power,
i.e.,
65.
weirs will
this
is
effective weight, is
a
were the case.
subject to no uplift, but the fore apron
See section 52 and also the later sections on
Sand", Part
if
is,
impaired by flotation. ''
Submerged Weirs
in
II.
Examples of Existing Weirs.
now be
given.
Fig. 48
is
a
Some examples of existing LaGrange over-
profile of the
DAMS AND WEIRS
84 fall
dam
No less
at the
head
of the
Modesto and Tuolumne
than 13 feet depth of water passes over
canals, Fig. 49.
its crest,
2 feet being
£1.315
f£!EVO
Fig. 48.
Profile of
LaGrange
Overfall
Dam at
\CA4iAL HEAD
Head
of
LWL
/99.0
Modesto and Tuolumne Canals
added to allow for velocity of apIt is built on a curve of
proach.
300 feet radius.
The
analj^sis of the section
graphical
shows that
the resultants (R.E.) and (R.F.)
drawn on the
profile fall within
f-\S'^o
the middle third.
In this process
the reverse pressure due to
water
has
effect will
been
neglected.
Location Plan of LaGrange Weir
whether
the reverse pressure actually exercised is that
Fig. 49.
Its
be small.
It is a doubtful point --'~.:?6o
tail
due to the
of the tail water.
causes a disturbance and probably more or less of
full
depth
The overflow a vacuum at
the toe of the weir wall, besides which the velocity of impact causes
a hollow to be formed which must reduce the reverse pressure.
some
instances^ as in the case of the Granite Reef
dam,
In
Fig. 55, the
DAMS AND WEIRS effective
deptb of the
tail
water
is
assumed as only equal to that
This appears an exaggerated view.
the film of OTerflow.
dam, the
a
that
can well be neglected altogether.
it
water
rises to f or
more
of
How-
effect of the reverse is often so small
ever, in
iiigh overfall
85
In cases where the
of the height of the
dam
its effect
tail
begins to
be considerable, and should be taken into account.
"Ogee"
Objections to
66,
seems
now
Overfalls.
curved base of the fore slope which
American
Professional
opinion
to be veering round in opposition to the "bucket" or
overfall
dams.
is
so pronounced a feature in
Its effects are
undoubtedly mischievous,
as the destructive velocity of the falling water instead of being
reduced as would be the case is
if it fell
direct into a cushion of water,
conserved by the smooth curved surface of the bucket.
lately constructed
Bassano hollow dam
the action of the bucket
is
(see Figs.
In the
84 and 85, Part H),
sought to be nullified by the subsequent
addition of baffles composed of rectangular masses of concrete fixed
on the curved
slope.
The
following remarks in support of this
view are excerpted from ''The Principles of Irrigation Engineering"
by Mr. F. H. Newell, formerly Director of United States Reclamation Service. "Because of the difficulties involved by the standing wave or whirlpool at the lower toe of overflow dams, this type has been
made
many
in
cases to depart from the conventional curve
and to
drop the water more nearly vertically rather than to attempt to shoot
it
away from the dam
in horizontal lines."
Folsam Weir. Fig. 50 is of the Folsam weir at the head of the canal of that name. It is remarkable for the great depth of passing the flood water over crest which is stated to be over 30 feet 67.
deep.
The
stress lines
have been put on the
profile
with the object
of proving that the reverse pressure of the water, although nearly
due to the flat incKnawhich has the effect of water on the toe where it is least wanted
40 feet deep has a very small
effect.
This
is
tion given to the lower part of the weir,
adding a great weight of
and thus the salutary neutralized.
effect of the reverse pressure is
omy would undoubtedly equiangular profile,
and to drowned but very
more than
The section is not too heavy for requirements, but econ-
little
were canted forward to a nearly and this applies to all weirs having deep tail water,
weirs.
result
if it
It will
be noted that a wide crest allows
consequent reduction in the base width in any case.
DAMS AND WEIRS
86
The stress diagram in Figs. 50 and 50a are interesting as showmethod of combining the reverse pressures with the ordinary
ing the
Haessler's diagram of the direct water pressure. ,The profile
is
divided into three parts as well as the direct water pressure, whereas the reverse pressure which only extends for the two lower divisions is
in
i?2 is
two
The
parts.
stress
force 1" before
it
reaches
its
H£:AD
Fig. 50.
pressure of Rz,
is
diagrams present no novel features objective 3^
WA TER H+
ifs
its
effect of the reverse
Graphical Analysis of Folsam Weir
to deflect the direction of the resultant in the direction
which
latter, as
continues
reverse line
The
d
shown
in the force polygon, Fig. 50a,
resultant of 1'^ set out from the point
ant
till
This force on the profile comes in contact with reverse
reached.
till it
meets
drawn upward
3^.
6,
The
and
of R^.
resultant of
is
the
3^ is
the
the dotted line joining the termi-
nation of 3S i.e., (a) with that of 1". Following the same method the resultant R\ 2",
Ri and
is
result-
to meet the vertical force 3, parallel to
reciprocal in Fig. 50a, which
downward to meet
The new
which
latter in the force
is
next drawn
polygon
is
set out
DAMS AND WEIRS from the ternination of the vertical 2" is the final R^.
This
secting the base at B.
line is
If
87
The
(3).
and
resultant of Ra
drawn upward on the
profile inter-
the reverse pressure were left out of con-
R2 would continue on to its intersection with and thence the reverse recovery line drawn to meet (3) will be
sideration, the force 3^
This reverse
ba (not drawn) in the force polygon.
parallel to
the line
will intersect
line
same spot
(3) in the profile almost at the
as before.
The
final line will
and
in Fig. 50a)
will cut
be parallel to
its reciprocal
ca (not
the base outside the intersection of
drawn
To
i?5.
prevent confusion these lines have not been drawn on; this proves that the effect of the reverse pressure of the wall, except in the profile
were
tilted
detrimental to the stability
is
If the
matter of the inchnation of R^-
forward this would not be
If Pi the resultant
so.
water pressure at the rear of the wall be drawn through the to intersect the resultant of +ij2, this
all
point will be found to be the same as
producing the
profile
1+2+3+^)^ that obtained by
the vertical forces, viz,
R^ backwards to meet Pi.
final
Determination of Pi.
To
effect this,
the position of Pi has to
be found by the following procedure: The load line db, Fig. 50a, continued to
and
oCj oj,
so as to include the forces 3,
I,
ol are
Vi,
and
V2*
drawn; thus a new force polygon dol
is
The
is
rays
formed to
is made reciprocal. This decides the 1+2+3, viz, the center of pressure (R. E.) as also that of W+V1+V2 which latter are the reverse pressure loads. The location of Pi is found by means of another funicular polygon C
which the funicular. Fig. 50b, position of
W,ov
of
derived from the force polygon oad,
Pi
oe;
is
then drawn through the
resultant last mentioned at A.
with P5 on Fig. 50.
The
by drawing the rays profile intersecting
The
line
vertical fine
AB
through
is
A
oa, ofj
then coincident is
not
iV, i.e., is
not identical with the vertical in Fig. 50, for the reason that the resultant of tion
is
all
N
is
the vertical forces, whereas the vertical in ques-
the centroid of pressiu'e of
all
the vertical force less Wi, the
weight of water overlying the rear slope of the wall.
N
and
the vertical
The
location
found by drawing a horizontal P through the intersection of Pi with a line drawn through the e.g. of the triangle of water of
is
pressure Wj this will intersect the back continuation of
A
vertical
CD
through this point
will
BA
at
c.
correspond with that marked
DAMS AND WEIRS
88
N in Fig. 50.
The
profile. Fig. 51, is
a reproduction of that shown
Fig. 50 in order to illustrate the analytical
that
method
of calculation or
by moments.
The
Analytical Method.
68.
R
incidence of the resultant
is
required to be as ascertained on two bases, one the final base and
the other at a level 13 feet higher. is
divided into three parts
(3) of
838 square
mula
(7) to
feet.
:
The
section of the wall as before,
(1) of area 840 square feet, (2) of 1092,
The
position of the e.g. of (1)
and
found by for-
be 15.15 feet distant from a the heel of the base and will be
/r^r
FLOOD
is
LEVEL
Fig. 51.
Diagram
15.65 feet from
/FS7
^1.35
of
Folsam Weir
That of
h.
Illustrating Analytical
(2) is
Method
of Calculation
32.3 feet distant from its heel
reduced area of the water overlying the back down to h
is
Tk
h.
estimated
and by formula (6) to be .5 feet distant from 6. Again the reduced area of the reverse water overlying the fore at 26 square feet
slope
1)1
is
92 square feet and the distance of
its e.g.
from
6 is 55
18.5 "
= 48.8
feet.
with that of
3
The moments of all these vertical forces equated their sum (iV) about the point h will give the position
of iV relative to
6.
— DAMS AND WEIRS
89
Thus 840X15.65-13146 1092X23.3 =25443
(1) (2)
=
{w)
26 X
(vi)
92X48.8 = 4490
.5
=43092 -Moment
2050Xa;
=
X
:,
To
P=1257 and
trapezoid having
its
wall
by formula
is
calculated
base at
&,
and
above base
its e.g.
^_ ^'
8
is
120 square
For the lower base, the statement its
,
"2050"^^*^^*
2050
and
feet,
Consequently:
feet.
(1257X22.1)-(120X8) _ 26820 _.,^,
lows, Vz being 240,
Now
^^
Again the reduced
to be 22.1 feet.
is
^-tjt.
crest level with that of the
its
area of the reverse water pressure triangle pi
the height of
N
—
/=
i?,
of
nearly
feet,
the height of the e.g. of the
and
(6),
21
N
obtain the distance / between
the reduced area of
13
of
moments about c is by formula (6),
as fol-
distance 65 feet
(iV)2050X(21 + .3)=43665 (3)
838X32.3
=27067
{w)
lOX .15 240X65 3138XX
=15600
fa)
Total
86334
=
2
=86334
= 27.4
feet
3138
Now/i =
—
vi+v2
^^
^
j^ being the distance
between
iVi
and Rx,
iVi
The value
of Pi, the trapezoid of water pressure
27
feet,
that of
^o = 121
feet.
Then
(5) is
, *^'
is
to the base
its e.g.
^ (1747x27)-(285Xl2i) _ 47169-3514 ^43655^ ~ 3138
The
R
{pi+V^
down
by formula (19) or 285 square feet and its lever arm
1747 square feet and the height of
c, is
and
are lines
3138
N and iVi being obtained, the directions of drawn to the intersections o| the two verticals
positions of
i?i
3138
DAMS AND WEIRS
90
N and
with two
lines drawn through the c.g.'s of the trapezoid of by the mcnnent of the reverse pressure, if any, or by (P—p). This area will consist, as shown in the diagram, of a trapezoid superposed on a rectangle; by using formula (5) section 1, the iVi
pressure reduced
positions of the e.g. of the upper trapezoid
above the base at
a,
the rectangle, then
by taking moments
height of the
found to be 23.6
e.g. is
found, to be 12.58 feet
is
while that of the lower
is
at half the depth of
of these areas about
feet
above the base at
down
the height for the larger area [Pi—{'Pi+p2)]
to c
6,
is
the
while
6,
27
feet.
In the graphical diagram of Fig. 51a the same result would be obtained by reducing the direct pressure by the reverse pressure area. Thus in the force diagram the vertical load line would remain unchanged but the water-pressure load line would be shorter being P—p and Pi—(pi+p2), respectively. This would clearly make no difference in the direction of the resultants R and Ri and would save the two calculations for the c.g.'s of P and Pi.
This weir long,
in the
when
which
is
is
masonry it falls.
provided with a crest shutter in one piece, 150 feet
and lowered by hydraulic jacks chambered up by the gate This is an excellent arrangement and could be imiraised
of the crest so that they are covered
tated with advantage.
The
shutter
at base of lamina 2 of this weir
is
is
55
The width
5 feet deep. feet,
or very nearly
—-^, Vp
formula (16).
Dhukwa Weir.
69.
dam is
This overfall of the crest this
The
A
very similar work
is
the
is
of pentagonal section.
stress resultant lines
does not
formula
—Vp TJ
is
I
up
Owing
have been drawn on the
to half the height of the weir.
nil.
applicable in stage
profile,
The
tail
which water
Consequently the
63X— = 42
3.
The
effect of the tail
water
According to this formula the base width would
2
be
to the width
J
j^r- is
practically
weir
obviously the best outline.
prove the correctness of the base width adopted. rise
Dhukwa
which has been recently completed.
in India, Fig. 52,
feet,
which
it
almost exactly measures
—a further
o
demonstration of the correctness of the formula. should be, according to formula (18),
The
V50+Vl3 = ll
crest width feet.
The
DAMS AND WEIRS width of 17 feet adopted
is
necessary for the space required to work
These are of
the collapsible gates.
Ql
steel,
by batches by
are held in position
struts connected with triggers,
and can be released
chains worked from each end.
The
in
gates, 8 feet high, are only 10
This involves the raising and lowering of 400 gates, the
feet wide.
The arrangement adopted
weir crest being 4000 feet long.
Folsam weir
hydrauUc jacks operating long gates
of
An excellent feature in this design is the subway with
is
in the
far superior.
occasional side
chambers and lighted by openings, the outlook of which is underneath the waterfall, and has the advantage of relieving any vacuum under the
falling water. AFFLUX
El. 903
->-h
p
850
'^
ei 840
p
" ^
Fig. 52.
^f
^
-"
'"^^
'' i
MyM^M/zd^
Graphical Analysis of Profile of
The subway could be
Dhukwa Weir
in India
utilized for pressure pipes
and
for cross
communication, and the system would be most useful in cases w^here the obstruction of the crest
by
piers is inadvisable.
The
weir
is
4000 feet long and passes 800,000 second-feet, with a depth of 13 feet.
The
weir,
which
discharge is
very high.
feet per second. will
is,
therefore,
The
200 second-feet per foot run of
velocity of the film will be
With a depth
of 13 feet
still
—= -
15.4
w^ater, the discharge
be by Francis' formula, 156 second-feet per foot run.
To produce
a discharge of 200 feet per second, the velocity of approach must be
about 10 feet per second.
This
will
add
2.3 feet to the actual value
_
DAMS AND WEIRS
92 of
raising
rf,
it
from 13 to 15.3
feet
which
strictly
should have been
done in Fig. 52. 70.
Fig. 53,
of
American design,
the Mariquina weir in the Philippines.
It has the ogee
Mariquina Weir. is
Another high weir
curve more accentuated than in the LaGrange weir.
The
stress
have been dra\vn in, neglecting the effect of the tail water which will be but detrimental. The section is deemed too heavy at the upper part and would also bear canting forward with advanlines
tage,
but there are probably good reasons
why an exceptionally solid
A FFLUX ^h {A5SUMED)
Fig. 53.
crest
was adopted.
Profile of
Mariquina Weir
The ogee curve
in tlie Phi!ippine9
also is
a matter on which
opinion has already been expressed. 71.
Granite Reef Weir.
The Granite Reef weir over
the
Salt River, in Arizona, Figs. 54 and 55, is a work subsidiary to the great Roosevelt dam of which mention was previously made. It is founded partly on rock and partly on boulders and sand overlying rock. The superstructure above the floor level is the same throughout, but the foundations on shallow rock are remarkable as being founded not on the rock itself, but on an interposed
cushion of sand.
(See Fig. 54.)
Reinforced concrete piers, spaced
20 feet apart, were built on the bedrock to a certain height, to clear
DAMS AND WEIRS all inequalities;
these were connected
93
by thin
reinforced concrete
side walls; the series of boxes thus formed were then filled level with sand, and the dam built thereon. This work was completed The portion of the profile below the floor is conjectural. in 1908.
This construction appears to be a bold and commendable novelty. Sand in a confined space is incompressible, and there is no reason
why
it
A
should not be in hke situations.
suggested improvement
would be to abandon the piers and form the substructure of two long outer walls only, braced together with rods or old rails encased Fig. 55 is the profile on a boulder bed with rock below. in concrete.
Hydraulic Condi=
72.
The
tions.
levels
of the
afflux flood of this river are
obtainable so that the stresses
In most
can be worked out.
cases these necessary statis-
The
wanting.
tics are
flood
downstream has been given the same depth, 12 feet, as
^^
that of the film passing over
the crest.
This
erroneous.
The
is
clearly
velocity of
the film allowing for 5 feet per second approach,
is
quite
12 feet per second, that in 54. Section of Granite Reef Weir Showing Sand Cushion Foundations
the river channel could not
be
much
quently
over 5
Fig.
feet, conse-
would require a depth of
it
12X12 =
28
feet.
The dam
would thus be quite submerged, which would greatly reduce the stress.
As
previously stated, the state of
maximum
stress
would
probably occur when about half the depth of flood passes over the crest. However, the graphical work to find the incidence of the resultant pressure
on the base
given downstream flood level. given,
no
special
comment
reverse water pressure. into 2 straight lines
will
be made dependent on the
After the
explanations already
caUed for except with regard to the Here the curved face of the dam is altered is
and the water pressure
consists of
two
forces
DAMS AND WEIRS
94
having areas of 17 and 40, respectively, which act through their Instead of combining each force separately with the resultc.g.'s.
ant (R)
it is
more convenient to
that single force with (R.) the intersection of
its
find their resultant
and combine
This resultant Pi must pass through
two components, thus
if
their force lines are
—
:
DAMS AND WEIRS run out backward
till
95
they intersect, a point in the direction of Pi
Pi is then drawn parallel to its reciprocal in the force polygon which is also shown on a larger scale at the left of the profile. The final resultant is Ri which falls just within the middle third of the base. i?2 is the resultant supposing the water to be at crest level only. The water in the river is supposed to have is
found.
mud 01t
-n
^ water pressure
The
The base length
in solution with its s.g. 1.4. will
^u then u be
of the triangle
(H+d)X(p-l) = 32X1.4 = .„„^ 1o.od. ^r—. 2A p
—
other water-pressure areas are similarly treated.
curtain reaches rock the
dam
If the rear
should not be subject to uplift.
It
could, however, withstand sub-percolation, as the hearth of riprap
and boulders will practically form a filter, the material of the river bed being too large to be disintegrated and carried up between the interstices of the book blocks. The effective length of travel would being 20 feet, then be 107 feet; add vertical 52 feet, total 159 feet,
H
— works out to — H
--
= 8 which ratio is a
liberal
allowance for a boulder
20
bed.
The
fore curtain is wisely provided with
weep holes to
release
any hydrostatic pressure that might otherwise exist underneath the dam. The Granite Reef dam has a hearth, or fore apron of about 80 feet in width. A good empirical rule for the least width for a solid or open work masonry fore apron is the following
L = 2H+d
(20)
which H is the height of the permanent weir crest above floor, and d is the depth of flood over crest. In this case H = 20j d=12; The Bassano least width of floor should then be 40+12=52 feet. dam is 40 feet high with 14 feet flood over crest, the width of hearth according to this formula should be 94 feet, its actual width is 80 With a low submerged weir, feet which is admittedly insufficient. in
formula
(34),
Part
II, viz,
L = 3Vrfl^,
will apply.
Beyond the hearth
a talus of riprap will generally be required, for which no rule can well be laid down. 73.
Nira Weir.
Fig. 56 is of the Nira weir,
an Indian work.
Considering the great depth of the flood waterdown stream, the provision of so high a subsidiary weir
is
deemed unnecessary, a water is bed rock. The section of
cushion of 10 feet being ample, as floor
DAMS AND WEIRS
96 the weir wall
itself, is considered to be somewhat deficient in base Roughly judging, the value of H+dy on which the base width is calculated, should include about 3 or 4 feet above crest level. This value of d, it is beUeved would about represent the height of head water, which would have the greatest effect on the weir. The exact value of d could only be estimated on a knowledge of the bed slope or surface grade of the tail channel. The above estimate
width.
would make (ff+rf)=36
The top
width, 8.3,
and with p = 2l,
feet,
is
just
^ H+^d,
H+d = 24
feet.
in accordance with the
rule given in formula (18).
A
section
on these
lines is
shown dotted on the
profile.
pro\dsion of an 8-foot top width for the subsidiary weir
Fig. 56.
quite
Section of Nira Weir in India Showing Use of Secondary Weir
indefensible, while the base
which
is
The
width
is
made
nearly equal to the
For purposes of instruction in the principles of design, no medium is so good as the exhibition of plans of actual works combined with a critical view of their excellencies height,
or defects.
is also
excessive.
The former
is
obtainable from record plans in
technical works, but the latter
is
many Thus
almost entirely wanting.
an inexperienced reader has no means of forming a just opinion and to blindly follow designs which may be obsolete in form
is liable
or otherwise open to objection.
Castlewood Weir.
74.
The Castlewood
weir, Fig. 57,
is
of
remarkable construction, being composed of stonework set dry, enclosed in a casing of rubble masonry. section
is
any
less
It is doubtful if
such a
expensive than an ordinary gravity section, or
DAMS AND WEIRS much
than an arched buttress
less
dam
97 Shortly after
of type C.
which was stated to be due to faulty connections with banks of the river; but whatever the cause it had to be reinforced, which was effected by adding a solid constructioUj
bank
it
showed
signs of failure,
of earth in the rear, as
shown
in the figure.
This involved
In the overfall portion the bank must
lengthening the outlet pipes.
have been protected with riprap to prevent scouring due to the velocity of the approach current.
American Dams on Pervious Foundations.
75.
States a very large lating works,
number
of
up to over 100
feet in height
have been
Section of Castlewood Weir Showing Construction of Stone
Fig. 57.
In the United
bulkhead and overfall dams and regubuilt
Work
on foun-
Set Dry,
Enclosed in Rubble Masonry
dations other than rock, such as sand, boulders, and clay.
Most
of these, however, are of the hollow reinforced concrete, or scallop
arch types, in which a greater spread for the base
than would be the case with a wall
is
solid gravity
dam.
practicable
is
Whenever a
not run down to impervious rock, as was the
core
the Granite Reef Overfall dam. Fig. 55, the matter of sub-percolation
and
uplift require consideration, as is set forth in the sections
''Gravity
Dams" and "Submerged Weirs on Sand".
feet high is
and deep rear
piling will
it is
If
on
dam
clear that a very long
be necessary for safety.
down silt in suspension. When the overfall a high one with a crest more than 15 or 20 feet above river-
All rivers bring is
a
50 on sand or sand and boulders, of a quality demanding
a high percolation factor of say 10 or 12, rear apron
dam
cr.se in
DAMS AND WEIRS
98
bed
the deposit that
level,
obstruction
Avill
additional light stanching
comparatively
dam.
is
bound to take place
in rear of the
not be liable to be washed out by the current, and
still
be deposited in the deep pool of
silt will
water that must exist at the rear of every high
For a low weir however
this does
not follow, and
if
deposit
is
made it will be of the heavier, coarser sand which is not impermeable. The difficulty and expense of a long rear apron can be surmounted by the simple expedient of constructing only a portion of it
of artificial clay, leaving the rest to
To
itself.
voir
filled,
ensure safety the in
two or three
dam
stages,
be deposited by the river
should be constructed and reser-
with intervals between of
length to allow the natural deposit to take place.
sufficient
Thus only a
works are
in existence
which owe their safety entirely to the
nate but unrecognized
frac-
Many
tion of the protective apron need be actually constructed.
fortu-
circumstance of natural deposit having
stanched the river bed in their rear, and
many
failures that
have
taken place can only be accounted for from want of provision for the safety of the work against underneath scour or piping and also
The author
uplift.
himself once had occasion to report on the
ure of a head irrigation work which was designed as
if
fail-
on rock,
was on a pervious foundation of boulders. When it failed the designers had no idea of the real cause, but put it down to a "treacherous river", "ice move", anything but the real reason,
whereas
of
it
which they were quite ignorant.
Had
a rear apron of sufficient
width been constructed, the work would be standing to
Dam
this day.
The fore apron and base must be of one level throughout its The length, if the foundation is of any other material than rock. foundation core walls may have to vary more or less with the surface of the river bed, which is deep in some places, and shallow in 76.
of
Base of
an overfall
others, but the
dam
apron level should be kept at or about low water
level throughout.
built across a river in so,
it,
and Fore Apron.
or weir
When bed
it
a horizontal wall as an overfall obliterates the depressions
the discharge over the weir
is
the same at
all
dam
is
and channels
points or nearly
consequently the tendency will be to level the bed downstream by the hollows and denuding the higher parts. Under these conditions it is evidently sheer folly
filling
to step
up the
apron to coincide with the section of the river bed, as the higher
DAMS AND WEIRS bound to be
parts of the bed are
in time
99
washed out by the faUing
water and deposited in the deeper channels, and portions of the
dam may
easily
be undermined.
This actually occured in one case.
Section of Spillway of St. Maurice River
77.
Dam.
Fig. 58
is
a section of the spillway portion of the reinforced bulkhead gravity
dam,
illustrated in Fig. 44.
crest of Fig. 44, the
back
Owing
heavy
to the absence of the
of the spillway profile is
\\\\u\^\uiiMiY/ft/////,/.-/\/A/////////////A^/-
provided with
^^0-a
UlJ•Scaie inPeet
Fig. 58.
Diagram Showing
Profile of Spillway Portion of Saint (See Fig. 44)
Maurice River
Dam
double the amount of reinforcement shown in the former example.
One
half, viz, \\ inches,
extends right
other half stops short at El 280.
diagrams in the same final resultant
side the
way
This
down is
to the base, while the
arranged for in the stress
as explained in section 55, R^ being the
on the base.
The
line of pressure falls slightly out-
middle third in the upper half of the section.
The
effect
would be to increase the tension in the reinforcement somewhat above the limit of 8 tons per square inch. The adoption of a trapezoidal profile, would,
it is
as well as in the former.
deemed, be an improvement in this case
DAMS AND WEIRS PART
II
ARCHED DAMS General Characteristics.
78.
being arched in plan,
is
an arch under pressure. to the foundations
In this type, the whole dam,
supposed to be in the statical condition of As, however, the base
by the
freedom of motion nor
elasticity,
immovably
dam
cannot possess
and consequently must act more or
dam subject to oblique pressure. However this may be, experience has conclusively proved
the profile be designed on the supposition that the whole
elastic arch, this conflict of stresses
by the
practical
The
man.
probability
is
an
that both actions take
is
merging into transverse
near the base; the result being that the safety of the
enhanced by the combination of tangential and vertical
two
that
near the base can be neglected
place, true arch action at the crest, gradually stress
of
full
a gravity
less as
if
fixed
due to the weight
frictional resistance
the structure, the lowest portion of the
is
dam
stresses
is
on
planes.
In this type of structure, the weight of the arch
conveyed
itself is
to the base, producing stress on a horizontal plane, while the water pressure normal to the extrados, radial in direction, is transmitted
through the arch rings to the abutments.
The
pressure
is,
therefore,
distributed along the whole line of contact of the dam with the sides as well as the ground. In a gravity dam, on the other hand, the whole
pressure
is
Arch pressure
is
concentrated on the horizontal base.
The average unit expressed by the formula
Stress.
^^^RHw
h
=
M^
stress
developed by the water
"Short" Formula
(21)
"Short" Formula
(21a)
DAMS AND WEIRS
102 in
which
R is the radius of the extrados,
H the
center of the crest,
the unit weight of water or
sometimes measured to the
depth of the lamina, b
^
Into this formula
ton.
width, and
its
p,
gravity of the material in the arch, does not enter.
formula answers well for
all
w
the specific
This simple
arched dams of moderate base width.
When, however, the base width
is
considerable, as, say, in the case
dam, the use of a longer fornlula giving the maxibe preferred. This formula is derived from the same principle affecting the relations of s and 5i, or of the maximum and average stresses already referred to in Part I on "Gravity Dams".
of the Pathfinder
mum
The
stress (5) is to
expression
is
as follows, r being the radius of the intrados:
RHw 2R 2R '-'\R+r)~ b "^R+r or in terms of
R and b 2Hw
Long" Formula
(22)
K-i) also
>
79.
"Long" Formula (22a)
Theoretical and Practical Profiles.
to gravity is
= rIi~JiJl^\
dams, the theoretical
a triangle having its
In a manner similar
profile suitable for
an arched dam
apex at the extreme water
level, its
width being dependent on the prescribed limiting pressure. ful
examples have proved that a very high value for
stress,
can be adopted with
able use of
safety.
arched dams would be
If it
s,
the
Success-
maximum
were not for this, the profit-
restricted within the
of a short admissible radius, as with
base
narrow
limits
a low limit pressure the section
would equal that of a gravity dam.
The practical profile is a trapezoid, a narrow crest being necessary. The water pressure area acting on an arched dam, is naturally similar to that in a gravity
that there
is
as in a weir.
point
The
is
dam, the
difference being, however,
no overturning moment when reverse pressure occurs The difference or unbalanced pressure acting at any
simply the difference of the direct and the reverse forces.
areas of pressure on both sides, therefore, vary with the squares
of their respective depths.
DAMS AND WEIRS The water back and
its
103
pressure on an arch acts normally to the surface of radial in direction; consequently the true line of pres-
is
sure in the arch ring corresponds with the curvature of the arch and
has no tendency to depart from this condition.
no such tendency to rupture as
is
There
therefore,
is,
the case in a horizontal circular
arch subjected to vertical rather than radial pressure.
This prop-
erty conduces largely to the stability of an arch under
This condition
pressure. to*
not strictly applicable in
is
its
liquid
entirety
the case of a segment of a circle held rigidly between abut-
ments as the arch
is
The
then partly in the position of a beam.
complication of stress involved
is,
however, too abstruse for practical
consideration.
Correct Profile.
80-
profile of the
with a narrow outline
is
dam
arched
As we have already
With regard
crest.
acts
on a
different plane
due to the weight
The
in the arch ring.
which
may
be expressed as
This tends, when the extrados
back
most favorable
The
reason for
stresses in the arch
maximum
induced stress
vertical pressure produces a transverse expan-
WxEXtUj
cient of elasticity of the material
stress in the section;
the correct
of the arch, although
from the tangential
has a definable influence on the
ring, still
sion
to arch stresses, the
that with the back of the extrados vertical.
this is that the vertical stress it
seen,
a triangle modified into a trapezoid
is
in
m that
and
is vertical,,
which
E is
the
coeffi-
of transverse dilation.
to diminish the
whereas when the intrados
is
maximum
vertical
and the
inclined, the modification of the distribution of pressure
unfavorable, the trapezoidal profile tion exists.
A
maximum is
stress
being augmented.
When
is
the
equiangular, an intermediate or neutral condi-
profile
with vertical extrados should, therefore, be
adopted whenever practicable. In very high dams, however, the pressure on the horizontal plane of the base due to the weight of the structure, becomes so great as to even exceed that in the arch ring; consequently
it is
necessary to adopt an equiangular profile in order to bring the center of pressure at, or near to, the center of the base, so as to reduce the ratio of
maximum
As stated
pressure to average pressure to a
in the previous section,
when a
minimum.
vertical through the
center of gravity of the profile passes through the center of the base,
the
maximum
pressure equals the average, or s = Si,
DAMS AND WEIRS
104
Support
81 .
When
of
Vertical
Water Loads
in
Dams.
Arched
dam is incHned, the weight of the supported by the base, the horizontal pressure of
the back of an arched
water over
it is
the water alone acting on the arch and being conveyed to the
abutments.
In the case of inclined arch buttress dams, however, a
portion of the vertical load thrust above
what
is
due to overhang,
82,
Crest Width.
This
safely
made much
less
is
by the
carried
arch, increasing its
due to the horizontal water pressure
is
i.
e,,
when the
alone.
e.g. falls outside the base.
The crest width of arched dams can be than that of gravity dams and a rule of
i..|v7j
(23)
would seem to answer the purpose, unless reinforcement is used, when it can be made less.
EXAMPLES OF ARCHED DAMS The following actual examples of arched dams will now be given. 83. Bear Valley Dam. This small work, the most remarkable arched dam in and forms a valuable example of the enormous theoretical stresses which this type
Fig. 59,
is
existence
of vertical arch can stand.
The mean
radius
being 335 feet according to formula (21) the unit stress will be
— — = 60 r-
Fig. 59.
Section of Old Bear Valley Dam
tons, nearly
This section would be better actual
stress is
if
reversed.
The
probably half this amount.
This work has now been superseded by a new
dam
built
below
it,
Fig. 77, section 103. 84.
to a
Pathfinder
Dam.
radius of 150 feet
This immense work, Fig. 60,
measured to the center of the
crest.
is
built
That,
however, at the extrados of the base of the section is 186 feet and this quantity has to be used for the value of R in the long formula (22).
The
unit stress then works out to 18 tons, nearly.
in the lowest arch ring
is
undoubtedly
much
The actual stress
less, for
the reason
DAMS AND WEIRS
105
that the base must absorb so large a proportion of the thrust that
very Httle
is
The
transmitted to the sides of the canyon.
exact
determination of the proportion transmitted in the higher rings is
an indeterminate problem, and the only safe method
is
to assume
with regard to tangential arch stress that the arch stands clear of RAD. 150'
Fig. 60.
the base.
This
will leave
and enable the adoption
Section of Pathfinder
Dam
a large but indeterminate factor of safety of a high value for
5,
the
maximum
unit
stress.
The is
profile of the
dam is nearly equiangular in outline. This dam in order to bring the vertical resultants
necessary in so high a
(W) R, E. and
(iV)
R. F. as near the center as possible with the
object of bringing the ratio of
maximum
to
mean
stress as
low as
possible.
The
estimation of the exact positions of
analytically as below.
W and of N
is
made
DAMS AND WEIRS
106
There are only two areas to be considered, that of the water (v) and that of the dam itself {W). Dividing v hy 2^ (the assumed specific gravity of the material), reduces it to an equivalent area of concrete or masonry. overlying the inclined back
210X31.5 ^ 1470 =103 tons 2X2.25
Tr= :!^X210 = 10920 = 768
iV=12390 = 871 tons
Total, or
Using formula
(7),
tons
Part
I,
the
e.g. of
TF is 50.8 distant from the
toe of the profile, then q or the distance of the incidence of 94 = 3.8. the center point of the base is 50.8 —
W from
— Zd
The
value of
5i,
or the
mean
unit stress
m = l+^=l+^^ = 1.24;
and
then
94
^
W — b
is
768
or
,
——=8.1
tons
94
^=^ = 1.24X8.1 =10.1 b
tons.
For Reservoir
moments
Full, to find the position of iV,
will
be
taken about the toe as follows
Moment Moment
of of
?j
= 103x83.5=
Total iV = 871
then
a:
=
47614 -
= 9.26. By
r^a =54.6; whenceg = 54.6 ^, ^
formula
V
(9),
5
From
Part
I,
=47614
„.,
94
would equal W,
(7a),
c. g.
^i
is
= 9.26X1.48 = 13.7 a high figure.
tons
if
due to
be reduced by
It could
still
the back were vertical
Let this latter case be considered. The distance
of the profile
from the heel
will
then be by formula
Part I
and the value
871
= "r = "qT
m^W-^~-^\A%.
further inclining the back; on the contrary,
of the
.
iV
,
—^=/.6teetand
this it is evident that the unit stress in the base,
vertical load only,
iV
8600
TF = 768X50.8 = 39014
of q will
be
94 — -31.66 = 15.33 feet
DAMS AND WEIRS Si
as before
W
= ^- = 8.1
107
tons
Then Q2
m = l+^ = 1.98
and
5
= 8.1X1.98 = 16
tons
94
This stress
is
greater than that of
proves that the forward to reduce the
maximum
tilt
N in the previous working which dams
given to these high
is
necessary
unit stress on the base to a reasonable limit,
A more equiangular profile would give even better results. 85.
on
Shoshone Dam.
lines identical
The Shoshone dam,
with the last example.
being the highest
dam
Fig. 61.
in the world
Profile
and Force Diagram
Fig. 61, is designed
It has the distinction of
but has recently
for
Shoshone
lost
this
Dam
preeminence, as the Arrow Rock, quite lately constructed. Fig. 37, Part I, is actually 35 feet higher. This work is also in the United States,
The
incidents of the resultants Reservoir
Empty and
Reservoir Full, which will be explained later, have been shown graphically,
and the
vertical forces
analytical computation
taken from
left
is
given below.
The
to right are (1), area 6480; (2),
14,450; (3), water overlying back, reduced area 1880; total 22,810.
Taking moments about the toe (1) is
54
of (3)
is
feet, of (2) calculated
95
feet, roughly.
of the base, the distance of
by formula
(7),
Part
I, is
58.3,
and
DAMS AND WEIRS
108
Then (6480 .'.
a:
= 60
X
54)
+
X
(14450
58.3)
+
X 95) =22810Xa:.
(1880
feet, nearly.
The value
of q for
N then is 60-^ = 6 feet,
—N --^ =211, and by formula '^'^810
Now
51=
— —
(108+36) ——
„^^
^
„
=281 square ieet=
(9),
Part
I,
5= 211X
—— = ^_ 21 tons, nearly.
281X2.4
lUo
,
32
The maximum arch unit
stress
by formula
(22) is as follows: the
radius of the extrados of the base beine ^ 197 feet the fraction
-=- '=
.
=
H=
and
.55
245 therefore
^^^^
s=
R
— 197
2x245x1
H^-i)
.55X1.45X32
490 ,_„,tons. —-=19.2
Below the
-
ft.
above base, the
stress
on the arch does
than that due to vertical This base should undoubtedly have been widened,
N.
pressure
60
level
The arch
not increase.
stress is less
down
the battered faces being carried
to the base, not cut off by
vertical lines at the 60-foot level.
Center of Pressure
— New Graphical Method.
In order to find the
center of pressure in a case like Fig. 61, where the lines of forces (1)
and
(2) are close together,
the ordinary method of using a force and
funicular polygon involves crowding of the lines so that accuracy difficult
to attain.
Another method now
on the same principle as that for finding the
c. g.
In Fig. 61,
will
be explained which
is is
of the intersection of cross lines used
of a trapezoid.
first
the
c.g.'s of
the three forces are found
(1)
the
water pressure area divided by p or 2.4 which equals 1880 square feet, (2) the upper trapezoidal part of the dam area 14,450, and (3) the lower rectangular area 6480. line projected
From from
bj
bd
on one
side in
Then
a, ac is set off horizontally is
(1) is
any location as at
drawn equal to
joined to (2) and this b in Fig. 61a.
equal to
(1) or 1880; cd
is
(2)
or 14,450 and
then drawn and
its
intersection with ab at e gives the position of the resultant 1-2, which
can
now be
of the
projected on the profile at G.
components
(1-2)
with
(3)
To
the line G-3
is
obtain the resultant
drawn on the
profile
DAMS AND WEIRS and a
parallel to
it
drawn from
zontal through (3) at
/.
From
e
on
e,
eg
109
Fig. 61a, intersecting the horilaid off horizontally equal to
is
and from/, fh equal to (1+2) or 1880+14,450 = 16,330. hg is then drawn and its intersection with ef at j is the centroid of the three forces, which projected on the profile to Gi on the line G-3 gives (3)
or 6480
the location of the vertical resultant of
Sweetwater Dam.
^6,
The
in CaUfornia is given in Fig. 62.
-
^^-JtS-^
1+2+3.
profile of
The
the Sweetwater
original crest of the
dam dam
_ — ^^-^ >
/iddition
3-^3.^7bns Fig. 62.
Graphical Analysis of Sweetwater
Dam,
was at the
El, 220, or 95 feet
dam depended
California
above the base.
for its stabihty
sidered as a gravity
dam
on
its
Under these conditions arched plan.
vertical pressure area is the triangle a&e, here g = 16.5
out to 3.15
3.15X226 =
iV
=
226 tons and 6
15.5 tons
which
If con-
with allowable tension at the heel, the
is set
=
and
46 feet whence
down from a
to
s
m =
h.
46
The
tension at the heel
=
52
2^
15.5-9.82 = 5.7 tons
works
110
DAMS AND WEIRS
DAMS AND WEIRS When
Value of S
111
Heel Is Unable to Take Tention, If the heel
is
unable to take tension, the pressure triangle will then be adc in which ac
=3
times the distance of the incidence of
R
from the
toe, or
3X6.5 = 19.5 feet and s is obby the following formula
tained
4
_
(24)
4X226
= 23.2 tons 3X46-33 dam has lately been raised
here s = -
This
iVorl^F
to EZ240, or
by
20, feet
and by
the addition of a mass of concrete at the rear transformed into a
gravity dam.
The
diagram,
s
resultant due
Ri on the works out to 10.6
to this addition
tons and there
is
is
no tension at
the heel.
Any bond between
new
and the old has been
wall
the Fig. 64.
studiously
work
is
avoided.
Profile of Barossa
Dam
The new
reinforced with cross bars
superstructure. Fig. 63
is
and the rear mass
a plan of the
Fig. 65.
Site
Plan
of
dam
Barossa
tied into the
as altered.
Dam
87. Barossa Dam. This dam, Fig. 64, is an Australian work, and although of quite moderate dimensions is a model of good and bold design.
DAMS AND WEIRS
112
The back outline
on a
is
is vertical
and the
viz,
a square crest imposed
with the hypothenuse of the latter by The crest is slender, being only 4^ feet wide, but is strength-
triangle, the face joined
a curve.
ened by rows of 40-pound iron
rails,
The maximum arch
concrete.
fished together, built into the
stress
works out to 17^ tons, the
corresponding vertical stress on base to 6f tons. plan of the work. 88. is
The
fore batter is nearly 1 in 2.7.
not trapezoidal but pentagonal,
Lithgow Dam.
the Lithgow dam, No.
a
site
Another example very similar to the
last
2, Fig. 66.
The arch
Fig. 65
stress in this
out by the short formula to nearly 13 tons; the radius feet,
is
is
works
only 100
the vertical stress works out to 7 tons.
Arched dams abut either on the solid else on the end of
rocky banks of a canyon or a gravity dam.
In cases where a narrow deep
central channel occurs in a river, this portion
can advantageously be closed by an
arched dam, while the flanks on which the arch
can be
abuts
gravity
dams aligned The dam
tangential to the arch at each end. will
thus consist of a central arch with two
inclined straight continuations.
The plan
the Roosevelt dam, Fig. 26, Part Fig. 66.
Profile of
Dam
Lithgow
an idea of
is
shown the
for domestic
The
profile of
Dam.
In
a temporary reinforced arched
dam
water supply at Barren Jack, or Burrin Juick, Australia.
reinforcement consists of iron
rails.
at the base works out to 21 tons, nearly.
nent dams
of
will give
this class of work.
Burrin Juick Subsidiary
89-
Fig. 67
I,
down
to the base
rode in time and cause
is
The
Reinforcement of perma-
not desirable, as the metal
failure,
although the possibility
The main Burrin Juick dam
stoutly denied.
unit arch pressure
is
may is
cor-
often
given in Part
I,
Fig. 36. 90.
Dams
just described,
with Variable Radii.
is
The
use of
dams
of the type
generally confined, as previously noted, to narrow
gorges with steep sloping sides in which the length of the level of the crest.
bed of the canyon
The
is
dam
at the
but a small proportion of that at the
radius of curvature
is
usually fixed with regard to the
DAMS AND WEIRS
113
length of chord at the latter level, consequently at the deepest level,
the curvature will be so shght that arch action will be absent
and the lower part
of the
dam
will
be subject to
to tension as well as compression.
beam
stresses,
In order to obviate
i.e.,
this, in
some recent examples the radius of curvature at the base is made less than that at the crest, and all the way up, the angle subtending the chord of the arc, which is variable in length retains the same measure throughout.
This involves a change in
the radius corresponding to the variable
span of the arch. is
The
further advantage
obtained, of reduction in the unit stress
and in rendering the more uniform throughout. In
in the arch ring stress
very high dams, however, the base width cannot be
much reduced as otherwise due to the
limit stress will
be exceeded.
varying radii
is
the
vertical loading
This arrangement of
somewhat
similar to that
used in the differential multiple arch given later.
MULTIPLE ARCH OR
HOLLOW ARCH BUTTRESS DAMS
Multiple Arch Generally
91.
Useful
Than
Single Arch
evident that a
Dams.
dam which
More It is
consists of a
single vertical arch is suitable only for a
narrow gorge with rock
sides
on which the
Fig. 67.
Profile of Burrin Juick
Subsidiary
arch can abut, as well as a rock bed; consequently
its
Dam
use
limited to sites where such conditions are obtainable.
foundation
is also essential for
is strictly
A
rock
gravity dams, the unit compression
on the base of which
The
is too high for any material other than rock. adA^antages inherent in the vertical arch, which are con-
can however be retained by use of the so-termed multiple or scallop arched dam. This consists of a series of vertical or
siderable,
inchned arches, semicircular or segmental on plan, the thrust oi
DAMS AND WEIRS
114
which tical
carried
is
by
The arrangement
buttresses.
with that of a masonry arched bridge.
is,
in fact, iden-
If the latter
be con-
sidered as turned over on its side, the piers will represent the but-
In the case of a wide river crossing, with a bed of clay,
tresses.
boulders, or sand, the hollow buttressed
and
dams The wide
slab buttressed
are the only ones that can well be employed with safety.
spread that can be given to the base of the structure in these two types enables the unit pressure on the base to be brought as low as
from 2 to 4 tons per square
foot.
As has already been noticed
in section 78, the arch is peculiarly
This
well suited for economical construction.
that the liquid pressure to which the arch to the surface and
radial in direction.
is
The
is
due to the fact
subjected
is
normal
pressure lines in the
interior of the arch ring correspond with its curvature
and con-
sequently the arch can only be in compression; thus steel reinforce-
ment
is
unnecessary except in a small degree near the crest in order
to care for temperature stresses.
the deck forced.
is
composed
The
In slab dams, on the other hand,
of fiat slabs
which have to be heavily
spacing of the buttresses for slabs
to 20 feet, whereas in hollow arch limit to the spans which
may
dams
be adopted.
there
is is
rein-
limited to 15 practically
Another point
is,
no
that
the extreme compressive fiber stress on the concrete in deck slabs is
limited to five hundred to six hundred and fifty pounds per square
inch; in an arch,
pression which
is
on the other hand, the whole section
is
in
com-
much greater area. For the type now under consideration should
thereby spread over a
reasons above given the arch
be a cheaper and more
scientific construction
than the slab type in
spite of the higher cost of forms.
Mir Alam Dam. The first example given is that of the Mir Alam tank dam, Fig. 68. This remarkable pioneer structure was built about the year 1806, by a French engineer in the service of H. H. the Nizam of Hyderabad in Southern India. The ahgnment of the dam is on a wide curve and it consists of a series of vertical semicircular arches of various spans which abut on short buttress 92.
The spans vary from 83 to 138 feet, the one in Fig. The maximum height is 33 feet. Water feet. overtop the crest. The length of the dam is has been known to
piers. Fig. 69.
68 being of 122 over 3000
feet.
DAMS AND WEIRS On
account of the inequality of the spans, the adoption of the
semicircular form of arch for the reason that
no
115
is
an arch
evidently a most judicious measure,
of this
lateral thrust at the springing.
form under liquid pressure exerts
The water
pressure being radial
in direction, cross pressure in the half arches in the line of the spring-
ing is
is
balanced and in equilibrium.
not in the direction of the axis of the
piers.
On
the other hand,
the terminal thrust
two
resolved in
is
component acts along the
when
axis of
Plan of One Arch of oMir Alam Dam built in 1806, and consisted of 21 such arches.
dam was
This has to be met, either by the abutment,
if it is
an
by the corresponding thrust of the adjoining half The other component is carried by the buttress; therefore,
end span, or
if
exerted
intermediate between the two axes, and
Fig, 68.
arch.
is
but that of the buttress
the arches were segmental in outline
directions one
This remarkable pioneer
the dam.
if
Whatever thrust
dam
else
segmental arches are used, the spans should be equal in order to
avoid inequality of thrust.
The whole
of this
work
is
Longer buttresses
will also
built of coursed rubble
be requisite.
masonry in lime
mortar; the unit stress in the arch ring at the base, using the short 1 roi^ lormula (21),
r
(^Hw) — — -
dam,
,
_
works out to
b
therefore, forna^
68X33X1 — -——— = 5 .
14:Xo2
an economical design.
,
i
tons, nearly.
rp,
The
DAMS AND WEIRS
116
The
buttress piers are
AB
being taken through piers are
shown
in section in Fig. 70, the section
In this work the buttress
of Fig. 68.
very short, projecting only 25 feet beyond the spring Hne
and being altogether only 35, feet long. This length and the corresponding weight would clearly be inadequate to withof the arches,
Fig. 69.
stand
Plan
of Entire
Mir Alam
Dam
immense horizontal thrust which
the
is
equivalent
to
E Iw = 332X1X146 = 2500 tons, nearly. 2X32 It is evident that
arches behind
it
must
if
the buttress pier slides or overturns, the
which reason the two half arches and
follow, for
the buttress pier cannot be considered as separate entities but as
and consequently the effective length of the base must extend from the toe of the buttress right back to the extrados of the two adjoining arches. At, or a little in the actually forming one whole,
^-/0-0*i L^,,
Section Py-B
Present Surfocei
Fig. 70.
Section of Buttress Pier of
Mir Alam
rear of the spring line, the base
continuations.
The weight
of
Dam Taken through AB of Fig.
is split
up
into
these arms,
68
two forked curved
i.e.,
of the adjoining
half arches, has consequently to be included with that of the buttress proper
or sliding
is
a^
hen the
estimated.
stability of the structure against overturning
DAMS AND WEIRS Stresses in Buttress.
93.
taken through
CD
117
In the transverse section, Fig. 71,
of Fig. 68, the graphical calculations establish
R intersects the base, thus lengthened, center; the direction of the resultant R is
the fact that the resultant line at a point short of its
also satisfactory as regards the angle of frictional resistance. J?i
is
the resultant on the supposition that the buttress
is
on the base proves that the arch
is
nonexistent.
Its incidence
stable without the buttress,
With regard
which
is
to sHding on the base,
therefore actually superfluous.
P = 2500
and TF = 6828
tons.
^-^^^Tons
Fig. 71.
The
Transverse Section of Mir
Alam
Dam
Taken through CD,
Fig.
coeflicient of friction being .7 the factor of safety against sliding
If the arch were altered on plan from a semicircle segment of a circle, the radius would of necessity be increased, to a and the stress with it; a thicker arch would, therefore, be required. This would not quite compensate for the reduced length of arch, is
nearly 2,
but on the other hand, owing to the effective
being depressed, the
cfoa^ti
base width would be reduced and would have to be made
good by lengthening the buttress position of arch
piers.
What
particular
and buttress would be the most economical
disis
a
DAMS AND WEIRS
118
matter which could only be worked out by means of a number of trial designs.
J to |.
The
ratio of versed sine to span should
vary from
Arcs subtending from 135 to 120 degrees are stated to be
the most economical in material. 94.
Belubula Dam.
There are not as yet very
many modern
examples of arch buttress dams, but each year increases their numr^'^"
Fig. 72.
ber.
Profile Sections
and Force Diagram
for Belubula
The Mir Alam dam has remained
resting
Dam, New South Walea
on
its laurels
a rival for over 100 years, but the time has come is
being largely adopted.
segmental panel arch dam. Wales.
same
The arch
without
this type
shows an early example of a
Fig. 72
It is the
crest is 37 feet
as in the last example.
when
Belubula
dam
in
New
South
above the base, very nearly the
The arches, which are
inclined 60 degrees
to the horizontal are built on a high solid platform which obliterates
DAMS AND WEIRS This platform
inequalities in the rock foundation.
high, so that the total height of the
119
dam
is
over 50
16 to 23 feet
is
The spans
feet.
are 16 feet, with buttresses 12 feet wide at the spring line, tapering
The buttress
to a thickness of 5 feet at the toe they are 40 feet long. ;
piers,
which form quadrants of a
by
thickness
diminish in
steps from the base up, these insets corresponding
These steps are not shown drawn as if in one straight batter. form, and the spandrels are filled up
with similar ones in the arch in the drawing; the arch also
The arches
circle in elevation,
are elliptical in
itself.
is
flush with the crown, presenting a
flat
surface toward the water.
vSome of the features of this design are open to objection: Firsts the
filling in
of the arch spandrels entirely abrogates the advantage
accruing to arches under liquid pressure.
The
direction of the water
but normal to the rear
pressure in this case is not radial
slope, thus
exactly reproducing the statical condition of a horizontal arch
The
bridge.
haunches and
pressure, therefore, increases is
The
to allow of a radial
arches
and the spandrels left pressure which partly balances itself.
should have been circular, not
empty
from the crown to the
parabolic, not circular, in curvature. elliptical,
Second, the stepping in of the intrados of the arch complicates the construction. in concrete.
toe
is
A
plain batter would be easier to build, particularly
Third, the tapering of the buttress piers
toward the
quite indefensible; the stress does not decrease but with the
center of pressure at the center of the base as in this case, the stress will
be uniform throughout. Inclination of Arch to Vertical.
95.
The
inclination of the axis
of the arch to the vertical is generally a desirable, in fact, a necessary
feature
when segmental arched panels
carried
is
are used; the weight of water
of value in depressing the final resultant line to a suitable
angle for resistance to shearing stress.
As noted
in section 90,
the weight of the water overlying the arch does not increase the unit stress in the arch ring. axis can
Consequently,
be adopted without in any
way
any
inclination
of
increasing the unit stresses
due to the water pressure.
When an
arch
vertical it
water pressure is conveyed to the abutments and the weight of the arch to its base. When an arch hes horizontally under water pressure both is
is
clear that the
all
the weight of the water and that of the arch
itself
are conveyed
DAMS AND WEIRS
120 to the abutment;
when in an intermediate
position part of the weight
and part to the abutments. With regard to water pressure, the thrust being normal to the extrados of the arch the whole is carried by the abutments. In the case of arches which do not overreach their base the weight In of water overlying the inclined back is conveyed to the base. of the arch is carried to the base
any case the unit
cannot exceed that due
stress in the arch
The
to horizontal thrust.
total water pressure is greater with
inclined back, as the length of surface acted
on
the diagram, Fig. 72a, the vertical load line
W
represents the
weight of one unit or one cubic foot of the arch ring which
This force
to wp.
n=W sin
vertical
force
n
and
'p
and the other
6,
=W
cos
the
The
6,
also radial in direction is
mean
unit stress
The
normal to the former.
and
is
by the radial by the water pressure which is
unit stress developed
RiU] but
the radius in this case,
i?i,
The
radius, the pressure being internal, not external.
Si will
then be
= Riwp sin 6
S\
When
n,
6 being the inclination of the arch axis to the
similar to that produced
is
equal
is
resolved in two directions, one p^ parallel
is
to the axis of the arch, force
= ^; when
e is 30° sin
It will easily
an In
increased.
is
45°, sin 6
(25)
= —,
be understood that this unit stress due to n does is the same at the first foot depth of the arch
not accumulate, but as
it is
it.
at the bottom; the width of the lamina also does not affect
However, the component p does accumulate, and the expression should be multipUed by the inclined height Ex, lying
lop cos B
above the base under consideration. compressive stress at the base will be
s
would equal 96.
details of
Hwp
The Ogden dam,
which are shown
arch and buttress type. is less
,
in
which
h is the mean
simply.
Ogden Dam-
of the arches
j—^
unit
arch were a rectangle, not a trapezoid,
If the
width of the arch.
As Hi=Hsec.9j the
the profile and sectional
in Fig. 73, is a notable
Its height is 100 feet.
than | to
1,
example
The
of the
inclination
or about 25 degrees to the vertical.
DAMS AND WEIRS The
equinangular except for a small out-
profile of the buttress is
throw of the
On
toe.
design, but could be
the arch
is
from 6 to 2
121
the whole
improved
it
must be pronounced a good For example,
in several particulars.
unnecessarily thick at the crest, and could well be reduced feet,
thus effecting considerable economy.
The designers
were evidently afraid of the concrete in the arch leaking, and so overlaid the extrados with steel plates.
arch causing
it
by
could have been provided
SECTlOti Ofi
AA
The
greater thickness of
increasing the span
""
and radius
5ECTI0H Oti B3
Fig. 73.
arches.
A
to possess less liability to percolation under pressure,
Profile
and Sections
of
Ogden
of the
PL AH OVERALL
Dam
design consequently would be improved
by adopting
larger spans, say 100 feet; buttresses, say, 25 feet thick, their length
being dependent on the width of base required to provide sufficient
moment
of resistance;
and
further, the inclination of the arches
might require increasing to bring the center of pressure to the center of the buttress.
arch forming a roadway
is
dam;
preferable
overflow provided.
an
at, or close
by another arrangement, and is well
finish of the crest
excellent
for an overfall, on the other hand, the on account of the increased length of The stress diagram shows that the value of the
suited for a bulkhead
curved crest
is
The
DAMS AXD WEIRS
122
vertical load A' i3 155,000 cubic feet or 10,598 tons, p being
at 2J,
The
whence
q
R
incidence of
= Oj and by
on the base,
fomiida
taken
5 feet from the center,
is
Part I
(9),
^140
10598
„„,
-fxC4) = 110X16-X-~:=8.91
^ tons
110
the dimensions of
.1,
the area of the base, being
pressm'e on the arch ring at the base
by
110X16
The
feet.
the short formula works
—
„ .^ 24X100 —- = 9.4 out^^to -^ tons.
hX32
The
contents of the
dam ^per
2,177 cubic feet; that of a g^a^Hty
nm, making a 30 per cent. With a
foot run
— —=
amounts to
'-^
4S
dam would
be about 3,500 cubic arched t^pe of
feet per foot
5a\-ing in favor of the
nearly
better disposition of the parts as indi-
cated above, the sa\'ing would be increased to 40 or 50 per cent.
Actually the sa^'ing amoimted to only 12 per cent; this was owing to the steel covering which, as
we have
seen, could
have been
dispensed with.
Design for Multiple Arch Dam.
97,
Fig. 74 is a design for
a segmental arch panel dam, or rather, weir. crest is
The
height of the
64 feet above base with 5 feet of water passing over; the
apex of the triangle of water pressure
The
the base.
will
inclination given the axis,
that of the spring line and the intrados,
is
then be 69 feet above
which
is
coincident with
60 degrees with the horizon.
In designing such a work, the following saHent points
first
require consideration.
Width of Span. This, it is deemed for economical reasons should be not less than the height of crest unless the state of the foundation requires a low imit stress. In the ^lir Alam dam (1)
the span case
is
it will
(2)
over four times the depth of water upheld.
be made the same, that
is,
two
(3)
In the present
feet,
As with bridge piers, the accommodate the skew-backs
Thickness of Buttress Piers.
^sidth should be at least sufficient to of the
64
arches; a ^\idth of 12 feet or about \ span will effect this.
Radius and Versed
feet; this allows
Siiie.
The
radius will be
a versed sine of J span, or 16
feet,
made 40
which
is
con-
sidered to be about the flattest proportion to afford a good curva-
DAMS AND WEIRS
124
ture, the greater the length of the arc, the
more
condition will
its
approximate to that of a circular arch, under liquid pressure. (4)
This must
Thichiess of Arch.
thickness depends on
the value assigned to
be assumed, as
first
its
the radius of the extrados, as well as on
i?,
This latter
the limiting pressure.
Sij
will
be fixed at below 15 tons, a value by no means excessive for arches
under liquid pressure.
With a base width of 7 feet, the radius The base will be considered, not 64 feet below crest, but at the point marked
of the extrados will be 47 feet.
at the extreme depth of
where a
Z),
line
normal to the base of the inclined intrados cuts // will, therefore, be
the extrados of the arch.
The
for the reverse pressure.
stress
60
feet, allowing
due to the water pressure,
using the short formula (21), section 78, will be
47X60X1
RHtv
To
,...,
must be added that due to the weight of the arch ring from being 30° and its sine = i), (25), Si = Riwp sin 6 (the angle 43.5X3 =1.6 tons, the total stress being which in figures will be this
formula
Z X4U
a
trifle
p
is
The
over 14 tons.
taken as 2.4 and
this pressure is
ivp
7-foot base width will then be adopted,
= j\
ton.
The depth
taken as 60, not as 65, feet which
being that the reverse pressure due to the
of water producing
is
tail
(
H+d),
be at least level with the water cushion bar wall, effective depth to 60 feet, during flood conditions. 98.
Reverse Water Pressure.
pressure of water pressure alone
is
is
much more
exerted than
is
The
the reason
water, which must will
reduce the
influence of the reverse
considerable
when
hydrostatic
the case with overturning moment.
In the case of an upright arch acting as an overfall weir the pressure of the tail water effects a reduction of the pressure to the extent of its area.
Thus
if
A
be the area of the upstream water pressure,
and a that of the downstream, or will
be their difference, or
respective depths.
When
A — a,
tail
and
water, the unbalanced pressure will
overturning
vary as the square of their
moment
is
concerned, the
areas have to be multiplied by a third of their depths to represent the moment on the base. The difference of the two will be in
that case as the cubes of their respective depths.
DAMS AND WEIRS 99.
The
Crest Width of Arch.
ing to formula (23), should be
made
width of the arch, accord-
|V// = 3J
feet, nearly.
It will
if
this falls
The
below 2
feet,
be
The
3 feet, with a stiffening rib or rim of 3 feet in width.
width could be made proportional to the base width, say
crest
and
crest
125
.36,
reinforcement will be required.
length of the pier base
is
measured from the extrados
of
the arch, the two half arches forming, as already explained in section 92, a forked continuation of the buttress pier base.
The
battering of the sides of the pier would clearly be a correct
procedure, as the pressure diminishes from the base upward.
combined batter of
The
of 5.6 feet.
determined by
1
in 10
is
length of the pier base, as also
trial
A
adopted, which leaves a crest width its outline, w^ere
maneu-
graphical processes, with the object of
vering the center of pressure as near that of the base as posequalize the
sible, so as to
much
as possible.
dence of the
final resultant
measured by
the
mean
unit stress as
sho^^^l
by the
on the elevation of the buttress
The
Pressure on Foundations.
ICO. is
maximum and
This has been effected, as
N in the force diagram,
and
total is
imposed weight
equivalent to 150,000
cubic feet of masonr}', which at a specific gravity of 2.4
150,000X3 —
n 9rn =11,^50
. tons.
inci-
pier.
T^i.
Ine average pressure
•
is
is
equal to
.i.-
this quantity
the quotient being 7| tons, nearly.
125X12 = 1500 square feet, The maximum pressure will
be the same owing to the incidence of
R
divided by the area of the base, or by
This 7i tons if
is
at the center of the base.
a very moderate pressure for a hard foundation;
excessive, additional spread should be provided or else the spans
N greatty exceeds W. This is due to the added weight of water represented by the inclination given to the force line P, which represents the water pressure. reduced.
It will be noticed that
Economy
of Multiple
run work out to
Arches.
The
cubic contents per foot
—^— =850 cubic feet, nearly, the
denominator in
the fraction being the distance apart of the centers of the buttress piers. o
The top is
contents of a gravity weir with base width f o
^ H+d, works out
{H+d) and
to 1,728 cubic feet; the saving in material
therefore over 50 per cent.
•
DAMS AND WEIRS
126
Differential Arches.
101.
buttress arch weir. in the stress
The
Fig. 75
is
a study of a differential
principle of the differential arch consists
radius increasing with the height of the arch, the unit
thus kept more uniform, and
is
sponds more closely with the trapezoidal to be adopted, than
is
the case
the stress area corre-
profile that
when a uniform
has necessarily
radius
is
adopted as
in Fig. 74.
The
arches are supposed to stand on a concrete or masonry
platform ten feet high above the deepest part of the river bed, so that sluices is
required could be provided below L.W.L, which
if
The
identical with the floor or fore apron level.
The depth
to crest level.
and the
at 5 feet
analyses will be level
The
and the
reciprocal depth of tail water
made
at
two
river channel
stages, first,
below
is
is
12 feet.
when water
empty, second, at
inclination given to the intrados of the arch
The
horizontal.
height
is
of film passing over the crest is
"buttresses are placed 31
is
35 feet
assumed
Graphical is
at crest
full flood.
3 vertical to 2
feet centers,
allowing
a span of 25 feet at base, here they are 6 feet wide, tapering to 2 feet at crest.
25 to 29
feet.
The span of the arch thus gradually widens from The versed sine of the arc is made 5 feet at base The radii at these positions are therefore 18.1 crest.
and 2| feet at and 43.3, respectively, measured to the intrados of the arch. These radii are horizontal, not normal to the intrados as in Fig. 73, and thus vary right through from 18.1 to 43.3 corresponding to the altered versed sine which decreases from 5 to 2^ feet, that half
way up
being 22
The
feet.
thickness of the arch at base
Arch Unit
made 2
is
feet.
Taking the base radius as
Stress.
unit stress due to water pressure will be
by formula
18.1, the base (21), s
=
——
adding that due to the transmitted weight of the arch, formula s
= Rwl
18. Iw s
\-p sin
—
|-(2.4X.6)
-
I
= 10.4
tons, a
The
sin
1,
,
6
being
it
the
p being taken at 2.4,
expression
w?
(25),
becomes
at 3^ ton,
whence
I
moderate
stress for a vertical arch.
real thickness of the arch
as properly
.6,
-
should be measured
is
more hke 2^
feet
than 2 feet
horizontally, not normally.
DAMS AND WEIRS
127
upper Pari io be Reinforced for Temperature
-
uj
H^
.
/ \
cfc'
I
I
m/er"^\'\\
\
onffrch
y
/393 To //
/ //
^l
i/ //
"^^nil-'fr^ssHrea
...^
ffrch,
35kewback::^'^
iv=d?'8
Shock
'
X \
§
k'
J
onBase
Doe to R
/
p=ro
Fig. 75.
Design Diagrams for Differential Buttress Arch Weir
:
DAMS AND WEIRS
128
Load Line. weights:
five
In the force polygon the load
(1)
that of
line is
made up
33x24x35 =
13860 cubic
145 tons;
the contents of the pier underlying the arch
(3)
equivalent to 433 tons;
feet,
by taking the contents of the whole as if the and deducting the pyramid formed by the side contents of the whole
batters.
A2) in which
by the prismoidal formula comes
A
Ai and
2
is
found
Thus the
=2310, that of the pyramid
is
are areas of the ends
^i = 0,
^;^ =
— X— = 52.5,
to
(5)
the contents
—6 liAi-^- 4^^ +
Am
and
is
the weight of
(4)
the horizontal arch of the crest of the weir, 12 tons; of the buttress,
the arch
(2)
were vertical
sides
=513, difference 1797, or 135 tons;
^
of
the overlying water has a content of
of the middle
and ^2 = 35X4 = 140;
section.
Here
therefore
(5)=-^X35X[0+(4X52.5) + 140]=2042
cubic feet equiv-
alent to 153 tons.
The
P
total load foots
up to 878
tons.
= u'—-- X = — X 772
the horizontal water pressure
I
(35)^
I
X 31 = 593
tons.
The That of of 2
is
position of the several vertical forces 1,
a triangular curved prism
found by formula
on to the plan.
The
(7),
Part
I,
is
obtained as follows
at J its horizontal width; and by projection of this level is
position of 3 has to be calculated
by moments
as below.
The
lever
arm
of the whole
mass including the battered
sides
at J width from the vertical end of 7^^ feet while that of the pyramidal batter is at J the same distance, or 5J feet. is
The statement
is
then
2310X7J = (1797Xa:) + (514X5.5)
The
whence
a:
= 7.84
position of 5, the battered sloping buttress
is
feet
obtained by
taking the center part 2 feet wide and the outer side batters separately.
The
35 — the length, — = 11§ from the 1
e.g. of
the former
is
at
o
o
DAMS AND WEIRS
129
352
vertical end,
The weight
and
contents are
its
of the whole
is
feet
= 92
tons.
153 tons, so that the side batters will
weigh 153 — 92 = 61 tons, and be
Taking moments about the
-^X2 = 1225 cubic
— = 8.75 feet distant from the end.
vertical end,
we have
153a;=(92XlL67) + (6lX8.75)
x=
1606
.r^rfleetX =10.5
15o Therefore, the incidence of the resultant on the base line meas-
ured 6.5 feet upstream from the center point. TV
In Fig. 75a, iV = 878 tons and -^ = 14; g = 6.5 feet,
foot wide,
—
divided
being 63 feet and
= 1.62 and the stress on the buttress, =14x1.62 = 22.7 tons. The compression at
whence
??i
2N 5= — (28 22.7) =5.3
=-7
b
only
1
the toe
These quantites have now to be
tons.
by the base widths
if
to obtain the unit stresses, which are as
22 7
follows: at heel,
14 5 3 = 3.8; at center, —- = 2.3; at toe, '-—-= -^ 2.6 tons. 2 6 6
This stress area
is
shown hatched
This stress diagram incidence of
R
is
in Fig. 75f.
useful as showing that
toward the toe of the buttress, consequently
ishes
owing to the
being behind the center point the total stress dimin-
tapered on plan, as has been done.
In Fig. 74
it
should be
it
has been shown
that the stress being uniform by reason of the incidence of the center point of the base, the buttress has been in plan at its base.
which
is
The
Here the
by 140 tons, the
vertical load line
resultant being
12 feet of water results in
Ri. is
Pi
The
is
now 763
is
by the
now be
increased
arch.
The
tons and iV=1018, their
reverse pressure due to
a depth of
70 tons, this combined with Ri, in Fig. 75a,
R2 the final resultant. The value of d is 35° 15' which is As q scales 5 feet, the unit stresses work out as
satisfactory.
follows:
N in Fig, 75e
additional weight of water carried
horizontal water pressure
at
bad foundation.
Flood Pressures. The second, or flood stage, will
investigated.
R
rectangular
indicated unit stresses are very light
a great adv^tntage on a
102.
made
<^Mm^
''^^^zmT^^/^
DAMS AND WEIRS
131
At heel 3.9 tons, at center 2.7, and at toe, 4.2 tons. The stress in the arch under a head of 38 feet comes to 11.5 tons. Thus the stresses in stage 2 are higher than is the cage with stage
1.
At the end of a series of these scallop arches near either abutment the thrust of the arch resolved axially with the weir has to be met either by tying the last two arches by a cross wall and reinforcing rods, or abutting the arch on an abutment supported by wall or a length of solid dam. This design would, it is considered, be improved if the versed sine of the arcs were made somewhat greater, as the arches are too flat near the crest.
The
men-
following remarks bear on the curvature of the arch
When
tioned in section 101. spring line
is
a segmental arch
is
inclined,
the
at a lower level than the crown, consequently the
water pressure
is
also greater at that level.
should vary with the pressure which
it
But the
thickness
This
does not in this case.
proves the advisability of making the circular curvature horizontal,
then a section at right angles to the inclined spring line will be an ellipse,
The
while a horizontal section will be a segment of a
reverse occurs with arches built in the ordinary way.
circle.
There
appears to be no practical difficulty in constructing forms for an
on this principle. Big Bear Valley Dam. Fig. 76 is a plan and sectional 103. elevation of the new Bear Valley reinforced concrete multiple arch dam which takes the place of the old single arch dam menThe following description is taken from tioned in section 83. "Engineering News", from which Fig 78 is also obtained.
inclined arch
The new dam top, abutting
consists of ten arches of 30^ feet, clear span at
on eleven
363 feet on the crest;
is
92
feet (in
buttresses. its
The
maximum
total length of the
dam
height from crest to base
is
a pocket at the middle buttress only), although, as the
elevation in Fig. 76 shows, the average height of the buttresses
much less than that figure. The water face of the structm-e and the rear edge of the buttresses are given such slopes as to bring the resultant of the water-pressure load and that of the structure is
through the center of the base of the buttresses at the highest portions of the dam.
up
Fig.
79.
to within 14 feet of the top
The is
slope
for
the water face
36° 52' from the vertical, and
DAMS AND WEIUS
132
from that point to the crest
is vertical.
stream edges of the buttresses
is
2 on
1
The
slope of the
down-
from the bottom to the
top,
the vertical top of the face arches giving the piers a top width of 10
from the spring
feet
back edge. The buttresses are 1.5 and increase in thickness with a batter of
line to the
feet thick at the top
0.016 feet per foot of height or all
heights.
down
The arch
1
in 60
on each side to the base
to the bend, from which point they are increased in thick-
ness at the rate of 0.014 f^et per foot to the base, or
Fig. 77.
The
View
of Big
Bear Valley
Dam
with Old
Dam Shown in
arc of the extrados of the arch ring
is
1
in 72.5.
Foreground
140° 08' from the
top to bottom the radius being maintained at 17 feet and the at
11.22 feet.
The
uniform throughout, 16
feet,
the arc
extrados all
is,
therefore,
rise
a cylindrical surface
changes in dimensions being
Thus at the 145° 08', and the
intrados of the arch. is
for
and
rings are 12 inches thick at the top
made on
the
top, the radius of the intrados rise 11.74 feet.
At 80
feet
from
the top. Fig. 79, the thickness of the arch ring will be 2.15 feet,
the radius of the intrados 14.85 feet (the radius of extrados less the thickness of the wall), the arc 140° 48' and the rise 10.59
DAMS AND WEIRS In
feet. rise of
all
133
cases of arch-dam design the clear gpan, radius,
and
the intrados decrease from the top downward.
t3
=3
O
>
Strut-tie
members
are
provided between the buttresses to
stiffen and take up any lateral thrusts that might be set up by
seismic disturbances
or vibrations,
these
consisting of
T-beams
DAMS AND WEIRS
134
and supporting arches all tied together by heavy steel reinforcement. The T-beams are 12 inches thick and 2.5 feet wide, with a 12inch stem, set on an arch 12 inches square at the crown and thickening to 15 inches toward the springing
with two spandrel on each side connecting the beam and arch, all united into one piece. There are provided copings for the arches and the tops lines,
posts
of the buttresses
with 9-inch projections, making the arch cope wide and that on top of the buttresses 3 feet wide. The beam slab of the top strut members is built 4 feet wide to serve 2.5 feet
as an extra stiffener, as well as a comfortable footwalk across the dam. This footwalk is provided with a cable railing on both sides
Fig. 79.
to
make
it
Profile
and Sections
of
Big Bear Valley
a safe place upon which to walk.
Dam
To add
to the archi-
tectural effect of the structure, the arches of the strut
terminate in imposts, built as part of the buttresses. are reinforced with twisted steel rods, all
all
members
The
struts
being tied together and
being continuous through the buttress walls.
The ends
entering
the buttresses are attached to other reinforcement passing cross-
by which the stresses beams may be transmitted to and distributed in the buttress walls. The ends of the strut members are all tied onto the granite rock at both ends of the structure by hooking the reinforcement rods into drill holes in the rock. The buttresses are not reinforced, except to be tied to the arch rings and the strut members, their wise into the buttress walls, forming roots
in the
DAMS AND WEIRS
135
shape and the loads they are to carry making reinforcement super-
The arch
fluous.
ribs are reinforced with |-inch twisted rods hori-
and variably to the rods protruding from the
zontally disposed 2 inches from the inner surface
These rods were tied For reinforcing the extrados of the arch ring ribs of HXlJX A-inch angles were used, to which "ferro-inclave" sheets were clipped and used both as a concrete form for the outer face and a base for the plaster surface. spaced.
buttresses.
Stress Analysis-
!04. is
shown
On
79 a rough stress analysis
Fig.
As
depth of water.
for 80 feet
be seen the resultant
will
R
cuts the base just short of the center point.
is
estimated at 4100 tons, the area of the base
sq. feet
—
The
... RHw weight 13 —7
stress
on the arch, 80
16X80X1 —— ^7T~ = ^^ 20
—
feet deep,
^
1
tons, nearly,
of
N
= 110x4. 2 = 460
N 4100 whence -— = — = 9 tons nearly, evenly distributed
taken as unity). its
The value yl
(vi
being
neglecting
u ims shows
rri,-
— or —3 of this
.i
the
1
necessity for the reinforcement provided to take
The tangent This better.
is
of (9=4-t = iV
= -78. 7T^ 4100
.'. 6'
= 39°.
a large value, 35 degrees being the usual
If the
stress.
8
2i
limit,
33 degrees
arch thickness were doubled, reinforcement would not
be necessary except near the crest and the additional load of about 320 tons would bring Q down to 35 degrees. If not, a greater inchnation given to the arch would increase the load of water on the extrados. It is quite possible that
be actually cheaper.
due to
its
own weight
a thicker arch without reinforcement would
The downward is
on a
thrust acting on the arch
different plane
Its effect is to increase the unit stress to
from the arch
a certain extent, as
thrust. is also
the case with the combination of shearing and compressive stresses in the interior of
a
dam
however, be neglected.
as explained in Part
A
I.
This increase can,
considerable but undefined proportion
of the water pressure near the base
is
conveyed to
it
and not
to
the buttresses; this will more than compensate for any increase
due to
The
and consequently it can be ignored. form an excellent provision for them against buckUng and vibration and are universally
vertical compression
ribs connecting the buttresses
stiffening
DAMS AND WEIRS
136
employed
The
in hollow concrete dams.
buttresses in this instance
are not reinforced.
HOLLOW SLAB BUTTRESS DAMS main
Construction
deck
is
is
dam and
a class of
is
introduced by the Ambursen Hydraulic
first
Company
weir
which
principles to the arch buttress type
believed to have been
flat
There
Description of Type.
105.
similar in its
In place of the arch an inclined
of Boston.
substituted, which has necessarily to be
made
of rein-
For this reason, the deck slabs cannot exceed a moderate width, so numerous narrow piers take the place of the forced concrete.
A
thick buttresses in the former type.
further development
is
a
thin deck which covers the downstream ends of the buttresses or
forming a rolhvay.
piers,
The
enclosed box thus formed
sionally utilized as a
power house
which purpose
well suited.
for
The
it is
occa-
is
for the installation of turbines,
inclination given to the flat deck
is
such that the incidence
of the resultant (R.F.) will fall as near the center of the base as pos-
and at the same time regulate the
sible
inclination of the resultant
to an angle not greater than that of the angle of friction of the material,
30 degrees with the
i.e.,
tendency to
slide
on the foundation
Dam an
Ells^vorth
is
A
Example.
By
vertical.
this
means any
obviated.
good example
of this style of
given in Fig. 80 of the Ellsworth dam in Maine. In inclination of the deck is 45° or very nearly so; the design the this
construction
is
piers are 15 feet centers with
widened ends, so that the
clear span
of the concrete slabs is 9' 1" at the bottom.
The
calculations necessary to analyze the thickness of the
and the
slabs
steel reinforcement at
one point,
viz, at
EL
2.5, will
given. In this case the pressure of water on a strip of the one foot wide, the unsupported span of which is 9' 1", is HIw.
now be slab,
Here
H = Q7
feet
67X9.1X^=19 slab. is
As
and tons.
^
is
To
this latter lies at
partly carried
ton per cubic foot; therefore,
must be added the weight
an angle with the horizontal
by the base and
piers.
The diagram
of slab
w
is
ife
this
in Fig. 80c
resolved in
two
is
is
its
W=
of the
weight
not entirely supported by the
the triangle of forces.
directions, a
and
b,
The weight
respectively, parallel
DAMS AND WEIRS and normal to face
of slab.
The
137
angle being 45 degrees,
w
a^h——r^' V2
Consequently the thickness, 37 inches, can be considered as reduced QfJO± 0001
00£
to
^=
The
2.2 feet.
1.4
portion of the weight of
the slab carried to the piers will, therefore,
9.1X2.2X~ = 1.5
be
tons,
the weight of the con-
assumed at
crete being
the usual value of 150
pounds per cubic foot. The total distributed load in the strip will then
be 19
+ 1.5 = 20.5
tons.
Now the moment of on a uniformly
stress
beam with
loaded
in
ends
or
M=
279
inch-
——,
is
8
20.5X109
=
free
tons.
moment must be equaled by that of This
the
resistance
the
of
concrete slab.
Formulas for
106.
Reinforced Concrete.
For tions
purpose
the
showing
in detail,
leading
of
the calcula-
some
formulas con-
nected with reinforced
beams and
now be
slabs
exhibited:
will
M2 =
DAMS AND WEIRS approximate formulas cZ
= 35
inches
and
(31a)
and & = 12
(32a),
139
=8
/s
tonSj /c
= .3
ton,
inches; then
7
il^
= 8x3X-rX35 = 735
inch-tons
8 ^^^c
= :^X—X 420X35 = 735 10
As already noted the moment
the results being identical. is
The end
but 279 inch-tons.
inch-tons
D
shear
may
of stress
have governed the thick-
Testing for shear the load on a 12-inch strip of slab
ness.
tons of which one-half
is
supported at each end.
is
Allowing 50
20.5
lb.,
or
.025 ton, as a safe stress, the area of concrete required is 10.25 h- .025
= 410 square inches the actual area being 37 X 107.
it.
The
piers are not reinforced at
stresses are all compressive is
= 444 square inches.
The reinforcement
Steel in Fore Slope.
more a matter of judgment than of hardly any weight to support, as the is
of
12
water
falling all,
nor
and the inclination i.e.,
of the
30 degrees.
bay is estimated
the trapezoid of water pressure,
80 drawn through the
e.g. of
is
drawn upward
upstream deck
Fig. 80a is a force
EL
The
0.00.
at 783 tons while that of P,
1700 tons.
The force line P in
Fig.
the water pressure area intersects the
W below the base hne.
vertical force
shoot clear
\\ith the vertical
diagram of the resultant forces acting on the base at total weight of a 15-foot
deck having
will
necessary, as the
is it
such that the resultant pressure makes an angle
not greater than that of friction,
of the fore slope
calculation, this
From
this intersection
R
is
parallel to its reciprocal in the force polygon, cutting
the base at a point some 9 feet distant from the center point.
The maximum
107X2 = 214
sq. ft.;
A= m = i^^±^
stress will occur at the heel of the base.
= 9-34 tons; q being 9 4 = 7^ ^ ^ 214 .4
ft.,
107
'
= 1.5 and 5 = 9.34X1.5 = 14 tons. Formula (9), Part I. The horizontal component of P = 1200 tons. The base being 2 ft. ^nde, 5s=^7TT7j^
=
5.6 tons;
V49+31.4 = 16.5 shearing stress
therefore
by formula
(10),
Part
I, c
=
7
+
The usual Hmit to per sq. inch, equivalent to 7.2 tons per
tons, a decidedly high value.
is
100
reinforcement
lb.
therefore not necessary
and is not provided. There appears to be no reason why a steeper slope should not have been given to the deck so as to bring the center of pressure up sq. ft.,
is
DAMS AND WEIRS
140
to the center of the base
and thus reduce the unit
higher river stage has been allowed
The
for.
Possibly a
stress.
position of
W as well
as the weight of the structure were obtained from the section given Fig. 80
in Schuyler's Reservoirs.
of the so-termed "Curtain''
is
The "Half Apron" type,
type of dam.
Fig. S2c, is
sometimes used for
main section of Fig. 82 illustrating the "Bulkhead" type. Slab Deck Compared with Arch Deck Dam. The Ambur-
overfalls, the
108.
sen dam, wherever the interior space tion of turbines,
not required for installa-
is
undoubtedly a more expensive construction than
is
the multiple arch type.
This fact has at
last
been recognized and in
one of the latest dams erected, scallop arches were substituted for the
deck, thus obviating the expense of reinforcement.
flat
Fig. 81.
By
Section of Arch with 30-Foot Span
increasing the width of the spans, the piers, being thicker in like
proportion, will be in
much
better position for resisting compressive
a thick column can stand a greater unit stress than a thin Another point in favor of the arch is that the effective length of the base of the piers extends practically to the crown of the arch. stress, as
one.
The arch
itself
need not be as thick as the
radial pressure to
which
it is
subjected
slab.
it is
in a
Owing
to the liquid
permanent state
of
compression and does not require any reinforcement except possibly
dam. Here the arch is generally widened, as in Ogden dam. Fig. 73, and thus greatly stiffened at the
at the top of the
the case of the
point where temperature variations might develop unforeseen stresses. Fig. 81
is
a sketch illustrative of the saving in material afforded
by doubling the spans from 15 to 30 feet and conversion to multiple arch type. The radius of the extrados of the arches is 18.5 ft. H is 67 at elevation 2.50 and w= ^ ton; hence the thickness of the arch by formula (21) (si being taken as 15 tons), will be ,
1=
— — —
RIIw = 18.5X67X1 oAfteet^ _=2.6 s 32X15
DAMS AND WEIRS
141
than the reinforced slab of one-half the span, or 15 feet. The greater length of the arch ring over that The area of of the straight slab is thus more than compensated. It is thus actually thinner
is 35X2.6 = 91 square bracketing at juncof the =93, that 30X3.1
the arch, counting from the center of the pier, feet,
that of the slab
is
The saving
tion with the piers, 13, giving a total of 106 square feet.
EL3JF.0 60
P'O
SO SO
fOOfl.
pe Thick StSOSpillwaij
Crost^ Section of 5pillyva(^
Cresf Level
'^£1
Fig. 82.
Profile
and Detailed Sections
due to decreased length lower part of the
of
x_.
^^o.o
_
Guayabal Dam, Porto Rico
of the piers is 25 square feet.
of concrete is saved, also all the steel reinforcement. is
Thus
in the
dam over 40 cubic feet per 30' bay per foot in height If
a roUway
considered necessary in the weir, the deck could be formed
thin reinforced concrete screen supported on
by a
I-beams stretching
across between the piers. 109.
dam it is
Guayabal Dam.
Fig. 82 is a section of the
recently constructed in Porto Rico, its height
on a rock foundation.
The
is
Guayabal
127 feet and
following are the conditions govern-
-
DAMS AND WEIRS
142 ing the design;
maximum
pressure
on foundation 10 tons per square pounds per square inch or 21.6
foot; compression in buttresses 300
tons per square foot; shear in buttresses 100' pounds per square inch, or 7.2 tons per square foot; shear in .03
ton per square inch;
square inch;
fs for
deck slabs 60 pounds, or
deck slabs 600 pounds or
fc for
ton per
.3
deck slabs 14,000 pounds, or 7 tons per square
inch.
The
concrete in the slabs
buttresses 1:3:6;
n = -^
is
is
in the proportion of 1 2 4, in the :
taken as 15 and r='^* = 23.3.
Ec slab
is
:
The deck
Jc
55 inches thick at El. 224, d
taken as 53, allowing 2 inches
is
= By
for covering the steel, bd or the area of the section one foot wide
= 636
53 X 12
Now A
square inches.
P=
the area of the steel =pM.
-01044, hence the ,(,3.3)4',3.3x,5 required area of steel will be 636X.01044 = 6.64 square inches, provided d is of the correct value. The calculation will now be made [for the thickness of the slab which is actually 55 inches. formula
(28),
The load on
^(^)
a strip 12 inches wide
is
109X13 _ w Water pressure —^— — = 44.3 tons .
To
..
r
must be added a portion
this
(13X55\ 4 5 = 3.2 -W
weight of the slab
of the /'
3
\
tons must be added to the 44.3 tons above, 44.3+3.2 = 47.5
V2 tons,
rp.
^
1-
,
ihe bendmg moment
M.r
.
is
TrX-= 47.5X13X12 -— = 8
The depth
tons.
of the slab can be estimated
(27) or the
(26) or
of k
and
j will be fc
all
four will be worked out.
found by formulas (29) and
.
For the
First the values
(30).
= V.313+.0245-. 156 = 582^.156 = 0.426 .
j=ri=-^) = l-.142 = .858 By
formula
(26),
,
by using formulas
approximate ones (26a) and (27a).
purpose of illustration,
„^^ 927 mcn-
o
^^^J^f ,,^,^Tmxm =
''''
DAMS AND WEIRS (i
= Vl234 = 35.07 9
By
formula
inches
Q27 X 2
T/f
(27),
^^=i^-=i2X.3X.426X.858^^^^^ rf= 1406 = 37.5 inches approximate formulas will be used. By (26a)
Now the
^.-
8X927 ^1854^^210 7X12X7X.0104 1.53
= Vl210 = 34.8
by
1^3
inches
(27a)
^,^6X927^556_2^^342 12X.3 (i
3.6
= Vl542 = 39.3
inches
The approximate formulas (26a) and (27a) give higher results than (26) and (27). The result to select is 37.5 inches, formula (27), which is higher than by (26) The depth of beam would then be 40 or 41 inches. .
It is actually 55.
been given a
This discrepancy
s.g. in
The corresponding
1.5, or
With regard
to direct shear
which half acts at each
on the
is
be
" round rods 1 A
inches.
the criterion.
^=pM =
slab,
The shear
usually turned
up
224
is
=
23 7 '
=
.036
This figure exceeds the Umit of
is
over 20 tons.
With regard
These rods are
to shear in the buttresses, the
horizontal component of the water pressure as
FA.
safe resistance is
at their ends in order to care for the shear.
Shear in Buttresses.
is
=47.5 tons of
The deficiency is made up by adding the shear of the The sectional area of this reinforcement is 4.7 square
inches the safe shearing of which
diagram
0104X12X37.5
W as before The
tons, nearly.
ton = 72 pounds per square inch.
.
spaced 3 inches would answer.
pier, viz, 23.7 tons.
6dXS, = 12X55X.03 = 20
steel rods.
shear
steel area will
= 4.7 square
60 pounds.
mud
excess of unity, owing to the presence of
say of 1.3 or
in suspension,
be due to the water having
may
3400 tons.
The
marked on the
force
area of the base of the buttress at
138X3.2 = 441.6, the shearing
stress or 5,
then
=^^ = 8 441.6
tons per square foot, nearly.
The allowable stress being only 7.2 made good by reinforcing rods
tons the difference will have to be of
which two of f-inch diameter would
suffice.
DAMS AND WEIRS
144
Now
with regard to compressive stresses on the buttresses the
R strikes the base at EL 224 almost exactly at the center, the angle
graphical working shows that the resultant
^^^ —^^^^, ^
WfBPlI" / 1^ "^T
The value of N is mean and s the maximum
^ ^Iso is 30 degrees.
r^^l
^H
5650 tons
;
stress will
^i
the
both equal—; and
=
5650 =
12.78 tons.
442 pression on the foundation is
4 feet lower
Tsill
P
well as
the area sq. ft.;
Thecom-
itself,
not be any
although the base width
as —
,
138X3.2 = 442
of the base, equals therefore, 5
. 1
which
less for,
is greater,
N as
Thus the
are also increased.
c3
Q $
pressure on the foundation of the limit
extent
is
and widening to a further
The maximum
Part
I,
EL
internal stress
224, will be
c
in
4
f:
I
c,
by formula
Here 5 = 12.8 i5+J^+5,2. \
and 5s as we have seen rt
in excess
required.
the buttress at (10),
is
= 6.4+
164
is
8 tons, therefore,
+64 = 16.6 tons. The limit
compression in the buttress is 300 pounds per square inch, or 21.6 tons per square foot.
In the bulkhead portion of the dam,
shown in
Fig. 82b, every pier is run
up
14 inches thick through the deck to form
a support for a highway bridge, the spans of which are therefore 16 feet 10 inches
roadway is carried on which are supported by arches of
in the clear; the
slabs
The buttresses are by several double rehi-
reinforced concrete. laterally supported
sway beams, 16"Xl4'', and below the crest a through roadway is provided. The spillway section is shown on Fig. 82c. The
forced
,
DAMS AND WEIRS
145 "^"yfir^
a
O
o
fa
>
DAMS AND WEIRS
146
ground
EL
on a high bench at EL 295.
level is here
325, the
fall is
The roadway
30
The
feet.
spillway
is
The
crest being
of the "half apron type".
on four reinforced concrete girders, a very neat construction; the piers are run up every alternate span and are therefore at 36-foot centers; they are beveled on both faces The spillway will pass 70,000 secondto reduce end contraction. here
feet; its length is
carried
is
775
The bulkhead
feet.
section of the
dam
(see also Fig. 83)
has 51 spans
of 18-foot centers, total length 918 feet; that of the spillway consists
The whole
of 21 spans of 36-foot centers.
depth of the 20 feet and hollow
tail
water
its effect
dams ever
not known,
is
would be but
constructed.
or corbels of the buttresses
work
is
it
length
1674
is
feet.
The
would probably be about is
one of the largest
The arrangement
of the haunches
trifling.
This
a better one than that in the older
of Fig. 80.
Bassano Dam.
110.
Another important work
is
the Bassano
dam illustrated in Figs. 84 and 85. This is an overfall dam built over the Bow River at the head of the eastern section of the Canadian Pacific
Railway Company's
irrigation canal
and
is
estimated to pass
100,000 second-feet of water at a depth of 14 feet.
high nor so long as the Guayabal interest.
dam
First its foundations are
it
on a thick blanket
very hard blue clay of excellent quality.
this formation, is
that
so
presents several features of
twelve feet deep which overhes boulders and gravel. is
Though not
The
of clay
some
This material
great advantage of
which extends over 1000 feet upstream from the work, all uphft, or very nearly so, consequently no
precludes
it
have to be adopted, such as a long apron to would be necessary in case of a foundation composed of porous and loose materials. It has also disadvantages. The allowable pressure on the clay is limited to special precautions
ensure length of percolation, as
2| tons per square foot. This influences the design necessitating a wide spread to the buttresses, laterally as well as longitudinally.
The whole
of the
dam
is
an
overfall
and the general arrangeiaents
are very similar to those prevailing at Guayabal.
The hearth
or
horizontal fore apron, a provision not necessary in the last example, is
at
EL
2512.
The
crest is at 2549.6
a height of 37.6 feet above
the apron and corresponds with the level of the canal intake
floor.
by draw
gates
Wat^r
is
held up to eleven feet above crest level
DAMS AND WEIRS
147
"^Si q!
Qi
DAMS AND WEIRS
148
eleven feet high, and this full supply level
the estimated afflux, which
is
three feet below that of
is
fourteen feet above the crest.
For overturning moment the water-pressure area will be a truncated triangle with its apex at afflux level plus the height h or 1.5
—
57,
Part
to allow for velocity of approach, as explained in section I.
This, in the
Bow
about 12 feet per second;
River with a steep boulder bed
h therefore will equal 1.5
X
——
will
144
3.4 feet
be
and
64
the apex of the truncated triangle will be at a point 14+3.4 = 17.4 feet is
above the
crest level.
not known, the ratio
consequently with
(i
The depth
— with a steep bed slope
= 14, D will have a value
d being depth over crest and
moments
of the tail water at full flood
of
will
about 25 to 28
D that of tail water.
depths up- and downstream and the unbalanced
The upstream head
downstream head say 25
feet.
.5,
feet,
The overturning
and reverse can be represented by the cubes
direct
difference.
not be under
of the
moment by
37.0+14+3.4 = 55
their
and the Their cubes are 166,375 and 15,625 is
feet
the difference being 150,750, thus the reverse pressure will not have
much
above crest
will
This quantity
the afflux level
moment when water
is
full flood will
the flood level up- and downstream, which
water
is
ft.
channel empty.
when
is
be the difference of 30
feet, as
allowed access to the rear of the deck slabs. feet,
held up to canal
estimating
In the case of direct water pressure on
the deck slabs, the acting head at
is
tail
that which has to be considered
than the head, 49
cor-
held up to 11
than the 150,750 previously stated, consequently
the overturning moment.
water
is
be 49^ = 117,649, supposing the
is less
The
the stability of the structure.
effect in assisting
responding representative
the
This
tail
is less
which exists when the gates are closed and full
supply,
the head that has to be considered
i.e.,
is
to
EL 2560.6,
consequently
that at this latter stage.
Analysis of Pressures on Bassano
Dam.
With
this data the
design can be analyzed, the procedure being identical with that
explained in the last example, excepting that the reverse pressure
might be taken into account as it will modify the direction and incidence of i? in a favorable sense though not to any great extent.
The
limit stresses are those given in the last
example with the
fol-
— DAMS AND WEIRS
149
Footings, compression in bending, 600 pounds
lowing additions:
per square inch, shear, 75 pounds per square inch.
Some
explanation will
now be
given of the method of calcu-
lation of the footing to the buttresses
be referred
known
=41 138
This
tons,' nearly. ^
is
will
the base of the buttresses are
In section 109 the value of
quantities.
—= 6
the pressures on
to, as
and the Guayabal dam
N is 5650
tons and
the unit pressure per foot run
on the base of the buttress. Supposing the limit pressure on the foundation was fixed at 3 tons per square foot, then the requisite base width of the footing would be
41 —=13.7
The footmg con-
feet.
o sists of
two
cantilevers attached to the stem of the buttress.
moment
bending
foot wide will be
M
The
-^.
—'———^ = 13.7
of
The moment
buttress
the
be
side will
3.2
reaction on a strip one foot wide will be
15.75 tons.
Then
at the base will be
=
-
According to formula (27a), 28.8 inches.
5.25 feet.
5.25X3X1 =
in inch-tons about the edge of the section
be
will
1
buttress being 3.2 feet wide the pro-
on each
jecting length of footing
The
The
at the junction with the buttress of a strip
~
'
M^-^.
^
.-.
d=
= 498
inch-tons.
J^|^ = V830 =
= 28.8X12 = 346 and A the area of the steel 0104X346 = 3.61 inches, this in a 12-inch IJ-inch bars 4 inches apart. When the weight
fed
pM =
.
wide strip will take on the buttress is considerable the depth of footing slab thus estimated becomes too great for convenience. In such cases, as in Fig. 85, the
beam
will require reinforcement in
compression at the top.
This complicates the calculation and cases of double reinforcement are best
The
worked out by means of tables prepared for the purpose. shown in Fig. 85, were thus double reinforced, in fact
footings
through bars were inserted at each step, the lower being in tension the upper ones in compression. right through the base of the
footing
is
111.
The lower bars were continuous dam.
This reinforcement of the
not shown on the blue print from which Fig. 85
Pressure
on
Foundation
Foredeck.
A
is
derived.
great
many
DAMS AND WEIRS
150
Ambursen dams have been constructed on river beds composed of boulders and gravel, which require a pressure limit of about 4 tons per square foot. This can always be arranged for by widening the footing of the pier buttresses, the same can of course be done with arch buttressed dams. The base of the arch itself can be stepped out in a similar manner. In the Bassano dam the sloping fore deck is unusually thick and is heavily reinforced in addition; this
is
done with the idea of strengthening the structure ice, as well as from the falling water, and with
against shock from
the further idea of assisting the buttresses in carrying the heavy load of the piers and superstructure. lations can well
be made
for this;
It
it is
is
doubtful
if
any
calcu-
a matter more of judgment
than of estimation.
As with the Guayabal spillway, every alternate buttress is run up to form the piers of the superstructure, which latter consists of a through bridge which carries the lift gear for manipulating the draw gates. The so-termed blind buttresses that is, those that do not carry a pier are of thinner section and are apparently not reinforced. Both kinds of buttresses have cross-bracing as shown on the profile. In hollow dams the location of the center of pressm*e moves with the rise of water from the heel toward the Buttresses.
—
—
center within the upstream half of the middle third.
on the other hand, the movement
is
In solid dams,
along the whole of the middle
third division, consequently in hollow
dams
there
is
no tendency
to turn about the toe as with solid dams, rather the reverse, namely,
to upset backward.
This latter tendency must cause tension in
the buttresses which the cross-bracing Baffles.
As noted
is
intended to care
for.
already in section 66, baffles have been built
on the curved bucket with the object of neutrahzing this mischievous arrangement which it is hoped will soon ^become as obsolete in western practice as has long been the case in the East. Hearth and Anchored Apron. The dam is provided with a horizontal fore apron or hearth 76 feet long
and beyond
solid
this the
device of an anchored floating apron of timber 30 feet in length has been added. The apron is undoubtedly too short and should have
been made 100 feet or 2
below
it.
The wooden
(H+d)
in length, with cribbed
sheet pihng in the rear of the
sidered to be worse than useless;
it
work
riprap is
con-
merely breaks up the good clay
DAMS AND WEIRS blanket
by
cutting
it
A
in two.
wide
151
'
solid curtain of concrete,
not
would have been a superior below the bucket is provided to
so deep as to penetrate the clay blanket,
arrangement.
The
guard against
sliding.
inclined piling
This
Their capacity
openings.
dam is provided with
is
such that one half
a number of sluice will
pass ordinary
dam to be cut off from the river construction. On completion of the work
floods, allowing the other half of the
by
sheet piling during
these sluices were
closed from the inside
all
by
slabs of concrete
deposited in position.
SUBMERGED WEIRS FOUNDED ON SAND 1
Description of Type.
12.
There
or submerged diversion weir which
is
is
a certain type of drowned
wide rivers or
built across
streams whose beds are composed of sand of such depth that a solid foundation on clay
is
an impossibility.
Consequently, the weir has
to be founded on nothing better than the surface of the river bed,
with perhaps a few this class of weir
lines of
but one
the United States, viz, at the head of the
hollow curtain walls as an adjunct.
Of
beHeved to have been constructed in the Laguna weir over the Colorado River
Yuma
is
irrigation canals.
This type originated in India and in that country are foxmd
numerous examples of weirs successfully constructed across very large rivers of immense flood discharge. For instance, the Godaveri River in Southern India has a flood discharge of 1,200,000
second-feet
Not only
is
and the weirs across it are nearly 2J miles in length. the length great, but as will be seen, the width has to be
The Okhla weir,
very considerable. the Jumna below f mile long. The
Figs. 101
the historic city of Delhi
and is
102, situated
height of these submerged weirs
is
12 feet, their role being purely diversion, not storage.
more of
this type of
low diversion weirs
will in
on
250 feet wide and
seldom over
No
doubt
the future have to be
constructed in the United States or in Mexico, so that a knowledge of the subject is a necessity for the irrigation engineer.
Principles of Design.
The
principles underlying the successful
design of these works are a comparatively recent discovery. Designs
were formerly made on no fixed principles, being but more or less Fortunately some of these works
modified copies of older works. failed,
and
it is
from the practical experience thus gained that a
DAMS AND WEIRS
152
knowledge of the hydraulic principles involved has at
last
been
acquired.
A
weir built on sand
is
exposed not only to the destructive
influences of a large river in high flood which completely submerges
but
it,
foundation being sand,
its
is liable
to be undermined and
worked out by the very small currents forced through the underlying sand by the pressure of the water held up in its rear. In spite of these apparent difficulties, it is quite practicable to design a work of such outline as will successfully resist all these disintegrating
and remain as
influences,
solid
and permanent a structure as one
founded on bed rock.
Laws
113.
The
of Hydraulic Flow.
principle
which underUes
the action of water in a porous stratum of sand over which a heavy
impervious weight
is
imposed
is
analogous to that which obtains in
HeodWaler /I
Fig. 86.
Diagram Showing Action
a pipe under pressure.
a pipe is
line
Fig.
BC, leading out
of
Water Pipe Leading Out
of Reservoir
86 exemplifies the case with regard to
The
of a reservoir.
acting head (i?)
the difference of levels between A\ a point somewhat lower than
A, the actual summit
level
the outlet of the pipe. line
A\C
is
and C the
level of the tail
The water having a
the hydraulic gradient or grade
pressure in the pipe at any point
drawn from the center
is
water beyond
free outlet at C, the
line.
The
hydrostatic
measured by vertical ordinates
of the pipe to the grade line
form velocity of the water in the pipe
is
A\C.
The
uni-
dependent directly on the
head and inversely on the frictional resistance of the sides of the pipe, that is, on its length. This supposes the pipe to be straight, or nearly
so.
114.
Dam. We will now consider the embankment thrown across the sandy bed of a
Percolation beneath
case of an earthen
DAMS AND WEIRS stream. Fig. 87.
The
rally cause leakage
pressure of the
153
impounded water
beneath the impervious earthen base.
low depth of water impounded
it
leakage might be harmless; that
under current would be
may
natu-
will
With a
well be understood that such
the velocity of the percolating
is,
insuflBcient to
wash out the particles of When, however, the
sand composing the foundation of the dam.
head will
is
increased beyond a safe limit, the so-termed piping action
take place and continue until the
Governing Factor for
115.
dam is completely undermined.
Stability.
The main governing
factor in the stability of the sand foundation
evidently not the
is
superimposed weight of the dam, as the sand
is
incompressible;
although a load in excess of the hydraulic pressure must
e^sercise
a certain though possibly undefined salutary effect in delaying the disintegration of the substratum.
Diagram Showing
Fig. S7.
However
this
Effect of Percolation under Earthen
may
be, it is the
Embankment
across
Stream
enforced length of percolation, or travel of the under current, that is
now
recognized to be the real determining influence.
In the case of a pipe, the induced velocity
is
inversely propor-
In the case under consideration, the hydraulic condition being practically identical wdth that in a pipe, it is the tional to the^ length.
enforced percolation through the sand, and the resulting friction against its particles as the water forces its way through, that effects
the reduction of the velocity of the undercurrent, and this frictional resistance is directly proportional to the length of passage.
In the
case of Fig. 87, the length of enforced percolation is clearly that of the impervious base of the earthen dam. The moment this obstruction is passed the water is free to rise out of the sand
and the hydro-
static pressure ceases.
116.
Coefficient of Percolation.
This length of enforced per-
colation or travel, which will be symbolized
by i, must be some
DAMS AND WEIRS
154
multiple of the head H, and
if
reliable safe values for this factor
can
be found, suitable to particular classes of sand, we shall be enabled to design
any work on a sand foundation, with
its stability.
If
perfect confidence in
by c, then
the percolation factor be symbolized
i, or
the length of enforced percolation, will equal c H, II being the head of water.
The factor
Fig.
c will vary in value with the quality of the sand. 88 represents a case similar in every respect to the last
dam
except, instead of a
of earth, the obstruction consists of a
termed the weir or drop
vertical wall
wall,
having a horizontal
impervious floor attached thereto, wliich appendage
is
necessary to
prevent erosion of the bed by the current of falling water.
At the level
stage of
with the
crest,
maximum
and the
pressure the head water will be
level of the tail
water that of the
HB, which
consequently the hydraulic gradient will be
is
piezometric line and as in the previous case of the pipe
-..-'.-, Fig. SS.
-'.
^QpiJ
Diagram Showing Design
86, the ordinates of the triangle
of Profile to
HAB
line, Fig.
-,.,-.-.
Reduce Percolation
will represent the
hydrostatic pressure on the base of the weir wall and of the 117.
structure
Criterion for Safety of Structure. is
moment
The
upward floor.
safety of the
evidently dependent on the following points:
First, the weir wall
ing
floor;
also the
must be dimensioned to
of the horizontal water pressure.
with in a previous section.
resist
the overturn-
This has been dealt
Second, the thickness,
i. e.,
the weight
must be such that it will be safe from being blown up or fractured by the hydrostatic pressure; third, the base length, or that of the enforced percolation L, must not be less than cH, the product of the factor c with the head H. Fourth, the length of the masonry apron and its continuation in riprap or concrete book blocks must be sufficient to prevent erosion. It is evident that the value of this factor c, must vary with the of the apron or floor
nature of the sand substratum in accordance with
its qualities of
DAMS AND WEIRS fineness or coarseness.
135
Fine light sand will be closer in texture,
passing less water under a given head than a coarser variety, but at
the same time will be disintegrated and washed out under less
on which the design mainly depends, can only be obtained experimentally, not from artificial experiments, but by deduction from actual examples of weirs; among which the most valuable are the records of failures due to insufficiency in ReUable values for
pressure.
length of percolation.
L
relation of
c,
From
these statistics a safe value of the
to H, the factor
which
c,
also the sine of the
is
hydraulic gradient, can be derived,
Adopted Values of Percolation
118.
Coefficient.
The
follow-
ing values of c have been adopted for the specified classes of sand.
River beds of light
Class I:
passes a 100-mesh
silt
and sand,
of
which 60 per cent
sieve, as those of the Nile or Mississippi; percola-
tion factor c = 18.
Fine micaceous sand of which 80 per cent of the
Class II:
grains pass a 75-mesh sieve, as in
Himalayan
rivers
and
in such as
the Colorado; c = 15. Coarse-grained sands, as in Central and South
Class III:
India; c = 12.
Boulders or shingle and gravel and sand mixed;
IV:
Class varies
from 9 to
In Fig. 88 of percolation
if
the sand extended only
would be CD, the
rise
up to the
D
from
to
B
level C, the length
not being counted
In that case the area of hydrostatic pressure acting beneath
in.
the floor would be the triangle
from
A
The
step
of
c
5.
to
C
HAB,
As, however, a layer of sand
interposes, the length will be
HHi
occurring in the outline
is
ACD, and
outline IlyB.
due to the neutrahzation
head symbolized by h, effected in the depth AC. Supposing to be 6 feet and the percolation factor to be 12, then the step the pressure area, equal to h, will be 6-^12 = 6 inches. The
^C in
resulting gradient
HxB
will,
however, be
flatter
than
1
in 12; conse-
quently the termination of the apron can be shifted back to
HiBiy being parallel to
HB;
J?iZ)i,
in which case the area of hydrostatic
IIiABu The pressure at any point on the base is by the ordinates of the triangle or area of pressure. Thus the upward pressure at E, below the toe of the drop wall, where the horizontal apron commences, is represented by the line pressure will be
represented
DAMS AND WEIRS
156
FG.
Supposing the head
HA to be
length
CB
.1
percolation between
H and G
is
the base width of the drop wall
= 8f
feet,
The neutralization of head,
i .
h=-^ = l\
10
The upward
feet.
then the total required This is the feet.
c^ = 12X10 = 120
length of percolation will be
by the enforced represented by (?J, and supposing
CE
h,
effected
to be 9 feet,
FG
pressure
is
AC
being 6
feet,
= 10-li
(i7-/t)
feet.
The stepped upper noted in Part level to
I, is
line bounding the pressure area as has been termed the piezometric line, as it represents the
which water would
It is evident
rise if pipes were inserted in the floor. from the above that when no vertical depressions
occur in the line of travel that the piezometric line will coincide
with the hydraulic gradient or virtual slope; when, however, vertical depressions exist, reciprocal steps occur in the piezometric
which then
falls
line is naturally
of the floor at
sure
is
below the hydraulic grade always parallel to the
E is
always a
critical
The piezometric The commencement
line.
latter.
point in the design as the pres-
greatest here, diminishing to zero at the end.
common
In the same way that by the head producing it, the
Simplifying the Computations.
119.
the water pressure
represented
is
factor Wy or the unit weight of water,
nated from the opposing weight of the
masonry, therefore, as the pressure, its
line,
weight.
represented
is
by
its
and if t be the thickness
Now the
floor
floor.
may
be
elimi-
The weight
of the
also
thickness in the
same way
of the floor, tp will represent
hes wholly below low water
level.
Con-
sequently, in addition to the external hydrostatic pressure repre-
sented by H, due to the head of water upheld, there
due to immersion.
The
measured by HC, not in the floor the water
is
actual pressure on the base
HA. Thus would
rise
if
the buoyancy
CDx
is really
a vertical pipe were inserted
up to the piezometric
line
and be
depth the ordinate of the pressure area plus the thickness of the floor. But it is convenient to keep the hydrostatic external pressure distinct from the effect of immersion. This latter can be
in
allowed for ture that
lie
by
reduction in the weight of these parts of the struc-
below L.
W.
Effect of Immersion,
L.
See sections 52 and 53, Part
When
a body
is
I.
immersed in a liquid
it
loses weight to the extent of the weight of the liquid displaced.
:
DAMS AND WEIRS Thus the
unit weight of a solid
weight will be w{p — l)*
As
is
^
wp.
is
157
When
immersed, the unit
a discarded factor, the unit weight being represented only by
the specific
p,
gravity, the weight of
the floor in question will g
be
I
We
t{p
— l)
immersed.
if
have seen that the
hydrostatic pressure acting at
meet
i^ is
8f
To
feet.
this the weight, or
of
thickness,
effective
the floor must be equal
+^
to 8f feet of water for
sym-
safety, or, in
bols,
t =
Assuming a value of 2, the
for p
thickness re-
quired
.
to
counterbal-
ance
the
hydrostatic
pressure will be ^
= 8fX^ = 11.6feet The
formula
for
thickness will then stand
I
-1(1^)
on
Uplift Apron,
It
<")
is
Fore
evident
that in Fig. 88 the long floor
is
subjected to a
very considerable uplift
measured by the area
HABy
the weight of the
apron also the ratio of p
:
(p — 1)
as
it lies
is
reduced in
below L.W.L., consequently
have to be made as already noted
of a
it will
depth of 11.6 feet which
is
a
DAMS AND WEIRS
158
The remedy is either to make the floor porous which case the hydrauhc gradient will fall below 1 in 12, and failure
quite impossible figure. in
will
take place by piping, or
reduce the effective head
else to
insertion of a rear apron or a vertical curtain wall as has
mentioned in section 53, Part
by the
been already
In these submerged weirs on
I.
large ri\-ers
and
in fact in
advisable.
The
length of this should however be limited to abso-
lute requirements.
most
overfall
dams a sohd
fore apron
is
This length of floor is a matter more of individual
judgment or following
successful precedent than one of precise
estimation.
The
following empirical rule which takes into account the nature
of the sand as well as the
head
of water is believed to be a
good guide
in determining the length of fore apron in a weir of this type,
L=3V^ In the case of Fig. 89, the head is 10 quently,
i-3Vl20 = 33
feet,
of floor has been inserted.
say
Now
times the head, or 120 feet = 10c 3c
36,'
(34)
and
assumed at
c is
12, conse-
In Fig. 89 this length
or 3c.
a total length of percolation of 12 is
by
required
hypothesis, of this
used up by the floor leaving 7c to be provided by a rear apron
is
and
feet
it is
Supposing the curtain
curtain.
is
made a depth equal
to IJc,
this will dispose of 3 out of the 7 (for reasons to be given later),
leaving 4 to be provided for
by the
rear apron, the length of which,
counting from the toe of the weir wall,
is
made
4c or 48 feet.
IV which
hydraulic gradient starts from the point
is
The
vertically
above that of ingress A, At the location of the vertical diaphragm of sheet piling, a step takes place owing to the sudden reduction of
head of 3 sides.
feet,
From
the obstruction being 3c in length counting both
here on, the line
pressure area
is
is
termed the piezometric
the space enclosed between
actual pressure area would include the floor
already allowed for in reduction of weight, unity instead of
The
uplift
this portion
and the
itself,
line
and the The
floor.
but this has been
its s.g.
being taken as
2.
on the weir wall
base and the piezometric if
it
line.
is
the area enclosed between
its
In calculating overturning moment,
were considered as having
lost
weight by immersion
would not quite fully represent the loss of effective weight due to uphft, because above the floor level the profile of the weir wall is it
DAMS AND WEIRS
159
is more nearly so. The foundation could be treated this way, the superstructure above AF being given full s.g. and the uplift treated as a separate vertical force as was the case in Fig. 40, Part I.
not rectangular, while that of the pressure area
Now
Vertical Obstruction to Percolation.
120.
to the vertical obstruction,
when water
with regard
percolates under pressure
beneath an impervious platform the particles are impelled upward
by the hydrostatic there
is
dam and
pressure against the base of the
The
a slow horizontal current do"vvnstream.
resistance is along the surface of
any
solid in preference to
course through the middle of the sand, consequently
also
line of least
a shorter
when a
vertical
obstruction as a curtain wall of masonry or a diaphragm of sheet
downward and the obstruction being passed it ascends the other side up to the base line which it again follows. The outer particles follow the lead of the inner as is shown by the arrows in Fig. 89. The value of a piling is encountered the current of
water
is
forced
vertical obstruction is accordingly twice that of a similar horizontal
length of base.
Valuable corroboration of the reliability of the
theory of percolation adopted, particularly with regard to reduction of
head caused by vertical obstruction, has been received, while this was on the press, from a paper in the proceeedings of the
article
American Society of Civil Engineers entitled ''The Action of W^ater under Dams" by J. B. T. Coleman, which appeared in August, 1915. The practical value of the experiments, however, is somewhat vitiated by the smallness of the scale of operations and the disproportion in the ratio of the
head
dam
H: L
experimented on should not be
less
The
length of base
than 50 feet with a
of 5 feet. 121.
Rear Apron.
termed the rear apron. subject to fore
to actual conditions.
The
extension of the floor rearward
Its statical condition is peculiar,
any upward hydrostatic pressure
apron or
floor.
as
is
is
not being
the case Mith the
Inspection of the diagram, Fig. 89, will show
that the water pressure acting below the floor is the trapezoid enclosed
and the floor level; whereas the downward pressure is represented by the rectangle HiAiHA, which is considerably larger. Theoretically no weight is required in the rear apron, the only proviso being that it must be impervious and have a water-tight connection with the weir wall, otherwise the between the piezometric
line
DAMS AND WEIRS
160
H may fall between the rear apron and the rest of the
incidence of
work, rendering the former useless. occurred.
It
Such a case has actually
however, considered that the rear apron must be
is,
of a definite weight, as otherwise the percolation of water under-
neath
would partake
it
of the nature of a surface flow,
and
so pre-
vent any neutralization of head caused by friction in its passage through sand. Consequently, the effective thickness, or rather t
(p
— 1)
should not be less than four
same as that of the
apron or
fore
feet.
Its level
In
floor.
need not be the
fact, in
some cases
it
has been constructed level or nearly so with the permanent crest
llsr^^ 'ijl? f ^ ?
_^^\j.f^ J
-
'^
*
^
^rlS^^^fli
--
^„^;'vi:n'
ii^ -^^ ^--^(ffi^-^^,
^^
View of Grand Barrage over Nile River
Fig. 90.
But
of the drop wall.
.. -'^-
this disposition has the effect of reducing
the coefficient of discharge over the weir and increasing the afflux or head water level, which is
is
open to objection.
The
best position
undoubtedly level with the fore apron.
free
Another point in favor of the rear apron is the fact that it is from either h^'drostatic pressure or the dynamic force of falling
water, to which the fore apron structed of
wet,
i.e.,
more inexpensive
puddle,
is
is
subject;
material.
it
can, therefore, be con-
Clay consolidated when
just as effective in this respect as the richest
cement masonry or concrete, provided where necessary by an
overlay of
it is
protected from scour
paving or riprap, and has a
reli-
DAMS AND WEIRS
161
and the rest of the work. In old works these properties of the rear apron were not understood, and able connection with the drop wall
the stanching of the loose stone rear apron
commonly provided, was
by the natural deposit of silt. This deposit and is of the greatest value in increasing the statical stability of the weir, but the process takes time, and until complete, the work is liable to excess hydrostatic pressure and an insufficient length of enforced percolation, which would allow piping to take place and the foundation to be gradually undermined. 122. First Demonstration of Rear Apron. The value of an impervious rear apron was first demonstrated in tlie repairs to the Grand Barrage over the Nile, Fig. 90, some time in the eighties. This old work was useless owing to the great leakage that took place whenever the gates were lowered and a head of water applied. left
to be effected
e^^entually does take place
In order to check this leakage, instead of driving sheet
Fig. 91.
Section Showing Repuira IMado on
piling,
which
Grand Barrage
the;
was feared would shake the foundations, an apron of cement masonry 24U feet wide and 3.28 feet thick, Fig. 91, \vas constructed
it
over the old
floor,
extending upstream 82 feet beyond
proved completely successful. drilled in the piers, all
By means
of pipes
it.
This
set in
holes
cement mortar was forced under pressure into
the interstices of the rubble foundations,
filling
up any hollows
that existed, thus completely stanching the foundations.
So
effec-
was the structure repaired that it was rendered capable of holding up about 13 feet of water; whereas, prior to reconstruction, it w^as unsafe with a head of a Kttle over three feet. The total length of apron is 238 feet, of which 82 feet projects upstream beyond the original floor and 44 feet downstream, below the floor tually
itself,
the latter having a width of 112 feet.
13 feet, and 18,
which
is
L
being 238
feet, c,
The head
or the percolation factor
H
being
is
—^ =
the exact value assigned for Nile sand in Class
7,
DAMS AND WEIRS
162
This value was not originally derived from the Grand
section 118.
Barrage, but from another work.
The
utility of this barrage
has been further augmented by the
construction of two subsidiary weirs below
two branches of the Nile delta. and enable an additional height of ten across the
it,
see Figs. 92
These are ten feet
to be held
the gates of the old barrage, the total height being
The
increased rise in the
additional head on the
the
moment
tail
106,
up by
now 22 J
feet.
water exactly compensates for the
wort as regards hydrostatic pressure, but on the base of the masonry piers
of the water pressure
BlVi
and
feet high
DAMS AND WEIRS TABLE
I
Showing Actual and Calculated Values Formula
(35),
163
of Li or Talus
L,=lOc^J^X^jf^
Width
DAMS AND WEIRS
164
a weir 13 feet in height; where the height
more or
is
less,
the
width should be increased or reduced in proportion to the square root of the height
and that
of the factor
c.
Beyond the impervious floor a long continuation of riprap or packed stone pitching is required. 123.
Riprap to Protect Apron.
The width
of this material
clearly independent of that appro-
is
priate to the floor, and consequently will be
same
starting point as the
The formula
floor,, viz,
measured from the from the toe of the drop wall.
for overfall weirs is
L^-lOc^^X^j^^
(35)
^75
'10
For sloping aprons, type B, the coeSicient of
c will
be 11
Then
.-n,4M: This formula
is
founded on the theory that the distance of the
toe of the talus from the overfall will vary with the square root of
the height of the obstruction above low water, designated
by
H],,
with the square root of the unit flood discharge over the weir crest g,
and
directly with
the percolation factor of the river sand.
c,
standard being these values;
Narora weir.
This height, Ih
and is manent masonry crest
H
The
10, respectively, in
when
W.
there are no
L. below the per-
This formula, though more or
remarkably in consonance with actual
will, it is believed,
form a valuable guide to design.
Table I will conclusively prove this. this class
and
equal to
of the weir.
less empirical, gives results
and
is
10, 75,
always the depth of L.
crest shutters,
value,
viz,
As nearly
all
the weirs of
have been constructed in India, works in that country
are quoted as examples. 124.
in type
A
Example will
of
now be
an actual work,
viz,
Design Type A.
Another example of design
given, Fig. 93, the dimensions being those of
the Narora weir over the Ganges River, the
design being thus an alternative for that work, the existing section
which is shown in Fig. 95 and discussed in section 125. The data on which the design is based, is as follows: sand, class 2; peror difference between head and low water, colation factor c = 15; of
H
the latter being always symbolized
by L. W.
L., 13 feet, unit dis-
DAMS AND WEIRS
165
charge over weir g = 75 second-feet, the total length of the impervious
apron and
^iiiiii-
vertical obstructions will, therefore,
have
tobeL-cH=15Xl3 = 195feet. The
first
point to be determined
is
the length of the floor or fore apron.
Having
L
fixed this length, the balance of
have to be divided among the
will
rear apron It is
and the
vertical sheet piling.
minimum
essential that this
be not exceeded, as tage to put as
it is
much
length
clearly of advan-
of the length into
the rear apron as possible, owing to the
inexpensive nature of the material of
which
it
can be constructed.
According
X = 42 feet, which is nearly
to formula (2),
equivalent to 3c, or 45 feet, there thus
remains 10c to be proportioned between the rear apron and the vertical curtain. If
the latter be given a depth of
feet the length of travel
absorb 4c, leaving rear apron.
-
t7£-
J-^-
6c, or
30
90 feet for the
The measurement
taken
is
The from the toe of the drop wall. neutralization of the whole head of 13 feet is thus accomplished.
L+r/-
2c, or
down and up mil
A second cur-
tain will generally be desirable
at the
extremity of the fore ^apron as a precautionary measure to form a protec-
downstream washed out or sinks. This curtain _must have open joints to
tion in case the loose riprap
from the apron
is
offer as little obstruction to percolation
as possible. area,
^i
drawn
that
The is
outline of the pressure
the
as follows:
piezometric
cH = 195 feet
line, is
is
meas-
ured horizontally on the base line of the pressure area, that
is,
at L.
W.
L. from a
DAMS AND WEIRS line
^
through
167
The point B
to 5.
is
A
on the head water level at the commencement of the rear then joined with
apron.
The
thus be
1
this line
hydraulic gradient will
The
in 15.
BA
through the
intersection of
with a vertical drawn
first line
of curtain
two
location of a step of
is
the
feet equal to
the head absorbed in the vertical travel
Another
at this point.
Q
line parallel to
the hydraulic gradient
now drawn
is
to the termination of the fore apron, I
j. (;
this completes the piezometric line or
the upper outline of the pressure area.
With regard to the at the toe of the S
^
d
of h, or loss of
floor thickness
drop wall, the value
head due to percola-
tion under the rear apron, is 6 feet,
from the rear curtain, 4
H—h
feet.
The
is,
10
— 10 = 3.
thickness of the floor according
{H — h)=^
to formula (33) where
^1
feet; total
therefore, 13
feet
comes to | X 3 = 4 feet, the value of p being assumed at 2. The floor naturtoward
ally tapers
its
end where the
The thickness at this made 3 feet which is about the minimum hmit. There remains now uplift is nil.
7^-
point
is
the talus of riprap, Eh
formula (35)
its
is
L = 10caJ^\ ^ = 150 The four of
length from
feet
thickness of the talus
—often
five feet
— and
is is
= 10c
generally
a matter
judgment considering the nature
of
the material used. 125.
Discussion of Narora Weir.
The Narora weir
itself, Fig. 94,
forms
;
DAMS AND WEIRS
168
a most instructive object lesson, demonstrating
what
the least
is
correct base A^idth, or length of percolation consistent with absolute safety, that
can be adopted for sands of
class 2.
The system
of
analyzing graphically an existing work with regard to hydraulic gradient
exemplified in Fig. 95 under three separate conditions;
is
as the
first,
work
originally stood, with a hydraulic gradient of 1 in
11.8; second J at the time of failure,
On
riprap blew up.
this occasion
when the floor and the grouted owing to the rear apron having
been washed away by a flood the hydraulic grade
to
fell
1 in 8;
the extension of the rear apron and curtailment of the fore apron had been effected. Under the first conditions the horizontal component of the length of travel or percolation L from A to E is 123 feet. The total length is made up of three parts: First, a step down and up in the foundation of the drop wall of 7 feet second, a drop down and up of 12 feet either side of the downstream third, after
The
curtain wall; third, the horizontal distance 123 as above. at the
end of the
floor is neglected.
123+7+12 + 12 = 154. This point C.
ylC
is
is
set
The
total value of
out on a horizontal
L
is
rise
then
line to the
then the hydraulic gradient.
This demonstrates that the hydraulic gradient was originally
something under
1 in 12,
The
deficient in.thickness.
toe of the drop wall tf)
of only 5 feet.
2, as
it
mortar.
is
and in addition to
8
The
submerged.
To meet
feet.
value of p —
In spite of
hydrostatic pressure on the floor at the this the floor has a value of
specific gravity of the floor will
was mostly formed
The
this the floor is very
not exceed
of broken brick concrete in hydraulic
1 will,
this,
therefore, be unity, the floor being
the work stood intact to
all
external
appearance for twenty years, when a heavy freshet in the river set up a cross current which washed out that portion of the rear apron nearest the drop wall, thus rendering the rest useless, the connection
having been severed.
On
this occurrence, failure at once took place, as the floor
doubtless been on the point of yielding for some time. state of affairs
had been suspected, as
In
had
fact, this
holes bored in the floor very
shortly before the actual catastrophe took place showed that a large
Thus the it, full, not of sand, but of water. was actually held up by the hydrostatic pressure; otherwise it must have collapsed. The removal of the rear apron caused this space existed below
floor
DAMS AND WEIRS pressure to be so
much
169
increased that the whole floor, together with
the grouted pitching below the curtain, blew up.
The It will line is
hydraulic gradient
EC
is
that at the time of the collapse.
be seen that it is now not shown on the diagram.
reduced to
The
1 in 8.
piezometric
In restoring the work the rear apron was extended upstream as shown dotted in Fig. 95, to a distance of 80 feet beyond the drop wall, and was made five feet thick. It was composed of puddle covered with riprap and at its junction with the drop wall was provided with a solid masonry covering. The puddle foundation also was sloped
to the level of the floor base to form a ground connec-
down
tion with the drop wall.
was driven to a depth
The grouted
At
its
upstream termination sheet piling
of twelve feet
below
floor level.
pitching in the fore apron
was
relaid dry, except
which was rebuilt in mortar, to form a continuation of the impervious floor. Omitting the mortar has the effect of reducing the pressure on the floor. Even then the uplift would
for the first ten feet
have been too great, so a water cushion 2 feet deep was formed over the floor by building a dwarf wall of concrete (shown on the section) This adds 1 foot to the effective value of
right along its edge.
It will be seen that the hydraulic gradient
now works
out to
1
tp.
in 15.
A
value for c of 15 has been adopted for similar light sands from which that of other sands, as Classes I and III, have been deduced.
be noticed that the crest of this weir
It will
shutters which are collapsible
hand or by
a traveling crab that
shutters as
it
20 feet long.
proceeds.
They
are
is
furnished with
when overtopped and moves along the
are raised
crest, raising the
The shutters are 3 feet deep and some held up against the water by tension rods
hinged to the weir, and at about J the height of the shutter, at the center of pressure. 126.
Sloping Apron Weirs, Type B.
designated B, will
now be
vertical drop, the fore
the crest to the L.
being also on a
W.
by
i.e.,
Another type of weir,
discussed, in which there
is
no
direct
apron not being horizontal but sloping from L. or to a
flat slope
little
above
it,
the talus beyond
or horizontal.
In the modern examples of this type which the height of the permanent masonry weir wall
will is
be examined,
greatly reduced,
with the object of offering as Uttle obstruction as possible to the
DAMS AND WEIRS
170
The canal level is maintained by means of In the Khanki weir, Fig. 96, the weir proper, 7 feet high above L. W.L., while the shutters
passage of flood water.
deep crest shutters. or rather bar wall,
are 6 feet high. as
is
It, therefore,
holds up 13 feet of water, the same
was the case with the Narora
The tion in
wet foundations, as most
above L.
weir.
object of adopting the sloping apron
W.
L.
striction of the
of it
The disadvantage
is
to avoid construc-
can be built quite in the dry
of this type lies in the con-
waterway below the breast
wall,
which causes the
With a on the other hand, a depth of 7 feet for water to churn in would be available at this point. This would check the flow and the increased area of the waterway rendered available
velocity of overfall to be continued well past the crest. direct overfall,
Fig. 96.
Profile of
Khanki Weir, Showing Restoration Work Similar to that Narora Weir
should reduce the velocity.
For
of
this reason, although the action
on the apron is possibly less, that on the talus and river bed beyond must be greater than in the drop wall of type A. This work, like the former, failed for want of sufficient effective base length, and it consequently forms a valuable object lesson. As originally designed, no rear apron whatever, excepting a small heap of stone behind the breast wall, was provided. The value of feet;
L up
whereas
to the termination of the grouted pitching it
should have been
cH
or
hydraulic gradient, as shown in Fig. 96b, neglects the small vertical
15X13 = 195 is
only
component at the breast
of this deficiency in effective base width, the floor,
is
but 108
feet.
1 in 8.3.
wall.
The This
In spite
owing to good
workmanship, did not give way for some years, until gradually increased piping beneath the base caused its collapse.
DAMS AND WEIRS Owing to the
171
raised position of the apron,
high hydrostatic pressure.
At
its
not subject to
it is
commencement
it
ten feet
is
below the summit level and nine feet of water acts at this point. This
is
met by four
almost balances
it.
masonry unsubmerged, of s.g. 2, which Thus the apron did not blow up, as was the case feet of
with the Narora weir, but collapsed.
Some as
it
The
explanation of the graphical pressure diagram
some
offers
full
head, or
differing
peculiarities, -ff,
is
from the
required,
is
last
examples.
Owing, however, to the raised and
13 feet.
sloping position of the apron, the base line of the pressure area will
not be horizontal and so coincide with the L.
W.
but
L.,
will
be an
from the commencement a to the point b, where the sloping base coincides with the L. W. L. From b where L. W. L. is
inclined line
With a
reached onward, the base will be horizontal. the pressure
is
shaped but approximates to a rectangle. also properly rectangular in profile,
the profile 127.
is,
An
The
is
not wedge
apron, therefore,
is
whereas in the overfall type
or should be, that of a truncated wedge.
Restoration of Khanki Weir.
work the restoration was on very weir.
sloping apron
nearly uniform, the water-pressure area
After the failure of this
similar lines to that of the
Narora
impervious rear apron, seventy feet long, was constructed
of puddle covered with concrete slabs, grouted in the joints.
A rear
curtain wall consisting of a line of rectangular undersunk blocks
twenty
feet deep,
was provided.
These additions have the effect The masonry curtain having
of reducing the gradient to 1 in 16.
regard to
its
great cost
is
of doubtful utility.
A
further prolonga-
tion of the rear apron or else a line of sheet piling would,
have been equally
effective.
Reinforced-concrete
very suitable for curtain walls in sand and
is
ponderous and expensive block curtain walls
it is
deemed,
sheet piling
is
bound to supplant the which form so marked
a feature in Indian works. 128.
Merala Weir.
quite recently constructed
same
historic river, the
Another weir on the same principle and is the Merala weir at the head of the
Chenab, known as the *'Hydaspes" at the
time of Alexander the Great.
This weir, a section of which
upper reaches of the river and consequently
its
is
is
given in Fig. 97,
is
located in the
subjected to very violent floods;
construction has to be abnormally strong to resist
DAMS AND WEIRS
172
the dynamic action of the water. is
entirely a matter of
This
judgment and
no
definite rules can possibly be given which would apply to different condi-
From a hydrostatic point of view the two lower lines of curtain tions.
blocks are decidedly detrimental and
could well be cut out.
If this were done the horizontal length of travel
or percolation will
head
is
12 or 13
come to
feet.
If
The
140.
the
latter, c
having the value 15 as in the Khanki weir, the value of
The
feet. is
horizontal length of travel
140 feet and the wanting 55 feet
be just made up by the rear cur-
will
The
tain.
nr
L will be 15 X 13 = 195
hues
is
superfluity of the
The
hydrostatic requirements.
impervious sloping apron 'o
two
fore
thus apparent with regard to
is
long
a necessity
to prevent erosion.
t?^
It is a question
whether a
line of
steel interlocking sheet piling is equally efficient as
a curtain formed of wells
of brickwork
12x8 feet undersunk and
connected with filling.
^^
of solidity
former.
m
i^ ^w^-C .
The
of
piling
latter
and weight lacking
The system
undersunk blocks
India.
and concrete
has the advantage in the
of curtain walls is
peculiar to
In the Hindia Barrage, in
Mesopotamia, Fig. 115, interlocking sheet piling has been largely employed where well foundations would have been used in India. This change is probably due to the want of skilled in places
well sinkers,
who in India are extremely
expert and form a special caste.
DAMS AND WEIRS rear apron^ in the Merala weir
The
as the fore apron and
arrangement
is
built
on a slope
The
facilitates discharge.
is
173
of as soUd construction
right
up to
crest level; this
velocity of approach
be very great to necessitate huge book blocks of concrete
The
fore
apron extends for 93 feet beyond the
crest,
A
under
twice as long as would be necessary with a weir of type
The
normal conditions.
distance
L
of the talus is 203 feet against
182 feet calculated from formula (35a).
Khanki
is
6X6X3
behind the slope and beyond that a 40-foot length
feet being laid of riprap.
must
170
feet.
That
of the lower weir at
This shows that the empirical formula gives
a fair approximation.
The glacis.
of type
fore apron in type
B
will
extend to the toe of the slope or
on a sloping apron
It is quite evident that the erosive action
B is far greater than that on the horizontal floor of type A; Summit Level
B
/f
Toll
underneaih woiL Fig. 98.
the
Wafer Level
'
Diagram Showing
Effect of Percolation under a Built on Sand
uplift howevei^ is less, consequently the sloping
Wall
apron can be made
thinner and the saving thus effected put into additional length. 129.
Porous Fore ApronSo
with
is
of it
and the
type C.
As
it
The next type
of weir to
principles involved will be necessary.
A
be dealt
involves some fresh points, an investigation
and
The
previous
B
have been cases where the weir has as appendage an impervious fore apron which is subject to hydroThere is another very common type which will be static pressure. termed C, in which there is no impervious apron and the material examples of types
which composes the body of the weir
is not solid masonry but a porous mass of loose stone the only impervious parts being narrow
vertical walls.
In spite of this apparent contrariety
found that the same
it
will
be
principle, viz, that of length of enforced per-
colation, influences the design in this
type as in the others,
DAMS AND WEIRS
174
Fig. 98 represents a wall upholding water to its crest
on a pervious substratum, as sand, of
all
The
three materials.
pressure area
ACB, and
Unless this base length
Now
undermined.
resting
hydraulic gradient
AD;
is
the upward
CD is the travel of the percolation. equal to AC multiplied by the percolation
the base
is
by experiment, piping
factor obtained
and
gravel, or boulders, or a mixture
will set in
and the wall be
as shoAvn in Fig. 99, let a mass of loose stone
be deposited below the wall.
The weight
of this stone will evi-
dently have an appreciable effect in checking the disintegration and removal of the sand foundation. The water will not have a free
untrammeled egress at D; rise in
it will,
on the contrary, be compelled
the interstices of the mass to a certain height
by the extent
to
EE determined
to which the loose stones cause obstruction to the flow.
Summit Level
ProDoo € Fig. 99.
The but
LOL '3e or
^aler pariicies i,ndcrneoth wall
Effect on Percolation
Due
to Stones below
resulting hydraulic gradient will still
now be
Weir Wall
of Fig.
98
AE—flatter than
AD,
too steep for permanency.
In Fig.
10.0,
the wall
is
shown backed by a
rear apron of loose
stone, and the fore apron extended to F. The water has now to filter through the rear apron underneath the wall and up through
the stone
amount
During this process a certain washed up into the porous body and the sink until the combined stone and sand forms a
filling in
the fore apron.
of sand will be
loose stone will
compact mass,
offering a greater obstruction to the passage of the
percolating water than exists in the sand greater resisting of
water at
E
power to
disintegration.
itself
This
to rise until equilibrium results.
and possessing
far
will cause the level
When
this is the
If a is flattened to some point near F. body is p^o^'ided, the resulting gradient will be equal to that found by experiment to produce permanent equilibrium.
case the hydraulic gradient sufficiently long
DAMS AND WEIRS The mass
after the
175
been
sinking process has
made good up to the original prorock filling. At F near the toe of
finished is then file
by
fresh
the slope the stone offers but
by its weight or depth
either
;
so
little it is
resistance
evident that
the slope of the prism should be flatter than the hydraulic gradient.
The same
action takes place with the rear
apron, which soon becomes so
filled Avith silt,
as
to be impervious or nearly so to the passage of
But
water.
unless
silt is
deposited in the river
bed behind, as eventually occurs right up to level,
crest
the thin portion of the rear slope, as well as
the similar portion of the fore slope, cannot be
counted as
effective.
Consequently out of the
whole base length this part GF, roughly, about one-quarter, can be
deemed
inefficient as regards
length of enforced percolation.
As the
consoli-
dated lower part of the body of the weir gains in consistency, static pressure.
can well be subject to hydro-
it
Consequently, the value of
the mass should be in excess of that of as
tp of
H — h, just
was the case with an impervious floor. 130. Porous Fore Aprons Divided by Core /'///,
Walls.
In Fig. 101 a
effected
by the
still
further development
is
introduction of vertical body or
core walls of
masonry
fore apron.
These impervious obstructions mate-
rially assist
the stability of the foundation, so
much
in the pervious
mass
of the
a Si
so that the process of underscour and set-
tlement which must precede the balancing of the
opposing forces in the purely loose stone mass
need not occur at extent.
If
all,
or to nothing like the
same
the party walls are properly spaced,
the surface slope can be that of the hydraulic gradient is
itself
and thus ensure equilibrium.
clearly illustrated in the diagram.
passing underneath the wall base
CD
This
The water will rise to
.^ tt
'
DAMS AND WEIRS E being
th6 level F, the point
somewhat higher;
under the other walls in the substratum
The head
of water.
AC
177
will
will, therefore,
be
fill
similar percolation
all
split
the partitions
up
full
into four steps.
Value of Rear Apron Very Great, The value of water tightness in the rear apron is so marked that it should be rendered impervious
by a thick under
layer of clay,
imperfect surface
dams where a
and not
left entirely
to more or less
stanching, except possibly in the case of high
silt
formed in rear of the work, Okhla and Madaya Weirs. In Fig. 102 is shown a detailed section of the Okhla rock-filled weir over the Jumna still
settling pool is
131.
It is remarkable as being the first rock-filled weir
River, India.
not provided with any lines of curtain walls projecting below the base
line,
stability
of
on
ent
its
The
which has hitherto invariably been adopted.
its
sand foundation
weight and
its
is
effective
seen, the section is provided with
The
the breast wall.
consequently entirely dependbase
As
length.
two body walls
will
be
in addition to
slope of the fore apron is 1 in 20.
It is
believed that a slope of 1 in 15 would be equally effective, a hori-
making up the continuation, as has been done in the Madaya weir. Fig. 103, which is a similar work but under much zontal talus
greater stress.
The head of the water in the Okhla weir is 13 feet, with shutters up and weir body empty of water a condition that could hardly This would require an effective base length, L, of 195 feet; exist.
—
the actual
is
250
feet.
But, as noted previously, the end parts of
the slopes cannot be included as effective; consequently the hydraulic
gradient will not be far from tp,
feet.
tp
1 in 15.
The weight
exactly balances this head at the beginning, as If
of the stone, or
it is
10X1.3 = 13
the water were at crest level and the weir full of water,
would equal 8
feet,
or rather a
trifle less,
owing to the lower
level
body wall. The head of 13 feet is broken up The first is 3 feet deep, acting on a part of the
of the crest of the
into four steps.
rear apron together with 30 feet of the fore apron, say, 1 in 15; the rest are 1 in 20.
A
slope of 1 in 15 for the first party wall
cut the base at a point 40 feet short of the toe. fourth party wall
is
would
Theoretically a
required at this point, but practically the rip-
rap below the third dwarf wall
is
so stanched with sand as not to
afford a free egress for the percolation; consequently the slope
may
DAMS AND WEIRS
178
be assumed to continue on to
its inter-
As aheady noted, material would be saved section with the horizontal base.
in the
by adopting a
section
reliably
stanch rear apron and reducing the fore slope to 1 in 15, with a horizontal con-
was done
tinuation as
Madaya
in the
weir.
Economy is
of
Type C,
This type
only economical where stone
is
dant.
It requires httle labor or
work.
On
C
abun-
masonry
the other hand, the mass of
the material used greater, in fact,
is
very great, much
than is shown by the
sec-
This is owing to the constant sink-
tion.
ing and renewal of the talus which goes
on for many years after the
first
construc-
tion of the weir.
The is
action of the flood on the talus
undoubtedly accentuated by the con-
traction of the
/
waterway due to the high
The flood
sloping apron.
velocity 20 feet
below the crest has been gaged as high as 18 feet per second.
very materially reduced overfall ,
s
the
A type
were adopted, as the area
of of
•••
waterway at G€^:
\
This would be if
this point
than doubled. Dehri Weir. 132.
example of
would be more Another typical
this class is the Dehri weir
over the Son River, Fig. 104. of L,
stone
if
The
value
the apexes of the two triangles of
filling
are deducted
and the
cur-
tain walls included, comes to about 12 H,
12 being adopted for this class of coarse sand.
The curtain walls, each over 12,500
feet long,
costly.
must have been enormously
From
the experience of Okhla,
DAMS AND WEIRS a contemporary work, on a
much worse
179
class
of sand, curtain protection is quite unneces-
sary
if
horizontal base
sufficient
The head on
provided.
and the height
is
this weir is 10 feet,
of breast wall 8 feet, tp
1.3X8 = 10.1, which
therefore,
width
is
is,
sufficient,
considering that the full head will not act
The
here.
lines of curtains
dispensed with
of pressure
length; (2)
duced;
the following alterations
if
Rear apron to be reliably order to throw back the incidence
were made: stanched in
could be safely
(1)
and increase the effective base three more body walls to be intro-
(3) slope 1 in
12 retained, but base to
be dredged out toward apex to admit of no thickness
under
five
feet.
This probably
would not cause any increase in the quantities of masonry above what they now are, and would entirely obviate the construction
of
nearly five miles of undersunk curtain blocks. 133.
Laguna Weir.
The Laguna weir
over the Colorado River, the only example of
C
in the United States is shown in Fig. Compared with other examples it might be considered as somewhat too wide if regard is had to its low unit flood discharge, but the
type 105.
inferior quality
of the
sand of this river
probably renders this necessary.
The body
walls are undoubtedly not sufficiently numer-
ous to be properly effective. of
an impervious
rear apron
The
provision
would
also be
advantageous. 134.
"-S^
Damietta and Rosetta Weirs.
location of the Damietta
The
and Rosetta subsid-
iary weirs, Fig. 106, which have been rather
recently erected below the old Nile barrage, is
shown
in Fig. 92.
These weirs are of type
C, but the method of construction
is
quite
DAMS AND WEIRS
180
novel and
it is this
alone that renders this
work a valuable object
lesson.
The deep
foundation of the breast wall was built
without any pumping,
material having
all
been deposited in the water of the Nile First the profile of the base was
River.
dredged out, as shown in the section. the
core
Then was constructed by first a temporary box or enclosure
wall
depositing, in
by a few
piles,
loose stone from
barges floated alongside.
The whole was
secured
then grouted with cement grout, poured
through pipes
let into
the mass.
completion of one section
all
On
the
the appliances
were moved forward and another section built,
and so on
completed.
until the
and the
of the core wall
up by 5^
whole wall was
Clay was deposited at the base profile
then made
loose stone filling.
This novel s^^stem of subaqueous con-
I
struction has proved so satisfactory that in
many
cases
methods. tions in
bound to supersede
it is
older
Notwithstanding these innova-
methods
of rapid construction, the
open to the objection of being extravagantly bulky even for the type adopted, the base having been profile of the weir itself is
dredged out so deep as to greatly increase the fi-iudcoi^ I
w
or
mean depth
of the stone
filling.
It is open to question whether a row two rows of concrete sheet piles would
not have been just as efficacious as the deep breast wall, and would certainly have been
much [
1
less costly.
The pure cement
grout-
ing was naturally expensive, but the admixture of sand proved unsatisfactory as the
two materials of different specific gravity separated and formed layers; consequently,
DAMS AND WEIRS pure cement had to be used. here
is
much
less
It
may
L
be noted that the value of
than would be expected.
Here, with a value of
or 165 feet.
181
At Narora weir
c of 18, it is
it is
but 8^c, or 150 This
instead of 200 feet, according to the formula.
is
1
Ic,
feet,
due to the
low flood velocity of the Nile River compared with the Ganges.
The Paradox
135.
prevailing in type
types
A
and B
is
C
of
From
a Pervious Dam.
it is clear
that an impervious apron as used in
not absolutely essential in order to secure a safe
length of travel for the percolating subcurrent. free to rise
the conditions
through the riprap and at the same
If
the water
is
time the sand in the
prevented from rising with it, the practical effect is the same as with an impervious apron, "Fountaining", as spouting sand is technically termed, is prevented and consequently also "piping". This latter term defines the gradual removal of sand from beneath a foundation by the action of the percolating under current. Thus the apparent paradox that a length of filter bed, although pervious, is as effective as a masonry apron would be. The hydraulic gradient in such case will be steeper than allowable under the latter circumstance. Filter beds are usually composed of a thick layer of gravel and stone laid on the sand of the river bed, the small stuff at the bottom and the larger material at the top. The ideal type of filter is one composed of stone arranged in sizes as above stated of a depth of 4 or 5 feet covered with heavy slabs or book blocks of concrete; these are set with narrow open intervals between blocks as sho^vn in Figs. 96 and 97, Protection is thus afforded not only against scour from above but also from uplift underneath. Although the subcurrent of water can escape through a filter its free exit is hindered, consequently some hydrostatic pressure must still exist below the base, how much it is a difficult river
bed
is
matter to determine, and tion.
If
the
filter
bed
included in that of
L
is
it will
therefore be left out of considera-
properly constructed
Unless exceptionally deep, is not of respect.
its
or the length of travel.
much,
if
length should be
Ordinary riprap. any, value in this
The Hindia Barrage in Mesopotamia, Fig.
115, section 145,
is
bed consisting of a thick layer of stone 65.5 feet wide which occurs in the middle of the floor. The object of this is to allow the escape of the subcurrent and reduce the uplift on provided with a
the
dam and on
filter
that part of the floor which
is
impervious.
DAMS AND WEIUS
182
Crest Shutters- Nearly all submerged river weirs are 136. provided with crest shutters 3 to 6 feet deep, 6 feet being the height adopted in the more recent works. These are generally raised
of a traveling crane running
by means
the hinge of the gate.
When
on
raising of the shutters is effected
from a
trolley running
wires strimg over steel towers erected on each pier.
The
groins are 500 feet apart.
released
fall
In the case of the Merala weir. Fig. 97, the
over this railway.
held up
behind
rails just
the shutters are tripped they
on overhead
These
piers or
6-foot shutters are 3 feet wide,
by hinged struts which catch on to a bolt and are easily by hand or by chains worked from the piers. On the
Betwa weir the
shutters, also 6 feet deep, are automatic in action,
being hinged to a tension rod at about the center of pressure, con-
sequently
when overtopped they turn over and
hinged at the same height; they should not
The advantage
ease the flood gradually.
fall
Not
fall.
all
are
simultaneously but
of deep shutters is very
much lower than othermuch less obstruction to
great as the permanent weir can be built
wise would be necessary, and thus offer the flood.
dent
staff
The only drawback is that crest shutters of experienced men to deal with them.
The Laguna
weir. Fig. 105, has
discharge of the Colorado
is,
no
shutters.
require a resi-
The
unit flood
however, small compared with that of
the Indian rivers, being only 22 second-feet, whereas the Merala weir discharges 150 second-feet per foot run of weir, consequently shutters in the former case are unnecessary.
OPEN DAMS OR BARRAGES 137.
Barrage Defined.
generally designates
what
is
The term "open dam",
or barrage,
in fact a regulating bridge built across
a river channel, and furnished with gates which close the spans as required. They are partial regulators, the closure being only effected during low water.
are opened
and
When
the river
is
in flood, the gates
free passage is afforded for flood water to pass, the
floor being level
with the river bed.
Weir scouring sandy
are indispensable adjuncts to weirs built over practically to the
sluices,
which
rivers,
belong
same category as open dams, as they are
also
partial regulators, the difference being that they span only a portion of the river instead of the whole,
and further are subject to great
DAMS AND WEIRS when the
scouring action from the fact that
normal
raised above its is
empty or nearly
level
by the
river water
weir, the
is artificially
downstream channel
so.
The
Function of Weir Sluices, First, to train the
fold:
183
function of weir sluices
deep channel of the
course of which, is obliterated
by the
and the low
weir, past the canal head,
Othermse, in a wide river
to retain it in this position.
two-
is
the natural
river,
water channel might take a course parallel to the weir crest or else one distant from the canal head,
itself,
and thus cause the approach
channel to become blocked with deposit. Second, by manipulating the sluice gates,
silt
The
deposit in the slack water in the deep channel.
is
allowed to
canal
is
thereby
when the accumulation becomes can be scoured out by opening the gates.
preserved from silting up, and excessive, it
The
sill
of the w^eir sluice
placed as low as can conveniently
is
be managed, being generally either at L.
W.
L.
itself,
or
somewhat
higher, its level generally corresponding with the base of the drop
Thus the maximum
or breast wall. is
subjected
is
statical
head to which the work
the height of the weir crest plus that of the weir
shutters, or Hi,
The ventage provided of the river,
is
dry season discharge. the river low supply
by the low-water discharge more than the average that of the Laguna weir, where
regulated
and should be capable In one case,
is deficient,
of taking
the weir sluices are designed to take
the whole ordinary discharge of the river excepting the highest floods.
which
This
may
is
with the object of maintaining a wide, deep channel
be drawn upon as a reservoir.
This case
is,
however,
exceptional.
As the
object of a weir sluice
is
in order to scour out deposit for
work,
it is
to pass water at a high velocity
some distance to the rear
of the
evident that the openings should be wide, with as few
obstructions as possible in the
way
of piers,
and should be open at
the surface, the arches and platform being built clear of the flood level.
Further, in order to take
power of the current, which
is
full
at a
advantage of the scouring
maximum
at the sluice
itself,
diminishing in velocity with the distance to the rear of the work, it is
absolutely necessary not only to place the canal head as close
as possible to the weir sluices, but to recess the head as
little
DAMS AND WEIRS
184
Scale of Feet
ZO ""'
'
Sp 'Op ^0 6 20 'fTlevahonk refer to heojleyel
Fig. 107.
Plan
of
Laguna Weir-Scouring
Sluices
DAMS AND WEIRS as practicable behind the face
185
Hne of the abutment
end
of the
sluice vent.
With regard by these
effected
much
jected to a
to canal head regulators or intakes, the regulation is entire,
not partial, so that these works are sub-
greater statical stress than weir sluices,
and conse-
quently, for convenience of manipulation, are usually designed with
narrower openings than are necessary or desirable in the
The design
works
of these
latter.
however, outside the scope of the
is,
subject in hand, 138.
lent
Example
of
Weir Scouring
example of a weir scouring
Fig. 108.
weir, the profile
canal intake
is
View
of
sluice,
Fig. 107
Sluice.
is
an
that attached to the
excel-
Laguna
Yuma Canal and Sluiceway Showing Sluice Gatea under Construction
of which
was given
in Fig.
105.
The Yuma
placed clear of the sandy bed of the river on a rock
foundation and the sluiceway in front of
rock independent of the weir.
it is
At the end
also cut
through
of this sluiceway
solid
and
just
past the intake the weir sluices are located, consisting of three spans of 33| feet closed
by
steel
counterweighted
hoisted clear of the flood gates are clearly
shown
by
electrically
in Fig. 108,
at
EL
138.0, that of the canal intake
which can be
operated winches.
which
taken during the progress of the work. is
roller gates
The
from a photograph The bed of the sluiceway
sill
is
is 147.0,
and that
of the
DAMS AND WEIRS
1S6 weir crest 151.0
—hence the
up with deposit to a depth
Fig. 109.
Plan of Weir Sluices for Corbett
discharge of the canal, or
the sluiceway can be of the
draw
whole sluiceway can be allowed to
gates,
if
filled
i.e.,
fill
of 9 feet, without interfering with the
Dam
on Shoshone River,
Wyoming
the flashboards of the intake are lowered
up to EL 156 which
18 feet deep.
The
is
the level of the top
difference
between \high
DAMS AND WEIRS
187
188
DAMS AND WEIRS
S
DAMS AND WEIRS
189
water above and below the is
consequently
feet,
1
1
are lifted
sluice gates
when the
gates
immense scour must take place
and any deposit be rapidly removed.
The
sluiceway
in
is
fact
a large
silt
trap. 139.
The weir
Weir Sluices sluices of the
Corbett
Wyoming,
the Shoshone River,
Dam.
of Corbett
dam on
are given
inFigs. 109, 110,andlll.
The
canal takes out through a tun-
the head of which has necessarily to
nel,
be recessed far behind the location of the Unless special measures
weir sluices.
were adopted, the space between the
and the tunnel head would up with sand and deposit and block
sluice gates fill
the entrance.
To
obviate this a wall 8 feet high
A
built encircling the entrance.
wall
is
also run out
sluices, cutting
and
its
these
is
''divide"
upstream of the weir
them
approaches.
off
from the weir
The
space between
two walls forms a sluiceway which
draws the current of the river in a low stage past the canal head and further forms a large scoured out
trap which can be
silt
when
convenient.
Only a
thin film of surface water can overflow
the long encircling wall, then it runs a paved warped slope which leads
the head gates, the heavy
silt
down
it
into
in suspen-
sion being deposited in the sluiceway.
This arrangement
The is
is
admirable.
fault of the weir sluices as built
the narrowness of the openings which
One would be much more
consist of three spans of 5 feet.
span of 12 feet
DAMS AND WEIRS
190
In modern Indian practice, weir sluices on large rivers
effective.
are built with 20 to 40 feet openings.
Weir Scouring Sluices on Sand. Weir scouring sluices on pure sand on as large rivers as are met with in India are very formidable works, provided \vith long aprons and deep lines of curtain blocks. An example is given in Fig. 112 of the so-termed undersluices of the Khanki weir over the Chenab River in the Pun140.
built
The spans
jab.
are 20 feet, each closed
by 3 draw
gates, running
in parallel grooves, fitted with antifriction wheels (not rollers), lifted
by means in
power winches which straddle the openings which the grooves and gates are located. The Merala weir sluices of the Upper Chenab canal have 8 of traveling
spans of 31
feet, piers
Fig. 113.
5^ feet thick, double draw gates 14 feet high.
View
of
Merala Weir
Sluices,
Upper Chenab Canal
These are Ufted clear of the flood, which is 21 feet above floor, by means of steel towers 20 feet high erected on each pier. These carry the lifting apparatus and heavy counterweights. These gates, Uke those at
Laguna
Fig. 113
is
weir. Fig. 108, bear against Stoney roller frames.
from a photograph
of the
Merala weir
sluices.
The
a partial regulator, in that complete closure at high flood is not attempted. The Upper Chenab canal is the largest in the world
work
is
with the sole exception of the Ibramiyah canal in Egypt, charge being 12,000 second-feet.
Its
depth
is
13 feet.
The
its dis-
capacity
Ibramiyah was 20,000 second-feet prior to head regulation. 141. Heavy Construction a Necessity- In works of this
of the
description solid construction crete construction
is
a necessity.
Light reinforced con-
would not answer, as weight
is
required, not only
DAMS AND WEIRS
191
to witlistand the hydrostatic pressure but the flood water in violent motion.
dynamic
effects of
Besides which widely distributed
r
;.
I
weight
is
undoubtedly necessary for works built on the shifting
sand of a river bed, although this rules
can be formed.
is
a matter for which no definite
DAMS AND WEIRS
192
The
weir sluices at Laguna and also at the Corbett dam, are
solid concrete structures
without reinforcement.
In the East, generally, reinforced concrete is
not employed nor
is
even cement concrete except in wet foundations, the reason
being that cement,
and wood
steel,
for forms are very expensive
items whereas excellent natural hydraulic lime
is
generally avail-
able, skilled and unskilled labor is also abundant. A skilled mason's wages are about 10 to 16 cents and a laborer's 6 to 8 cents for a
12-hour day.
Under such circumstances the employment
forced cement concrete
has to be taken care
is
of rein-
where tension
entirely confined to siphons
of.
In America, on the other hand, the labor conditions are such that reinforced concrete which requires only unskilled labor and
is
mostly made up by machinery,
of
is
by
far the
most
suitable
form
construction from point of view of cost as well as convenience.
This accounts for the very different appearance of irrigation
Both are
works in the East from those in the West.
suitable under
the different conditions that severally exist.
Dams
Large Open
142.
across Rivers.
Of open dams
built
across rivers, several specimens on a large scale exist in Egypt.
These works,
like weir sluices are partial regulators
and allow
free
passage to flood water. Assiut Barrage.
In the Assiut barrage, Figs. 114 and 115,*
constructed across the
quality than
value of
This
Nile
above the
e
dam
is
met with
in the
adopted for the Nile
is
silt
18, against 15 for
of a worse
rivers.
The
Himalayan
rivers.
great Himalayan
holds up 5 meters of water, the head or difference of
levels being 3 meters.
Having regard to
difference of levels but
when
the piers,
canal head
Ibramiyah
Egypt, the foundations are of sand and
in lower
— 6
is
uplift,
the head
considering overturning
the moment,
H
is
the
moment, on
and h being the respective
6
depths of water above and below the gates.
It
is
believed that in
the estimation of the length of travel the vertical sheet piling was left
out of consideration. *In
Figa. 115
and IIG and
sions used in the plans of the
Inspection of the section in Fig. 115
problems in the text, the metric dimenworks have been retained. Meters multiplied by the factor
in the discussion of these
3.28 will give the proper values in feet.
DAMS AND WEIRS
193
shows that the foundation is mass cement concrete, 10 feet deep,
on which platform the
superstructure
built.
is
This
122 spans of 5
latter consists of
meters, or 16 feet, with piers 2
meters thick, every ninth being
an abutment pier 4 meters thick and longer than the rest. This is
a work of excessive solidity
the ratio of thickness of piers to the span being .48, a proportion of .33
S would,
sidered be better.
it is
con-
This could
be had by increasing the spans
20 feet right
to 6 meters, or
through, retaining the pier thickness as
it is
143.
at present.
General Features of
River Regulators. All these river regulators are built on the
general lines, viz,
same
mass foundaan arched
tions of a great depth,
highway bridge, with spring arch at flood
of
level,
then a gap
left for insertion of
the double
grooves and gates, succeeded by
a narrow
strip of
arch sufficient
to carry one of the rails of the
traveling winch, the other rest-
ing on the one parapet of the bridge.
The ter
piers are given a bat-
downstream
ter distribute
in order to bet-
the pressure on
the foundation. of the weight of
The
resultant
one span com-
bined with the horizontal water
DAMS AND WEIES
194 pressure
must
within the middle third of the base of the pier,
fall
the length of which can be manipulated to bring this about.
In
even with increase of the span to 6 meters.
this case it does so
This combined work
from a military point
of value considered
is
of viewj as affording a crossing of the Nile River;
the extreme solidity of
consequently
construction was probably considered a
its
necessity.
In some regulators girders are substituted for arches, in others as
we have noted with regard
to the Merala weir sluices, the super-
structure above the flood line
open
is
steel
work
of considerable
height.
The
Stability of Assiut Barrage.
144.
hydraulic gradient in
drawn on the profile and is the line AB, the horizontal distance is 43 meters while the head is 3 meters. The slope is therefore 1 in 14 J. The uplift is the area enclosed between AB and a horizontal through B which is only 1.4 meters at its deepest part near the gates. Upstream of the gates the uplift is more than balanced by the weight of water overlying the floor. The horizontal travel of the percolation is from Fig. 115, neglecting the vertical sheet piling,
^
to
The
jB
plus the length of the
horizontal travel 51 5 = 17.2. —^ o
c, is
The
ing in this case the uplift is
very
The
hne.
slight,
is
therefore
cB
the line be
is
similarly
This work
is
i.e.,
is
to
c; for
parallel to
the
B
Z),
BC
and
CD
-77,
or
being 8 meters
then the hydraulic gradient which
drawn up from is
meters and the ratio
obtained by adding the vertical to
is
from
Steps occur at points b and
AD,
5H
135.
two vertical obstructions. Their effect on the owing to the fore curtain which raises the grade
the horizontal travel,
AD
as explained in section
piezometric line has also been shown, includ-
slope in this case
each in length,
filter
is
first
which has the practical
AD B
instance the line
^B
is 1 is
in 23.
part of
drawn up from C, and the
line
forming the end step.
to be built with a filter downstream,
effect of
adding to the length of percolation
travel irrespective of the hydraulic grade. 145.
The Hindia
Barrage.
The Hindia barrage,
erected over the Euphrates River near Bagdad,
is
quite recently
given in Fig. 116.
This work, which was designed by Sir WilUam Willocks, bears a
DAMS AND WEIRS
v-f
^1
Uj
1}
L
±± -Ci* ..diiik-
:?§.
U311U
\
'
f^;:;
-^
195
DAMS AND WEIRS
196
A
from
B
to
including .50 meter due to the sheet piUng
AB
meters.
is
then the hydraulic gradient, which
is
is
36.50
1 in
36-^ 5 — 3.0
=1 to
The
in 10.4.
AB.
The
below the
area of uphft
line
DE
tion, leaving the area
is
however accounted
is
masonry situated below counted
DFC is drawn up from D parallel DGHEF; that part of the uplift
piezometric line
for
by assuming all by flota-
El. 27.50 as reduced in weight
DEF
as representing the uplift
still
unac-
for.
Beyond the
filter is
a 21-meter length of impervious apron con-
puddle covered by stone paving, which abuts on a
sisting of clay
masonry subsidiary weir. This wall holds the water up one meter in depth and so reduces the head to that extent, with the further addition of the depth of film passing over the crest at low water
which
meter, total reduction 1.50 meters.
is .5
This
is
the
first
or talus; its object
instance of the use of puddle in a fore apron, is,
by the introduction
of
an impervious rear
apron 21 meters long, to prevent the subsidiary weir wall from being
The head being
undermined.
1|
required, taking c as 18, will be
length of travel provided
is
meters,
the length of travel
18X1.5 = 27
meters.
The
actual
vertical 15, horizontal 41, total 56
more than double what is strictly requisite. The long, hearth of solid masonry which is located below the subsidiary drop wall is for the purpose of withstanding scour caused by the overBeyond this is the talus of riprap 20 meters wide and a row fall. The total length of the floor of this work is 364 of sheet piling. That of the Assiut barrage feet, with three rows of sheet piles. The difference in head is is 216 feet with two rows of sheeting. meters,
meter only, so that certain unknown conditions of flood or that of the material in the bed must exist to account for the excess. half a
146.
ing works
American it is
to locate the
vs. Indian
Treatment.
In American regulat-
generally the fashion where entire closure
draw gates and
head wall that sluice openings.
closes the
required
upper part of the regulator above the
Thus when the
behind the panel walls.
is
their grooves inside the panel or bulk-
gates are raised they are concealed
In Indian practice the gate grooves in the
piers are generally located outside the bulkhead wall; thus
hoisted, the gates are visible
and
accessible.
Fig. 117 is
when
from a
DAMS AND WEIRS photograph of a branch head, illustrating
197
this.
The work
is
of
reinforced concrete as can be told from the thinness of the piers.
In an Indian work of similar character the pier noses would project
beyond the face wall
well
raised in front, not behind
The
of the regulator
and the gates would be
it.
use of double gates
is
universal in Eastern irrigation works;
they have the following unquestionable advantages over a single First j less
gate:
one; second^
power
when
for each is required to
lift
hoisted they can be stacked side
two gates than by side and so
the pier can be reduced in height; third, where sand or suspension, surface water can be tapped
down
while the upper
is
raised;
silt is
by leaving the lower
amd fourth, regulation
is
made
in
leaf
easier.
Typical American Regulating Sluices in ReinforcedConcrete Weir
Fig. 117.
In the Khanki weir
sluices, Fig. 112, 3 rates
running in 3 grooves
are employed.
dams the spans as large as convenient, the tendency in modern is to increase the spans to 30 feet or more; the Laguna weir are 33| feet wide and the INIerala 31 feet. The thickness of
147.
should be design sluices
the piers
Length of Spans.
In designing open
made
is
a matter of judgment and
function of the span, the depth of water piers is regulated,
the
is
best expressed as
by which the height
some of the
forms another factor.
The depth of water upheld regulates the thickness more than length. The length should be so adjusted that the resultant
combined of the weight of one pier and arch, or of- the water pressure acting on one span falls within the middle third of the base. line of pressure
superstructure and
DAMS AND WEIRS
198
The
For example take the Assiut regulator, Fig. 115.
and span allowing
tents of one pier
The
meters of masonry, an equivalent to 1000 tons.
W
incidence of
about 2 meters from the middle third downstream boundary.
is
The moment
of the weight
2 = 2000 meter tons.
Let
H
about this point
is
therefore 1000 X
be depth of water upstream, and h
downstream, then the overturning moment is expressed by
Here H==5fh = 2 meters, orf
con-
390 cubic
for uplift is roughly
•
one span
is
w = l,l
(H^~¥)wl
tons per cubic meter, the length
(125-8)X1.1X7 = _„
',
7 meters; ,1. , 7 then ,1. the
I
moment =
150
6
meter tons. The moment of resistance is therefore immensely in excess of the
moment of water
The
pressure.
governed by the high flood
height of the pier
highway bridge. At full flood nearly the whole immersed in water and so lose weight. There
when the water
intermediate stage
is
however
the width by the necessity of a
level,
of the pier will be is
probably some
pressure will be greater than
that estimated, as would be the case
the gates were left closed
if
while the water topped them by several feet, the water downstream not having had time to rise to correspond. 148.
Moments
for Hindia Barrage.
In the case of the Hindia
barrage, Fig. 116, Ii = 5 meters, A = 1.5, then
^^^J12o-3.4)Xl.lX6.50^^^g
^^^^^ ^^^^
6
The weight
one span
is
about the toe of the base
is
The
factor of
The
long
base
of
estimated at 180 tons.
safety against of
these
Its
moment
about 180X6.5 = 1170 meter tons.
piers
overturning is
distributing the load over as wide
is
required
therefore for
tt^=8.
the purpose of
an area as possible in order to
reduce the unit pressure to about one long ton per square foot. This is also partly the object of the deep mass foundation. The
same
result could doubtless
be attained with
much
less material
by
adopting a thin floor say two or three feet thick, reinforced by steel rods so as to ensure the distribution of the weight of the superstructure evenly over the M'hole base.
the Assiut barrage with
its
It seems to the writer that
mass foundation having been a success
DAMS AND WEIRS
199
DAMS AND WEIRS
200
TABLE n Pier
Thickness— Suitable
SPAN
for
Open
Partial Regulators
and Weir Sluices
DAMS AND WEIRS 152.
Upper Coleroon Regulator.
Fig. 119
201 is
from a photo-
graph of a regulating bridge on the upper Coleroon River in the
Madras Presidency^ southern India.
Originally a weir of type
constructed at this site in conjunction with a bridge.
The
A was
constric-
due to the drop wall, which was six feet high, and the piers of the bridge, caused a very high afi3ux and great scour on the talus. Eventually the drop wall was cleared away altogether,
tion of the discharge
the bridge piers were lengthened upstream and fitted with grooves
and
steel towers,
Fig. 119.
and counterweighted draw gates some 7
feet
deep
View of Regulating Bridge oq the Upper Coleroon River, Southern India
took the place of the drop wall.
In the flood season the gates can
be raised up to the level of the bridge parapet quite clear of the flood.
The work was thus changed from one of a weir of type A, The original weir and bridge were constructed
to an open dam.
about half a century ago.
Andrew's Rapids Dam. Another class of semi-open dam consists of a permanent low floor or dwarf weir built across the river bed which is generally of rock, and the temporary damming up of the water is effected by movable hinged standards being 153.
St.
lowered from the deck of an overbridge, which standards support
202
DAMS AND WEIRS
DAMS AND WEIRS down
either a rolled reticulated curtain let
203
them
to cover
or else a
mounted on rollers. The St. Andrew's Rapids dam, Fig. 120, a quite recent construcThe object of the dam is to raise tion, may be cited as an example. Manitoba, to enable steamboats to navithe water in the Red River, steel sliding shutter
from Winnipeg City to the lake of that name. To effect this the water level at the rapids has to be raised 20 feet above L. W. L. and at the same time, on account of the accumugate the
ri\'er
lation of ice brought do\\Ti
The Red River rises where the thaw sets
by the
a clear passage
river,
in the South, in the State of in
sequently freshets bring
is
a necessity.
North Dakota
much earlier than at Lake Winnipeg, condown masses of ice when the river and lake
are both frozen.
Camere Type of Dam, The dam is of the type known as the Camer6 curtain dam, the closure being effected by a reticulated wooden curtain, which is rolled up and do^n the vertical frames thereby opening or closing the vents.
having been
first
dam
movable
French invention,
It is a
The
constructed on the Seine.
principle of this
consists in a large span girder bridge,
from which
vertical hinged supports carrying the curtain frames are let drop
on to a low weir. When not required for use these vertical girders are hauled up into a horizontal position below the girder bridge and In
fastened there.
The
needle dam.
spans of 138
The
fact,
the principle
river is
is
very
much
like that of a
800 feet wide, and the bridge
is
of six
feet.
bridge
is
composed
of three trusses,
from internal cross-bracing, and carry tram ing apparatus of several sets of winches
two
lines
and
of
which are
with
all
free
the work-
hoists for manipuhiting
the vertical girders and the curtain; the third truss
is
mainly to
strengthen the bridge laterally, and to carry the hinged ends of the vertical girders. It will is
be understood that the surface exposed to wind pressure
exceptionally great, so that the cross-bracing
tial,
as
is
the pier
also the lateral support afforded
itself
above
ing
absolutely essen-
by a heavy projection
of
floor level.
In the cross-section
opening in the
is
pier.
and unwinding the
it will
be seen that there
This footbridge curtains,
and
is
will
is
a footbridge
carry winches for wind-
formed by projections thrown
DAMS AND WEIRS
204
out at the rear of each group of frames.
It will aflFord through communication by a tramway. The curtains can be detached altogether from the frames and housed in a chamber in the pier clear of the floodline.
The lower
part of the work consists of a submerged weir of
solid construction
feet 6 inches
to
which water
actually holds
5ide
which runs right across the
above L. W. L. at El. 689.50. is
upheld,
is
EL
river; its crest is 7
The top
of the curtains
703.6, or 14 feet higher.
The dam
up 31 feet of water above the bed of the river.
£!levaiion- Sectic
Half Front rieyation Lauchli Automatic Sluice Gate
Fig. 121.
This system
is
open to the following objections:
immense expense involved
in a triple
row of
First, the
steel girders of large
span carrying the curtains and their apparatus; and second, the large surface exposure to ^ wind which
must always be a menace
to
the safety of the curtains. It is believed that the raising of the water level could for a quarter of the cost
of the
if
not
much
less,
be effected
by adopting a combination
system used in the Folsam weir. Fig. 50, with that in the
Dhukwa
weir. Fig. 52, viz, hinged collapsible gates
which could be
pushed up or lowered by hydraulic jacks as required. lower part of the
dam
The
existing
could be utilized and a subway constructed
DAMS AND WEIRS through
it
ment which
communication and accommodation for the
for cross
pressure pipes, as is
is
205
the case in the
quite feasible would,
Dhukwa
it is
weir.
This arrange-
deemed, be an improvement
on the expensive, complicated, and slow, Camere curtain system. 1 54. Automatic Dam or Regulator. Mr. Lauchh of New York, writing for Engineering News, describes a
new
design for automatic
regulators, as follows:
dam
In Europe there has been in operation for some time a tj^e of automatic or sluice gate which on account of its simpUcity of construction, adapt-
Fig. 122.
View
ability to existing structures, its
Dam
of Lauchli Automatic Which Has in Successful Operation in Europe
Been
for Several
Years
exact mathematical treatment, and especially
successful operation, deserves to attract the attention of the hydraulic engineer
connected with the design of hydroelectric plants or irrigation works. Fig. 121 shows a cross-section and front elevation of one of the above-mentioned dams now in course of construction, and the view in Fig. 122 gives an idea of a small automatic dam of the same type which has been in successful operation for several seasons, including a severe winter, and during high spring floods. Briefly stated, the automatic dam is composed of a movable part or panel, resting at the bottom on a knife edge, and fastened at the top to a compensating roller made of steel plate and filled with concrete. This roller moves along a track located at each of its ends, and is so designed as to take, at any height of water upstream, a position such as will give the apron the inclination necessary for discharging a known amount of water, and in so doing will keep the upper pool at a constant fixed elevation. With the roller at its highest position the panel lies horizontally, and the full section is then available for discharging water. Any debris, such as
DAMS AND WEIRS
206
pass over the dam without any difficulty, even during excessive floods, as the compensating roller is located high above extreme
trees, or ice cakes, etc., will
flood level.
The dam now in course of construction is located on the river Grafenauer Ohe, in Bavaria, and will regulate the water level at the intake of a paper mill, located at some distance from the power house. The dam has a panel 24.27 ft. long, 6.85 ft. high, and during normal water level will discharge 1400 cu. ft. per sec, while at flood time it will pass 3,530 cu. ft. per sec. of water. As shown in Fig. 121, the main body of the dam is made of a wooden plank construction laid on a steel frame. The panel is connected with the compensating roller at each end by a flexible steel cable wound around the roller end, and then fastenedat the upper part of the roller track to an eyebolt. A simple form of roof construction protects the roller track from rain and snow. The panel is made watertight at each extremity by means of galvanized sheet iron held tight against the abutments by water pr,essure. This type of construction has so far proved to be very effective as to watertightness. It
may be
needless to point out that this type of
to the crest of overflow of
dam
of ordinary cross-section,
dam
can also be fitted fulfill the duty
and then
movable flashboards.
The
probability
the future. roller
is
that this type will became largely used in
A suggested improvement would be to
having instead separate
working independently. the span adopted.
There
rollers
will
abolish the cross
on each pier or abutment,
then be no practical limit to
INDEX PAGE
A Aprons 71
decrease uplift, rear aprons
base of dam and hearth and anchored porous fore
98 150 173, 175 159,161 164
fore,
,
rear
,
riprap to protect
169
sloping uplift, affect
70
^
Arched dams
101
characteristics
-
crest width
_'_
examples Barossa Bear Valley Burrin Juick subsidiary Lithgow Pathfinder
Shoshone Sweetwater profiles
correct theoretical
and
practical
variable radii, with vertical
water loads, support of
Arrow Rock dam Assiut barrage
Automatic
dam
or regulator
101
104 104 111 104 112 112 104 107 109 103 103 102 112 104 67 192, 194 205
B Barossa dam Barrages Bassano dam Bear Valley dam Burrin Juick subsidiary
111 182
146 104
dam
112
C Castlewood weir
96
D Damietta and Rosetta weirs Dams and weirs
179 1
INDEX PAGQ
Dams and
weirs
— continued
arched
101
definition
1 *
gravity
2
gravity overfall
^
hollow slab buttress multiple arch or hollow arch buttress
open dama or barrages submerged weirs founded on sand Dehri weir
Dhukwa
J
75 136 113 182 151 178 90
weir
E Ellsworth
dam
136
F Folsam weiv
85
G 92
Granite Reef weir
Gravity dams
2
design
method
analytical
broken
line profiles,
18, 34,
treatment for
method of high and wide
calculation,
11
crest,
13
crest
9
width
failure
by
31
sliding or shear, security against
method method
graphical Haessler's
36,
influence lines
maximum
stress, formulas for elementary profile, application to limiting height by pressure area in inclined back dam, modified equivalent
pressure distribution
pressure limit,
maximum
pressures in figures, actual profile, practical
27 28 29 37 23, 25, 26 27 34 8 4
profile, theoretical
39 10 30 37 22
curved back rear widening profiles,
shear and tension, internal
stepped polygon vertical
16 42 13 31
height, variation of
component
2
discussion .
4 43 41
graphical calculations
2
"middle third" and hmiting stress pressure of water on wall
3 .
2
INDEX PAQB
— continued
Gravity dams discussion
stress limit, compressive
examples
Arrow Rock Assuan Burrin Juick
Cheeseman Lake Cross River and Ashokan
New
Croton
Roosevelt foundations, special
aprons affect uplift aprons decrease uplift, rear gravity
ice pressure,
dam
reinforced against
rock below gravel "high" base of dam,
silt
against
partial overfall
pentagonal profile to be widened pressure, ice
dam,
toe of
filling
against
Gravity overfall dams or weirs American dams on pervious foundations analytical
method
.
base width, approximate characteristics
'
crest width, approximate depth of overfall, calculation examples
of
Castle wood
Dhukwa Folsam _' Granite Reef
^
Mariquina Nira "Ogee" St. Maurice River fore apron, base of dam and graphical process
hydraulic conditions pressm-e,
water
moments
1
of
96 90 85 92 92 95 85 99 98 78 93 81
level, pressures affected
Guayabal dam
4 ^^ 67 59 65 53 65 58 56 69 70 71 73 72 43 50 52 47 51 50 75 97 88 77 75 77 82 83
by
79 141
^
H Haessler 's method
Hindia barrage Hollow slab buttress
3G 42 194 198
dam
136
INDEX FAQE
Hollow slab buttress
dam— continued
description
examples Bassano
136 146 146
j -
Ellsworth
^
Guayabal
139 149
fore slope, steel in
foundation f oredeck, pressure on baffles
150 150 150 137 140
:
buttresses
hearth or anchored apron reinforced concrete, formulas for slab deck
136 141
compared with arch deck
'^
K Khanki weir
171
L Laguna weir
179 112
u
Lithgow dam
M 92
Mariquina weir Merala weir
171 114
Mir Alam dam Multiple arch or hollow arch buttress
dams
arch, crest width of
113 125
arches, differential
126
design
122
examples Belubula Big Bear Valley
_
118 131 114 ___.120
-
Mir Alam Ogden
119
inclination of arch to vertical
pressure flood
foundation, on
_
water, reverse stresses
value
129 125 124 117 113
N Narora weir Nira weir
167
95
INDEX PAGE
o "Ogee' gravity overfall '
dam
85 177 182
Okhla and Madaya weirs
Open dams
or barrages
advantages
American
vs.
Indian treatment
automatic -Corbett dam, weir sluices of '_
definition
.---
200 196 205 189 182
examples 192, 194
Assiut
Hindia North Mon Upper Coleroon St. Andrew's Rapids
194,198 200 201 201 203 190 198 200 201 193 192
.
Cam^r^type heavy construction moments for Hindia barrage piers, thickness of
regulator,
Upper Coleroon
river regulators, features of rivers, across
197 185 190
spans, length of
weir scouring sluice, example of weir scouring sluice on sand
P Pathfinder
dam
104
S Andrew's Rapids dam Shoshone dam Stepped polygon Submerged weirs on sand
201 107
St.
37 151
apron, rear
159, 161
:
computations, simplifying crest shutters
description
•
156 182 151
examples
Damietta and Rosetta Dehri
Khanki Laguna Merala Narora Okhla and Madaya fore aprons, porous
core walls, divided
by
179
178 171 179 171 167 177 173 175
INDEX PAGE
— continued
Submerged weirs on sand
hydraulic flow, laws of
152
percolation coefficient of
153
dam
152
values of coefficient of
155 159
beneath
vertical obstruction to
pervious dam, paradox of a
181
riprap safety, criterion for
164 154
sloping apron
169
stability,
Sweetwater
_.
governing factor for
153 109
dam
T Tables
200
pier thickness
values of L^ or talus width
•
163
U Upper Coleroon regulator
201
