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A COMPARISON OF THE VOLUMES
OF A GRAVITY DAM WITH AN
ARCH DAM

Thesis for the Degree of B. S.
MICHIGAN STATE COLLEGE

Sooran A. Yavruian

I942

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A Comparison of the Volumes

of a Gravity Dam with an Arch Dam

A Thesis Submitted to

The Faculty of
MICHIGAN STATE COLLEGE
of
AGRICULTURE AND APPLIED SCIENCE

.3:

Sooran A: Yavruian
Candidate for the Degree of

Bachelor of Science

July 1942

THESIS

INTRODUCTION
Many types of dams have been built using less material
and yet retaining the safety and permanence of a “gravity“
dam. The most successful of these is the single arch dam
where conditions permit its use; and so the purpose of this
thesis is to<ietermine the difference in material, e.i. the
amount of concrete, under a given set 0! conditions, between
a gravity dam and an arch dam.
The given conditions are:
l. A vertical walled canyon, 100 ft. in width,
capable of resisting the arch thrust.
2. A good foundation, non—porous rock-~n0 uplift.
3. Coef. of friction- concrete on concrete .6
concrete on rock .75

4. Height of dam 100 ft.

0|
0

Top width 6.5 ft
6. No wind or ice pressure
7

. Non-overflow type dam

143107

DESIGN OF NON-OVERFLOW
SOLID—GRAVITY DAM
The general requirements for the stability of a gravity
type dam are:
1. There shall be no tendency toward overturning
2. That the maximum pressure on any plane will
not exceed a certain prescibed safe working
stress. Mflflw
5. That there will be no sliding on horizontal
sections.
The first condition is met by keeping the resultant
pressure in the middle third of the base of the section, ,X
._..,.
thus making the factor of safety against overturning 2 or
greater. The second condition is met by spreading out the
base from that required to fulfill the first condition, if
necessary. The third condition is almost always met by the
requirements of the first and second conditions.
Nomenclature-~The symbols used will have the following sig-
nificance:
W = The weight of the masonry of the dam above any horizontal

section.

The height of the dam subjected to water pressure.

The horizontal water pressure on the upstream face.

your."
u

The resultant of P and W

0
II

The eccentricity of center of pressure on horizontal

section.

‘4
II

Unit compression on upstream face.
Unit compression on downstream face.

The factor of safety against sliding

' The factor of safety against overturning

The length of base

The unit shearing stress

The distance to R from the heel of the base of the
horizontal section

The distance of I from the heel of the base

The distance of P from the heel of the base

Determination of the Cross Section

(2657' 6.5'

 

F““~

 

 

bu- hx\' 3-4 = 100 x 64.5 = 64.5 IO'
150 say 65 ft. 6-5

 

 

Add triangular piece 4.5 x 90'

 

to upstream face to counteract

the effect of the top with Jfifi

 

of 6.5 ft., on the center

of gravity. 7o

 

ptt'fz/ “/ow (ea-.57" /OOI

51-675"

«I

,05

5/0/9!

 

Area of the cross section equals % (65 x 100 - 6.5xlO - 4.5x90)

= 3,485 sq. ft.

The volume of dam = 5,485x100 - 648,500 cu. ft. or 12,910 cu. yd

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H.Ndhom

Section 0' to 10' Reservoir full

L——- 65“-———4

 

 

 

 

 

VJ
.— & Q
_P__r_..
a
I
IC— 2 P”
R.
w = 6.5xlel50= 9750#
2 2
P : !§_ 3 62. x10 = 3,120

”heel ' 0

3130 x 10/5 1 9750x5.25 = 97502 = o

 

j /
z = 10J4OO+ 31,690 -
9750 ‘ 4°53.
e = 4.55- 3.25 = 1.08'
I , 4
I I I x ‘
p a 9760 # 6x1.08 _
6.5 (1 _B:3——_) _ 1,500 (1 i 1)
p : 5000“] .. 28 P5,,
P : 0

r3 = 9750x.6 = 1.88 J
5120

f- 8
O-M: 3..
62.4h 2é1?OX6§5 - 2,03

s=_2..=6120 =6
A 6.5x144

Res ervoir empty

9760
-——- (1 *i’ng-Qfl)=1,5oo (1 to)

6.5
. 1,500“ = 14 P...
- 1,500” = 14 n“

Section 10' t025'

 

Reservior full

 

 

 

 

 

 

 

 

 

 

 

8504-14,625 +10,975 = 26,450

 

 

 

 

T X (ha ' h?) = 31.2X(352 - 102) = 169380 a"

26,450 k

k = 6.58'

16,380X + BIZOXIO = 19,500x25x2
3

X

17.94'

y = 25.00 - 17.94 = 7,06'

”heel = O

16,580x7.06 + 5,120X15 4 26,425x6.58

850x-2§§§_ .14.625x4 '10,925x10.5

+ 9,750x4 - 56,2002: 0

z : 115J700 + 46,800 + 174,000 + 59,000 = 10.37:

 

 

56,200
e = 10.67 - 8.53 1-87
= 56.200 1* 6xl.87 - v t 6.6
p _1§___.( __i§__9 _ 2160 (1 I
p - 5540M a gsrej.
P 2 95-540.: 6'5 e“
f8 = W _
19,500 - 1'11
r - lefiflxll.fl§ -
0‘ 62.4x25 ’ 3'22
8 = 194500 = 8
17x12x13

Reservoir empty
9750x4 + 26,535x6.58 . 56,300x

X = 5.89

(D
II

805 ”5.89 z 2.61

p =_§§$ZOO (1': §z§$§l_I = 2150 (1 i .92)

p = 170‘°'= 1 r4».

0

”U
ll

4,100“ '= as f...

The Computations for the suceeding sections are done

in a simalar manner.

DESIGN OF AN ARCH DAM
Hanna Method
from

Design 2: Dams by hanna and Kennedy

 

y
Arch dams differ from gravity dams principally in that if

3

.~ 1)?
they transmit the water load to the sides, rather than to

the bottom of the canyon.

In order that the material of the gravity dam and arch
dam shall be similar, we shall assume the maximum allowable
compressive stress in the concrete eoual to the maximum
stress in the gravity dam, which was 159 p.s.i., or 20,200
pounds per scuare foot, and the a110wable sheer 51 p.s.i.
In any arch dam there will be some tension at the crown or
at the abutments; so we will use the allowable tension of
50 p.s.i.

We will use a constant radius of extrados, 82' and a
constant central angle of 75°.

The arch ring will be considered 'thin' if the radius
of the extrados is more than 5 times the thickness of the
arch ring, below 5 the arch ring is considered 'thick'.

Nomenclature:

 

N Ma, V : thrust, moment and shear at the abutments.

a’ a

Nh, Mh. Vh - thrust, moment and shear at the hsunch

NC, Mo: Vc . thrust, moment and shear at the crown

Nomenclature:

 

p 2 unit water pressure on upstream face of any arch ring
p' — unit water pressure on neutral axis of any arch ring

R - radius of extrados

r . radius of neutral axis

re . radius of neutral axis thick arch ring

t 3 thickness of arch ring radially to the neutral axis

¢a.' one-half central angle of extrados for loaded arch ring

4’1

angle between the crown radius and the radius of extra-
dos through point g

g : any stress point on neutral axis

unit stress at extrados

(D
II

- unit shear at extrados

unit shear

"b
4
ll

Cross section of the Arch dam

 

 

 

5
CREST , 6
The thicknesses of the arch rings at the ,0-
5.5‘
various ellevations were found by trial and
error. The'thin cylinder' formula which

was used in the calculations of arch /‘
dams in the past gave erroneous results
in this case, and so the arch ring

thicknesses were found by trial and .5b'

 

28'
error.

The volume of any section = ,
15

 

t - t' 4 2(rr1 - rn/2(2xrrx h)'75° 35'
5605

D‘Prf’ 6e/aw ((557 /00'

 

b'IZ'

The t-Otal volume Of arch 1"le ;-_- 810y60V2) (BB—6'5 ) -(16 .~ 6.5

 

2

16 6.5
(125) (82 - __4.+......._) .. (15 tea x 25)(82__16

(28 + 42 x 50)(82 - 16 4 28)
T".— "—7?

222,620 011. ft. : 8,550 cu.ydg.

)

 

2

» 28 )

 

4

,//7

 

 

//’

 

 

\\\\i \\\\\ ‘\\c ‘\\\s \\\\ ‘\\\
/‘ // /

 

 

Volume of the Arch wedge at abutments

 

 

 

x I t tan
Area of wedge 3 3:3- tan ‘v 1301381 91'88. g: 2 t2 tank; t2 tan
2
_ 767(6 :2 + 6.52
V01”me 0f wedge = ’ ‘” 2 sex 10 + (6.58 4 162) -7§Z x 15
*2

 

2 n
4 (102 + :8 ) (.767) x 25 +.i992 +2422) .vevx 50 a 60,995 cu. ft.

 

Total volume of arch dam = total volume of arch ring — the

volume of the wedges = 222:620 * 60:995' 285.615 CU- ft-

Horiz'l
plane ft.
below
crest

10

25

25

50

75

100

Horiz'l
plane ft
below
crest

10

25

25

5O

75

100

Radius
of
Extrados

82
82
82
82
82
82

Thick Type

of
Arch
Ring

6.5

16.0
16.0
28.0
58.0
42.0

of
Arch
Ring

thin
th in}
thick
thick
thick

thick

Abut'

49.5
22.0
21.0
85.0
46.0
55.0

Shearing Stress

t

P.s.i.
Haunch

26.5
11.0
11.5
18.0
26.0
29.5

Crown

0
O
0
0
0
0

Combined Compressive Stresses in Area P.s.i.

Crown

82.0
77.0
70.0
63.0
58.5
46.0

STRESS

In
14.0
-12.5
~12.5
-9.5

-11.0
-4.5

Haunch

Ex
60.5 57.
44.5 25.
59.5 25.
58.0 29.
29.5 27.
28.0 52.

In
0
0
5
0
O
0

TABLE FOR THE ARCH DAM

Abutment
Ex In

—15.0 114.5
-52.0 126.0
-46.5 157.0
-58.0 159.0
-54.0 158.0
-25.5 141.0

Arch ring 1 ft. high whose center is 25 ft below water surface

R/t z 5.01 arch ring can be considered
t - 16 ft.

as either 'thick or thin'
r = 74 ft.
We will use'thin'.

2 ~2
NC ‘ p'r x (1 "I2%_‘) = 25 x 62.4x74x (1 ‘i%%$%§l§—) -

127,900 (1- .415) = 74.900

 

 

Me = -r(p'r - N.) 221:1. - cos 4». = -74(53.ooo) W -cos 00
a -74(53,000) (.07) = 374:300

Vc = -(p'r - Ne) sin a 55,000xsin 0O - 0

Nh - p'r -(p'r -Nc) cos = 127,900-55,000xccs 18° 45': 77,700

Mb = -74(55 000) sin 570 50' o t -

: 57° 50' — cos 18 45 . ~74(55,000)(.017)
: €6,600
Vh = 55,000x sin is0 45' = - 17,000
N5 = 127,700-55,000x sin 57° 30' a 85.900
- - 3 o ' -

"a - 74(55:000) Bigvg 539—. - cos 87° 50' : -74(o5,000)(.157)
=-54e,000

Va =-55,000 x sin 37° 50' = 55,000 x609 .-52.100

Stresses at Abutments

f = N 6M -
1.. 1.Ez_ (compressive stress formula)

 

f . 85,900 - ~546,900x6
16 16x16

fe= 5550 - 12,850 =-7,500

fi = 5550 + 12850 4 18,200
r = Q! ._§£§§1129 = 5000 :22 .s.i.
u 33 2x16 p
at haunch
r = 221399 4 §§1§224§
10 16x16

fe = 4.250 r 1,550 = 6,400

f1: 4,850 - 1,550 4 5,500

r = £1. - 5x17,000
v 2t ‘ 2x16 2 1600 = 11 p.s.i.
at crown

f a 74,900 i 2241200x6
16 16x16

f1 = 4650 - 8450 = -1800--

H
0
ll

4,500 4 5,450 3 11,100

.714“ PtECe'Eo/MC (QLCULATION SIMILAI?

Arch ring 1 ft. high whose center is 25 ft below water surface

 

R : 82 ft. R/t = 5.01 arch ring can be considered
as either 'Ghick or thin”
t = 16 ft.
We will use 'thick'.
r = 74 ft.
r" =gf-aL--’ = lé—‘- ' 75 85'
lo 10 ’ °
R _ 5 Se 1.24

0 = 74.00 - 75.85 = .17

L = 02167 K. = 4097

rn

. NC = pH (1- K. ; 62.4x25x82 (1- 4.97 =
12 ) 12 ) 75,000

Me = -rn(pR _ Nb).§l%JE_. —cos¢)= -73,85 (52,900)(.07) £275,800
v0 = O
”h = PR - (9R -Nc) 00844. 127,900 -52,900x947= 77,700
”h = -75.85 (52,900))6017) - 55,400
Vh a -(pR-Nc) Bin 5 = —53,900 x .541 =-18,050
Na = 127,900 - 52,900x.795 = 88,000
M5 =~75.85 (52,900)(.157)=-1555.OOO

_ -52,900 X .609 -.32’300

.4
p
I

f1

 

Stresses at Crown

 

: 75,000x75.85 4 275,8

82x16

82

= 4200 + 5900 2 10,100

 

 

_ 75,000x75.82 - 225

66x16

= 5250 - 7050 - -1800

t
Nrn +Mrn( 2 + c)
Rt Eff
%
Mn‘ t —c)

Nrn - 2
(R —t)t (R-t)I
0

At the haunch

22,700x75.8§ * 66.400x75.85; .17
82x16 82x541

a 4,500 4 1,400 = 5:700

77,700x75.85 - 661400x75.85x7.85
66x541

66x16

= 5,400 — 1,700 . 5,700

5X18050 : .1, 700

2x16

At the Abutments

_86,000x75.85Ffs’5554000x75.82

F—82x16

. 4850 —11,500

82x541

-6,700

45:17 - 8,700

00x75.85x8.17

x165
12

,800x75.85x8.85

66x541

RESULTS AND CONCLUSIONS
The total volume of materials saved equals the volume of

the gravity dam minus the volume of the arch dam

: 548,500 - 285,515 = 84,885 cu. ft.

The % savino : 542385 a
b 548,500 = 18'6”

The arch dam , noionly represents a savings of 18.6% of
concrete, but also has the greater safety factor of the two
dams. And in conclusion, I believe that wherever possible

an arch dam can be used advantageously.

finis

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