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arch slice 1 ft. wide, with the middle of the crown at Elevation 5,
and with the middle of this slice at the springing line at Elevation
5 -f 7.424 = Elevation 12.424. With reservoir full to the crest, the
water load at the middle point of the crown would correspond to
5 X 62.5 = 312.5 lb. per sq. ft., whereas the water pressure at the
corresponding point at the springing line would correspond to
12.424 X 62.5 = 776.5 lb., or nearly 2.5 times more than at the crown.
At lower elevations this large difference (2.5 times) rapidly grows
leas, and becomes unimportant with crown elevations below Elevation
15. In the upper portion of the dam the radial component of the
arch weight borne directly by the arch itself tends to neutralize the
large difference in water pressure between the crown and the corre-
sponding abutment points, as this weight component has its largest
value at the crown, and its smallest value at the abutments. This
can be shown most clearly by a graphical method, and in Figs. 1 and 2,
two arch slices are drawn, each assumed to be 1 ft. wide. In Fig. 1
the middle of the crown is at Elevation 5, and then a width of 6 in.
is taken on each side; and in Fig. 2 the crown is at Elevation 10. The



MULTIPLE-AKCH DAMS



855




856



MULTIPLE-AECH DAMS




MULTIPLE-ARCH DAMS 857

arch is divided into twenty voussoirs, and the forces acting on these
voussoirs are calculated. These forces, though not distributed uni-
formly, are, of course, symmetrical on both halves of the arch, M^ith
respect to the center line between two buttresses.

On the right half of each diagram are shown the weights of the
different voussoirs resolved into their components. The weight of a
voussoir is partly transmitted to the base through the lower portion
of the arch, and partly supported by the arch. The last component is
equal to the total weight of the voussoir multiplied by cos. 50°, and is
set off vertically, acting through the center of gravity of each voussoir.
This vertical load is resolved into one radial component and one
perpendicular thereto. The radial component can now be added
directly* to the water load on the same voussoir, which, of course, is
also radial. The geometrical sum of the last named (perpendicular)
components represents the increased axial stress toward the abutments.

On the left half of each diagram the total radial forces acting on
the different voussoirs are indicated. A polygon of forces is drawn,
and, by using this, the line of pressure of the arch can be determined
by drawing lines parallel to the proper rays in the polygon of forces.
It is plainly seen that the line of pressure lies decidedly outside the
center line of the arch; especially is this true of the arch with its
crown at Elevation 5 (Fig. 1).

Using these diagrams, the correct shape of the wooden arch trusses
supporting the form work for these upper elevations can be ascertained.
The outside members of these trusses are made elliptical above Eleva-
tion 15, as shown in detail on Fig. 3, in such a way as to bring about
coincidence between the line of pressure and the center line of the
arch. At Elevation 15 the center line of the arch and the line of
pressure do not exactly coincide, but they are close enough together
to allow the circular shape of the arch to be used with entire safety.
Below Elevation 15 the arch is made circular in a plane perpendicular
to its sloping axis, and above Elevation 15 it is made elliptical in
accordance with the diagrams, Figs. 1 and 2, or a slight modification
thereof, depending on how far below the crest the maximum water
level is to be. It might not be out of place to state here that J. S.

* For greater accuracy, this force should be multiplied by the ratio,
length of mean radius ^^ ^^^ ^^^^ neglected in the present case, as this ratio

length of up-stream radius
Is very close to unity.



858



MULTIPLE-AECH DAMS



Length of up- stream f-



^



' i'' -■-■ [ ■■-•■■-'■- - ;i3%: .■■.; P;.- . vr ■ V ■" ^^7T>>2^(;f



UPPER ARCH TRUSSES
WITH FILLERS

SPAN OF ARCH 40 FT.




Dimensions of Trusses



Elevation
of Crown


Radius


Span


Height


1


2l'ni"


37'8"


U'VA"


5


21' 0"


37'6'/2"


wni"


10


20'l0'/2"


37'4'/s"


n'lU"



Pig. 3.



MULTIPLE-ARCH DAMS 859

Eastwood, M. Am. See. C. E., who has been very prominent in bring-
ing into actual use the multiple-arch tsrpe of dam, builds the top por-
tion of the arches vertical, whereby the circular shape can be used for
the entire arch.

Stresses Due to Temperature Changes.
As dams are generally built during the summer season, it is only
logical to assume that after their completion the individual arches are
\inder tensile stress most of the time when the reservoir is empty, and
decidedly so if it is empty during the cold season. The reinforcement
in the arch, therefore, has been placed with the sole purpose of taking
up these tensile stresses, which reach their maximum value near the
down-stream face at the crown, and near the up-stream face at the
abutments, under the conditions just stated. From the drawing at the
left of Fig. 4 it will be seen that the reinforcing steel is placed at a
distance equal to one-fourth of the total arch depth from the respective
faces subject to tensile stresses, at the crown and at the abutments.
The quantity of steel in the arch is perhaps not entirely sufficient to
take care of the maximum condition of temperature drop, but it is
believed that if tension cracks develop, the presence of the reinforce-
ment will cause them to be minute and well distributed, and that,
when the structure becomes loaded, the cracks will close tight. It was
not deemed advisable to put in the arches more steel than that shown
by Fig. 4, for the reason that it is of comparatively little use when the
reservoir is full. It was also kept in mind that a large change in tem-
perature is not likely to occur suddenly, as a time element of perhaps
weeks or months is generally interposed between the occurrence of
maximum and minimum temperature in a dam body. This time factor
can be depended on, to some extent, to prevent or minimize temperature
cracks. It gives the modvdus of elasticity time to adjust itself to the
new condition (colder or warmer concrete).*

The Stability.
After the buttresses have been given some preliminary dimensions,
the stability of the whole structure can be investigated, most con-

• That the deformation of concrete increases, under sustained load, has been
shown by tests and in practice. Two papers on the subject were read at the Twelfth
Annual Convention of the American Concrete Institute, in Chicago, one by A. H. Fuller,
M. Am. Soc. C. E., and C. C. More, Assoc. M. Am. Soc. C. E., entitled "Tests Showing
Continued Deformation under Constant Load", and one by Mr. Carl B. Smith entitled,
"The Flow of Concrete Under Restrained Load". This was abstracted in the Engi-
neering Record, March 4th, 1916, p. 329. See also "The University of Minnesota
Studies in Engineering, No. 3", by F. R. McMillan.



860



MULTIPLE-ARCH DAMS




MULTIPLE-ARCH DAMS



861




862



MULTIPLE-ARCH DAMS



veniently by a graphical method, as shown by Fig. 6, representing a
section through the crown of the arch. To facilitate the investiga-
tion, the dam is divided into horizontal sections, 10 ft. apart in eleva-
tion, and the forces acting on and above each section are shown to
scale in the location and direction in which they act.



DIAGRAM OF FORCES

ACTING ON BUTTRESS

OF MULTIPLE-ARCH DAM

50°Eace Slope
Buttresses 40' from Center to Center

All Weights given are in Tons




Horizontal Component of Water Pressure

Fig. 6.

The most important force acting on the structure is the water pres-
sure, and this, as usual, is assumed to be concentrated in a horizontal
plane which is two-thirds of the total depth below the water surface.

The horizontal plane in which the water pressure is assumed to be
concentrated intersects the up-stream face along an elliptical curve.
The point of application of a single force, representing the water pres-
sure on one total span of 40 ft., coincides with the center of gravity of



MULTIPLE-AKCH DAMS 863

this ellipse, at least, as long as the water pressure does not penetrate

the up-stream face skin, such as plastering, etc.

Due to the fact that the arches have been given a slope of 50°

with the horizontal, the water pressure will have a vertical com-

horizoutal component ^ , , .,.

ponent = ^ . io a very great extent, the stability

tan. o\)

of the dam depends on the presence and action of this component, as

it tends to hold the structure firmly down on its foundation.

The point of application of the water pressure, as already stated, is

taken at the center of gravity of the ellipse. From the crown this is

a little more than one-third of the total distance ( - — |- 5% for present

conditions) between the crown and the springing line.

Considering first the upper 10 ft. of the dam, the horizontal plane

in which the water pressure can be assumed to be concentrated is at a

2
distance of — ;— X 10 from Elevation downward, with the water surface

at Elevation 0, and the point of application, a, of the single force is

11.55



Vs ^ looj



5.78 ft.



sin. 50°
from the crown, measured horizontally.

The horizontal water pressure due to the 10 ft. of water on the
40-ft. span is

+ 625



2



X 10 X 40 = 125 000 lb., or 62.5 tons.



62.5

The vertical water pressure is equal to = 52.5 tons. This latter

tan. 50

force is now combined with the portion of the weight of the arch acting

vertically. The weight of the arch above Elevation 10 is equal to its

volume in cubic feet multiplied by the weight of 1 cu. ft. The center

of gravity of the section — a trapezoid — is found, and the center of

gravity, h, of the whole arch is then taken to coincide with the center

of gravity of an ellipse through the center of gravity of the section,

in the same manner as explained previously for the water pressure.

The weight of the arch is now combined with the vertical component

of the water pressure, and the location of their resultant is found by

taking moments around either point, a or I. The numerical value of

the resultant is equal to the sum of the two forces, 52 -j- 52.5 = 104.5

tons. Taking moments around the point, a, and scaling the distances.



864 MULTIPLE-ARCH DAMS

preferably on the sloping line between a and b for greater accuracy, we

52 X 2.75

have the equation — :; — = x = 1.367 ft., giving the location

104.5

of the resultant at a point 1.367 ft. from a along the line between a

and h. The weight of the buttresses (above Elevation 10) assumed

to be concentrated in the center of gravity, d, is calculated to be 14.5

tons; it is combined with the vertical load of 104.5 tons on the arch,

in the same manner as shown previously, by taking moments around

any point, say c. The location of the resultant is found to be at e,

and its value is 104.5 -j- 14.5 = 119 tons. This represents the total

vertical force, both as to size and location.

This force is now combined with the horizontal water pressure of
62.5 tons, and the resultant is drawn. It is the purpose of this dia-
gram. Fig. 6, to give the value of the resultant of all forces, and to
establish the point of intersection, g, between this resultant and the
base, in this case at Elevation 10. For convenience, the distance
from e to /, representing the total vertical force of 119 tons, may be
measured, and the distance, f-g, representing the 62.5 tons hori-
zontal water pressure, may be set oif on the base to the same scale.

Consider next the portion of the dam above Elevation 20 as a
whole. The place of application of the concentrated water pressure is

2
at a distance of -;- X 20 ft. below Elevation 0, and the point of applica-
tion of a single force representing the water pressure on the 40-ft.
span is located the same as before, 5.78 ft. from the crown toward the
springing line, measured horizontally. The vertical component of the

horizontal component ,. ^. i i .

water pressure = . actmg on the arch between

^ tan. 50°

Elevations and 20, is now combined with the weight of the arch

lying between Elevations and 20, and their resultant is draA\Ti in

the correct location, found by taking moments as shown previously.

This resultant is again combined with the weight of the portion of

the buttress lying between Elevations and 20. The shape of the

buttress is taken as that of an obelisk, and its volume and the location

of the center of gravity are found from ordinary rules applying to

such bodies. The weight of the struts and counterforts is to be

added to the weight of the buttress, whereby the location of the center

of gravity might be slightly changed.



MULTIPLE-ARCH DAMS 865

The resultant of all vertical forces acting on the dam above Eleva-
tion 20 can now be found, both as to size and location, and combined
with the horizontal water pressure, whereby the point of intersection
of the resiiltant with the base at Elevation 20 is determined.

The same method of procedure is followed for the remaining por-
tion of the dam, each time taking the base 10 ft. lower than in the
preceding calculations, and the whole diagram, Fig. 6, is completed.
If, now, the center line of the buttress, and also the two lines repre-
senting the middle thirds, are drawn, this diagram will point out very
clearly whether or not the load is distributed economically on the
buttress. For maximum economy, the resultant of all forces should
intersect the base in the center line of the buttress; then the load will
be distributed uniformly over the whole base. Toward the top this
is not quite possible, and is not important, as the material there can-
not be stressed very highly at any rate; but, toward lower elevations,
the down-stream slope of the buttress should be shaped so as to con-
form with this condition, viz., the resultant intersecting the base in
the center line, or approximately in the center line, of the buttress.
The total vertical load on the section shown in Fig. 6 is seen to be
6 332 tons, and the horizontal water pressure, 5 062 tons, both on a
40-ft. span. If the coefficient of friction is taken at 0.75, it is
seen that the actual shear along the base amounts to onlj
5 062 — 6 332. X O-'^'S = 313 tons. There is considerable steel in the
section to help take up this shear, and therefore it was not deemed
necessary to eliminate the shear entirely. There is no hydrostatic
uplift to amount to anything acting on a dam of this type, and water
could hardly find its way to lubricate the surfaces of possible cracks
in the buttresses. Wherever it would be desirable to eliminate the
shear entirely, the face slope should be made flatter, say 45, instead of
50 degrees. This, of course, adds to the material required for construc-
tion, but is the cheapest and best way of accomplishing the result.

Some difference of opinion may well exist as to the actual stress

per square unit of area of the buttress at any horizontal elevation.

The loads per buttress are given on Fig. 6. For instance, at Elevation

80, the vertical load is seen to be 4 977 tons, and the horizontal load,

4 000 tons. As the horizontal area is 360 sq. ft., the unit vertical stress

4 977
should be ^^^ = 13.82 tons per sq. ft., or 192 lb. per sq. in. The



866 MULTIPLE-ARCH DAMS

shear would be a little more than friction alone would take care of,
but steel is provided for the remainder. The resultant of the two
forces (horizontal and vertical) intersects the base 2 ft. down stream
relative to the center line, but on account of having the counterfort
(20 sq. ft.) on this side of the center line also, the stress is actually
distributed \miformly. Now, if, instead of the two principal forces,
their resultant is used, assuming it as acting on a number of steps
perpendicular to the direction of this force (the resultant), the appar-
ent imit stress will be much higher. Thus, the resultant (6 385 tons)
acting on an area equal to the svan of all the steps (360 X sin. 51°

6 385
13' = 280 sq. ft.) will produce a compression equal to = 22.8

tons per sq. ft., or 317 lb. per sq. in.

There is a great difference between 192 and 317 lb. per sq. in.
The actual unit compression will be somewhere between, undoubtedly
depending on the relative value of the modulus of elasticity of the
concrete for compression and for shear. It is seen that the first
method, using the principal forces, gives the maximum possible shear
that could occur, and that the other method gives the maximum pos-
sible unit compression that could occur. None of these methods is
very satisfactory, as none fixes the absolute value of stress within
narrow enough limits, but the writer knows of no better at present.

The reinforcing steel embedded in the buttress is put there for dif-
ferent purposes. Along the up-stream slope, the triangular steel con-
struction shown in Figs. 4 and 5 ties the adjacent arches into the
buttress. This is desirable on account of the fact that, in order to
facilitate construction, the buttresses are built first, and the arches
later. The hooping that interconnects the different bars is simply
left protruding through the concrete of the buttress at the time this
is built. Should one arch fail, this triangular girder would immedi-
ately take up the unbalanced thrust and pi'event adjacent arches from
collapsing, and this is its principal duty. Besides this, however, the
steel is active in preventing cracks in the buttress and in taking up
some shear. Toward the down-stream edge of the buttress, four ver-
tical rods are embedded in the concrete for the purpose of stiffening
it, preventing cracks, and taking care of wind pressure. Toward the
middle portion, reinforced counterforts are constructed for the same
purpose. These are still more effective in accomplishing this, due to



MULTIPLE-ARCH DAMS



867



the greater distance between the pairs of steel rods. All material in
the counterforts supports load the same as the buttress itself, but the
counterforts, at the same time, are most effective in stiffening the
large flat slab (buttress) so that fewer struts are required.

Toward the top of the buttress (at Elevation 15) two horizontal
struts are tied to the vertical reinforcement (Figs. 4 and 5). These
two struts are designed so that, besides their main purpose of holding
the upper portion of the buttress in place, they are capable of support-
ing a light roadway. At Elevation 45 another strut is placed near the
up-stream face, mainly in order to support the triangular girder,
should the latter ever be required to support any unbalanced arch pres-
sure. All struts are able to withstand tension as well as compression,
as can be judged from an inspection of the details on Figs. 4 and 5.

Elev. of bottom of Spillway 905O
Elev. of bottom of Spillway 90i8




^"^ MULTIPLE-ARCH DAM
LOWER GEM LAKE



Fig. 7.

Gem and Agnew Lake Dams.
The design sho%vn in the foregoing illustrations was made by the
writer for the Pacific Power Corporation, of Bodie, Cal., to be used
for the construction of their Gem and Agnew Lake Dams, on Push
Creek, Mono County, California. Fig. 7 is a plan of the Gem Lake
Dam and dam site, and Fig. 8 shows the Agnew Lake dam site. On
Fig. 7 is also indicated the outlet works, consisting of a concrete



868



MULTIPLE-AECH DAMS



chamber provided with a row of iron bars in front, and a 48-in.
steel pipe laid through a tunnel for about 300 ft. This pipe line
terminates in a power-house 1 808 ft. below (in elevation) the top
contour of the lake. The maximum height of any individual arch
of this dam is 84 ft., and the vertical distance from the deepest
point in the foundation to the crest is 112 ft. The length across the
dam site at the crest (Elevation 9 053) is 700 ft. The reservoir created
has a capacity of 17 000 acre-ft., and is capable of regulating Rush
Creek to yield an estimated average flow of 45 sec-ft. The drainage
area above the dam is 22^ sq. miles, and is on the eastern slope of
the Sierra Nevada Mountains. Its elevation ranges from 9 000 ft.



Agneta
Lake
Bottom 3405
^Water Line'




ULIPLE-ARCH
DAM
AGNEW LAKE



above sea level at the dam to 12 000 ft. along the crest of the moun-
tains. The precipitation at these elevations is still high, but decreases
very rapidly from the foot of the steep mountains out over the desert
immediately east, at elevations ranging from 6 000 to 7 000 ft. above
sea level.

The rock at the dam site, for the most part, was exposed bed-rock
worn clean by glacial action. Some excavation, however, was neces-
sary in the stream bed and through a rock slide at the north abutment.
The building material for the dam was foimd near-by. The sand
was taken from the shore of the natural lake. The rock had to be
hauled a short distance on a tramway, first from the outlet tunnel
dump (limestone), and later from a large rock slide (granite) about



MULTIPLE-ARCH DAMS



869





Fig. 13. — Gem Lake Dam.




Fig. 14. — GKM Lake Dam, Showing Section of Up-stream Form being Hoistkd
INTO Place, Also Main Upper Runway with Branches Along Buttresses.



So



55 O








A


^


^!T^?^^^HQH^M||^^^Sfe|








Wr


BE


Jr


f^^^^M'wm


ijftj^j.^i-*



Fig. 17. — Gem Lake Dam. Water Level 35 Ft. Above Tunnel Fxoor.




Fig. 18. — Gem Lake Dam, SHOwaNG Runways a>jd Concrete Elevator.



MULTIPLE-AECH DAMS 8T9

2 500 ft. away. All available materials in the neighborhood, especially
the different sand deposits, were tested before any particular material
was selected for construction. As the sand deposit along the shore
of Gem Lake was good, it was used. This sand was first pumped,
and later shoveled, from the lake, and transported to a storage pile
near the mixing plant. This lake sand, which contained 3J% of
clay and 1% of dirt, was mixed with the sand from the rock crusher
(all particles being less than i in. in diameter) in the proportion of
about three-fourths of lake sand to one-fourth of crushed rock sand.
This gave a very good combination, both as to strength and water-
tightness.

Compression tests on 6-in. cylinders were made as the work pro-
gressed, using Bear brand Portland cement. Gem Lake sand, and
crushed rock, in the proportions 1:2:4, and averaged about 900 lb.
per sq. in. for crushing at the age of 14 days.

A 1:2:4 mix was adopted for the arches and struts, and a 1:2^:5
mix for the buttresses. The actual proportions, however, were some-
times changed, but 1^ bbl. of cement for the arches, and li bbl.
for the buttresses were used always. The rock was crushed in a
gyratory crusher, and separated into three sizes through a revolving
screen having 1^, |, and i-in. meshes. The rejects from the screen
went into a jaw crusher, the jaws of which were set to give a maximum
size of 2 in.

The distribution of the concrete to the different arches and
buttresses was done with two-wheeled push carts and short chutes.

During the construction period, in the summers of 1915 and 1916,
tension tests were made of cement briquettes in a small field labora-
tory. Much attention was also paid to the sand used, and 1 : 2 and
1:3 mortar briquettes were tested frequently; silt analyses were made
as the work progressed.

The reinforcement placed in the dam consisted of high-carbon
steel bars, either corrugated or twisted. As the position and details
of these bars are fully shown in the figures, no further explanation
need be given.

The trees standing on the reservoir site were cut down, sawed
into lumber in a mill erected on the ground by the contractors, and
used for the forms. Large cone-bearing trees can be found in this
neighborhood up to Elevation 11 000 in fairly large quantities, but



OOU MULTIPLE-ARCH DAMS

not of the best quality. The lumber proved to be good enough for
form work, was used green, and on that account | by 4-in. battens
were placed on the under side of the sloping arch to prevent the
leakage of grout between the boards when shrinkage took place. These
battens also helped to give an even, circular shape to the outer | by
12-in. boards, constituting the form for the down-stream face. This
sheeting was supported on 2 by 6-in. studs, 2 ft. 6 in. apart, toe-nailed
to the wooden arch trusses shown in Figs. 3 and 9. These trusses
were 5 ft. apart in elevation, and a little more than 6J ft. apart,
measured along the arch slope. The form work for the up-stream



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